PART-I:
1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived
object,. calling of that virtual method will result in which method being called?
a. Base method
b. Derived method..
Ans. b
2. For the following C program
#define AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}
What is the output?
Ans. Area of the circle is 122.656250
Area of the circle is 19.625000
3. What do the following statements indicate. Explain.
-- Kernighan & Ritchie page no. 122
-- Schaum series page no. 323
4.
void main()
{
int d=5;
printf("%f",d);
}
Ans: Undefined
5.
void main()
{
int i;
for(i=1;i<4;i++)
switch(i)
{
case 1: printf("%d",i);break;
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}
Ans: 1,2,3,4
6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}
Ans: 6
7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i<j)
printf("less");
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}
Ans: less
8.
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}
1. 1000000
2. Overflow
3. Error
4. None
Ans: 4
9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
Ans: The first part of this question can be answered in at least
three ways:
1. char *(*(*a[N])())();
2. Build the declaration up incrementally, using typedefs:
typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */
3. Use the cdecl program, which turns English into C and vice
versa:
cdecl> declare a as array of pointer to function returning
pointer to function returning pointer to char
char *(*(*a[])())()
cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these complicated
C declarations "inside out" to understand them ("declaration
mimics use").
The pointer-to-function declarations in the examples above have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
10. A structure pointer is defined of the type time . With 3 fields min,sec hours having pointers to intergers.
Write the way to initialize the 2nd element to 10.
11. In the above question an array of pointers is declared.
Write the statement to initialize the 3rd element of the 2 element to 10;
12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}
What are the number of syntax errors in the above?
Ans: None.
13.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");
Ans: "one is defined"
15.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}
Ans: 20 20 20
16. There was question in c working only on unix machine with pattern matching.
14. what is alloca()
Ans : It allocates and frees memory after use/after getting out of scope
17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}
Ans: 321
18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}
Ans: anything is good.
19.
void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"
20. Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c<u)
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i<u)
printf("pass2");
else
printf("Fail2")
}
a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
22. What will the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
23. In the following code segment what will be the result of the function,
value of x , value of y
{unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
}
a) same, MAXINT, -1
b) not same, MAXINT, -MAXINT
c) same , MAXUNIT, -1
d) same, MAXUNIT, MAXUNIT
e) not same, MAXINT, MAXUNIT
Ans. (a)
24. What will be the result of the following program ?
char *gxxx()
{static char xxx[1024];
return xxx;
}
main()
{char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
}
a) The string is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these
Ans. (b)
25. Find the output for the following C program
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}
Ans. An empty string
26. Find the output for the following C program
main()
{
int x=20,y=35;
x = y++ + x++;
y = ++y + ++x;
printf("%d %d\n",x,y);
}
Ans. 57 94
27. Find the output for the following C program
main()
{
int x=5;
printf("%d %d %d\n",x,x<<2,x>>2);
}
Ans. 5 20 1
28 Find the output for the following C program
#define swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}
Ans. 10 5
29 Find the output for the following C program
main()
{
char *ptr = "Ramco Systems";
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}
Ans. Samco Systems
30 Find the output for the following C program
#include<stdio.h>
main()
{
char s1[]="Ramco";
char s2[]="Systems";
s1=s2;
printf("%s",s1);
}
Ans. Compilation error giving it cannot be an modifiable 'lvalue'
31 Find the output for the following C program
#include<stdio.h>
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}
Ans. RamcoSystems
32. Find the output for the following C program given that
[1]. The following variable is available in file1.c
static int average_float;
Ans. All the functions in the file1.c can access the variable
33. Find the output for the following C program
# define TRUE 0
some code
while(TRUE)
{
some code
}
Ans. This won't go into the loop as TRUE is defined as 0
34. struct list{
int x;
struct list *next;
}*head;
the struct head.x =100
Is the above assignment to pointer is correct or wrong ?
Ans. Wrong
35.What is the output of the following ?
int i;
i=1;
i=i+2*i++;
printf(%d,i);
Ans. 4
36. FILE *fp1,*fp2;
fp1=fopen("one","w")
fp2=fopen("one","w")
fputc('A',fp1)
fputc('B',fp2)
fclose(fp1)
fclose(fp2)
}
Find the Error, If Any?
Ans. no error. But It will over writes on same file.
37. What are the output(s) for the following ?
38. #include<malloc.h>
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}
main()
{
char *f();
printf("%c",*f()='A'); }
39. #define MAN(x,y) (x)>(y)?(x):(y)
{int i=10;
j=5;
k=0;
k=MAX(i++,++j);
printf(%d %d %d %d,i,j,k);
}
Ans. 10 5 0
40.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
->Why doesn't C have nested functions?
->What is the most efficient way to count the number of bits which are set in a value?
->How can I convert integers to binary or hexadecimal?
->How can I call a function, given its name as a string?
->How do I access command-line arguments?
->How can I return multiple values from a function?
->How can I invoke another program from within a C program?
->How can I access memory located at a certain address?
->How can I allocate arrays or structures bigger than 64K?
->How can I find out how much memory is available?
->How can I read a directory in a C program?
->How can I increase the allowable number of simultaneously open files?
->What's wrong with the call "fopen("c:\newdir\file.dat", "r")"?
PART-II:
C,C++
Questions1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived
object,. calling of that virtual method will result in which method being called?
a. Base method
b. Derived method..
Ans. b
2. For the following C program
#define AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}
What is the output?
Ans. Area of the circle is 122.656250
Area of the circle is 19.625000
3. What do the following statements indicate. Explain.
·
int(*p)[10]
·
int*f()
·
int(*pf)()
·
int*p[10]
Refer
to:-- Kernighan & Ritchie page no. 122
-- Schaum series page no. 323
4.
void main()
{
int d=5;
printf("%f",d);
}
Ans: Undefined
5.
void main()
{
int i;
for(i=1;i<4;i++)
switch(i)
{
case 1: printf("%d",i);break;
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}
Ans: 1,2,3,4
6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}
Ans: 6
7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i<j)
printf("less");
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}
Ans: less
8.
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}
1. 1000000
2. Overflow
3. Error
4. None
Ans: 4
9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
Ans: The first part of this question can be answered in at least
three ways:
1. char *(*(*a[N])())();
2. Build the declaration up incrementally, using typedefs:
typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */
3. Use the cdecl program, which turns English into C and vice
versa:
cdecl> declare a as array of pointer to function returning
pointer to function returning pointer to char
char *(*(*a[])())()
cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these complicated
C declarations "inside out" to understand them ("declaration
mimics use").
The pointer-to-function declarations in the examples above have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
10. A structure pointer is defined of the type time . With 3 fields min,sec hours having pointers to intergers.
Write the way to initialize the 2nd element to 10.
11. In the above question an array of pointers is declared.
Write the statement to initialize the 3rd element of the 2 element to 10;
12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}
What are the number of syntax errors in the above?
Ans: None.
13.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");
Ans: "one is defined"
15.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}
Ans: 20 20 20
16. There was question in c working only on unix machine with pattern matching.
14. what is alloca()
Ans : It allocates and frees memory after use/after getting out of scope
17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}
Ans: 321
18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}
Ans: anything is good.
19.
void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"
20. Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c<u)
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i<u)
printf("pass2");
else
printf("Fail2")
}
a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
22. What will the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
23. In the following code segment what will be the result of the function,
value of x , value of y
{unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
}
a) same, MAXINT, -1
b) not same, MAXINT, -MAXINT
c) same , MAXUNIT, -1
d) same, MAXUNIT, MAXUNIT
e) not same, MAXINT, MAXUNIT
Ans. (a)
24. What will be the result of the following program ?
char *gxxx()
{static char xxx[1024];
return xxx;
}
main()
{char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
}
a) The string is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these
Ans. (b)
25. Find the output for the following C program
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}
Ans. An empty string
26. Find the output for the following C program
main()
{
int x=20,y=35;
x = y++ + x++;
y = ++y + ++x;
printf("%d %d\n",x,y);
}
Ans. 57 94
27. Find the output for the following C program
main()
{
int x=5;
printf("%d %d %d\n",x,x<<2,x>>2);
}
Ans. 5 20 1
28 Find the output for the following C program
#define swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}
Ans. 10 5
29 Find the output for the following C program
main()
{
char *ptr = "Ramco Systems";
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}
Ans. Samco Systems
30 Find the output for the following C program
#include<stdio.h>
main()
{
char s1[]="Ramco";
char s2[]="Systems";
s1=s2;
printf("%s",s1);
}
Ans. Compilation error giving it cannot be an modifiable 'lvalue'
31 Find the output for the following C program
#include<stdio.h>
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}
Ans. RamcoSystems
32. Find the output for the following C program given that
[1]. The following variable is available in file1.c
static int average_float;
Ans. All the functions in the file1.c can access the variable
33. Find the output for the following C program
# define TRUE 0
some code
while(TRUE)
{
some code
}
Ans. This won't go into the loop as TRUE is defined as 0
34. struct list{
int x;
struct list *next;
}*head;
the struct head.x =100
Is the above assignment to pointer is correct or wrong ?
Ans. Wrong
35.What is the output of the following ?
int i;
i=1;
i=i+2*i++;
printf(%d,i);
Ans. 4
36. FILE *fp1,*fp2;
fp1=fopen("one","w")
fp2=fopen("one","w")
fputc('A',fp1)
fputc('B',fp2)
fclose(fp1)
fclose(fp2)
}
Find the Error, If Any?
Ans. no error. But It will over writes on same file.
37. What are the output(s) for the following ?
38. #include<malloc.h>
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}
main()
{
char *f();
printf("%c",*f()='A'); }
39. #define MAN(x,y) (x)>(y)?(x):(y)
{int i=10;
j=5;
k=0;
k=MAX(i++,++j);
printf(%d %d %d %d,i,j,k);
}
Ans. 10 5 0
40.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
Write a constant declaration that
declares constants to hold the number of days in a week and the number of weeks
in a year. In a separate constant statement declare a constant pi as 3.1415927.
Write declaration statements to
declare integer variables i and j and float variables x and y. Extend your
declaration statements so that i and j are both initialised to 1 and y is
initialised to 10.0.
Write C++ instructions to ask a
user to type in three numbers and to read them into integer variables first,
second and third.
Write C++ instructions to output
the value of a variable x in a line as follows:
The value of x is ......
Write C++ instructions to generate
output as follows:
A circle of radius .....
has area .....
and circumference .....
where the values of the radius,
the area and the circumference are held in variables rad, area, and circum.
Correct the syntax errors in the
following C++ program:
include <iostream.h>
main()
{
float x,y,z;
cout<< "Enter two numbers ";
cin >> a >> b;
cout << 'The numbers in reverse order are”<<
b,a;
}
Download copy to try it.
Show the form of output displayed
by the following statements when total has the value 352.74.
cout << "The final
total is: " << endl;
cout << "$"
<< total << endl;
What data types would you use to
represent the following items?
the number of students in a class
the grade (a letter) attained by a
student in the class
the average mark in a class
the distance between two points
the population of a city
the weight of a postage stamp
the registration letter of a car
Write suitable declarations for
variables in question 8. Be sure to choose meaningful identifiers.
->How
can I find the day of the week given the date? ->Why doesn't C have nested functions?
->What is the most efficient way to count the number of bits which are set in a value?
->How can I convert integers to binary or hexadecimal?
->How can I call a function, given its name as a string?
->How do I access command-line arguments?
->How can I return multiple values from a function?
->How can I invoke another program from within a C program?
->How can I access memory located at a certain address?
->How can I allocate arrays or structures bigger than 64K?
->How can I find out how much memory is available?
->How can I read a directory in a C program?
->How can I increase the allowable number of simultaneously open files?
->What's wrong with the call "fopen("c:\newdir\file.dat", "r")"?
What are the differences between
malloc() and calloc()?
malloc()
: allows to create memory dynamically and exact amount needed.
char
*str; str = (char *) malloc(10);
calloc()
: allocates main memory; allocates block
bytes and clears it to zero.
char
*str; str = (char *) calloc(10,
sizeof(char));
What are macros? what are its
advantages and disadvantages?
any
function or variable that is declared along with #define.
Difference between pass by
reference and pass by value?
pass
by reference is passing the address of the variable which is accepted by the
pointer variable..
pass
by value is creating the copy of existing values and doing operations on that
values, but they don’t effect the original values.
Where are the auto variables
stored?
defines a local variable and has
local lifetime and is used very rare.
Where does global, static, local,
register variables, free memory and C Program instructions get stored?
Global
variables can be used through out the program.
Static
variables are used to initialise only once.
local
means auto variables
register
variables stores tha variables declared in the CPU register
register
int i;
Difference between arrays and
linked list?
arrays
is collecion of similar elements, continuous memory allocation, and dynamic
memory allocation as specified by user, memory gets wasted, which allocates
memory in the begining, it conatins only one datatype.
linked
list creates dynamic memory in order to store the values, not continuous
storageie., may or may not be, no wastage of memory. and insertion and deletion
is easy in linked list, linked list contains collection of datatypes.
What are enumerations?
is
to invent your own data types and define what values the variable of this
datatype can take. or to define set of constants of type int. is used to
clarify the operation of a program.
Describe about storage allocation
and scope of global, extern, static, local and register variables?
Global
storage is ordinary and scope is anywhere it can be used.
extern
indicates the actual storage and initial value of the variable, or body of
function, and is defined in separate code.
static
storage is ordinary and can be used within a function or program, or block of
code.
local
is auto......
register
variables is to access the cpu variables
What are register variables? What
are the advantage of using register variables?
register variables used to access the CPU variables
What is the use of typedef?
to
create user defined datatypes.
typedef
int a; a b;
Can we specify variable field
width in a scanf() format string? If possible how?
***********
Out of fgets() and gets() which
function is safe to use and why?
Difference between strdup and
strcpy?
What is recursion?
Differentiate between a for loop
and a while loop? What are it uses?
What are the different storage
classes in C?
Write down the equivalent pointer
expression for referring the same element a[i][j][k][l]?
What is difference between
Structure and Unions?
What the advantages of using
Unions?
What are the advantages of using
pointers in a program?
What is the difference between
Strings and Arrays?
In a header file whether functions
are declared or defined?
What is a far pointer? where we
use it?
How will you declare an array of
three function pointers where each function receives two ints and returns a
float?
what is a NULL Pointer? Whether it
is same as an uninitialized pointer?
What is a NULL Macro? What is the
difference between a NULL Pointer and a NULL Macro?
What does the error 'Null Pointer
Assignment' mean and what causes this error?
What is near, far and huge
pointers? How many bytes are occupied by them?
How would you obtain segment and
offset addresses from a far address of a memory location?
Are the expressions arr and
&arr same for an array of integers?
Does mentioning the array name
gives the base address in all the contexts?
Explain one method to process an
entire string as one unit?
What is the similarity between a
Structure, Union and enumeration?
Can a Structure contain a Pointer
to itself?
How can we check whether the
contents of two structure variables are same or not?
How are Structure passing and
returning implemented by the complier?
How can we read/write Structures
from/to data files?
What is the difference between an
enumeration and a set of pre-processor # defines?
what do the 'c' and 'v' in argc
and argv stand for?
Are the variables argc and argv
are local to main?
What is the maximum combined
length of command line arguments including the space between adjacent
arguments?
If we want that any wildcard
characters in the command line arguments should be appropriately expanded, are
we required to make any special provision? If yes, which?
Does there exist any way to make
the command line arguments available to other functions without passing them as
arguments to the function?
What are bit fields? What is the
use of bit fields in a Structure declaration?
To which numbering system can the
binary number 1101100100111100 be easily converted to?
Which bit wise operator is
suitable for checking whether a particular bit is on or off?
Which bit wise operator is
suitable for turning off a particular bit in a number?
Which bit wise operator is
suitable for putting on a particular bit in a number?
Which bit wise operator is
suitable for checking whether a particular bit is on or off?
which one is equivalent to
multiplying by 2:Left shifting a number by 1 or Left shifting an unsigned int
or char by 1?
What are the advantages of using
typedef in a program?
How would you dynamically allocate
a one-dimensional and two-dimensional array of integers?
How can you increase the size of a
dynamically allocated array?
How can you increase the size of a
statically allocated array?
When reallocating memory if any
other pointers point into the same piece of memory do you have to readjust
these other pointers or do they get readjusted automatically?
Which function should be used to
free the memory allocated by calloc()?
How much maximum can you allocate
in a single call to malloc()?
Can you dynamically allocate
arrays in expanded memory?
What is object file? How can you
access object file?
Which header file should you
include if you are to develop a function which can accept variable number of
arguments?
Can you write a function similar
to printf()?
How can a called function
determine the number of arguments that have been passed to it?
Can there be at least some
solution to determine the number of arguments passed to a variable argument
list function?
How do you declare the following:
An array of three pointers to
chars
An array of three char pointers
A pointer to array of three chars
A pointer to function which
receives an int pointer and returns a float pointer
A pointer to a function which
receives nothing and returns nothing
What do the functions atoi(),
itoa() and gcvt() do?
Does there exist any other
function which can be used to convert an integer or a float to a string?
How would you use qsort() function
to sort an array of structures?
How would you use qsort() function
to sort the name stored in an array of pointers to string?
How would you use bsearch()
function to search a name stored in array of pointers to string?
How would you use the functions
sin(), pow(), sqrt()?
How would you use the functions
memcpy(), memset(), memmove()?
How would you use the functions
fseek(), freed(), fwrite() and ftell()?
How would you obtain the current
time and difference between two times?
How would you use the functions
randomize() and random()?
How would you implement a substr()
function that extracts a sub string from a given string?
What is the difference between the
functions rand(), random(), srand() and randomize()?
What is the difference between the
functions memmove() and memcpy()?
How do you print a string on the
printer?
Can you use the function fprintf()
to display the output on the screen?
How do you write a program which
produces its own source code as its
output?
How can I find the day of the week
given the date?
Why doesn't C have nested
functions?
What is the most efficient way to
count the number of bits which are set in a value?
How can I convert integers to
binary or hexadecimal?
How can I call a function, given
its name as a string?
How do I access command-line
arguments?
How can I return multiple values
from a function?
How can I invoke another program
from within a C program?
How can I access memory located at
a certain address?
How can I allocate arrays or
structures bigger than 64K?
How can I find out how much memory
is available?
How can I read a directory in a C
program?
How can I increase the allowable
number of simultaneously open files?
What's wrong with the call
"fopen("c:\newdir\file.dat", "r")"?
What is the output of
printf("%d")
What will happen if I say delete
this
Difference between "C
structure" and "C++ structure".
Diffrence between a
"assignment operator" and a "copy constructor"
What is the difference between
"overloading" and "overridding"?
Explain the need for "Virtual
Destructor".
Can we have "Virtual
Constructors"?
What are the different types of
polymorphism?
What are Virtual Functions? How to
implement virtual functions in "C"
What are the different types of
Storage classes?
What is Namespace?
What are the types of STL
containers?.
Difference between
"vector" and "array"?
How to write a program such that
it will delete itself after exectution?
Can we generate a C++ source code
from the binary file?
What are inline functions?
Talk sometiming about profiling?
How many lines of code you have
written for a single program?
What is "strstream" ?
How to write Multithreaded
applications using C++?
Explain "passing by
value", "passing by pointer" and "passing by
reference"
Write any small program that will
compile in "C" but not in "C++"
Have you heard of
"mutable" keyword?
What is a "RTTI"?
Is there something that I can do
in C and not in C++?
Why preincrement operator is
faster than postincrement?
What is the difference between
"calloc" and "malloc"?
What will happen if I allocate
memory using "new" and free it using "free" or allocate
sing "calloc" and free it using "delete"?
What is Memory Alignment?
Explain working of printf.
Difference between
"printf" and "sprintf".
What is "map" in STL?
When shall I use Multiple
Inheritance?
What are the techniques you use
for debugging?
How to reduce a final size of
executable?
Give 2 examples of a code
optimization.
What is inheritance?
Difference between Composition and
Aggregation.
Difference: Sequence Diagrams,
Collaboration Diagrams.
Difference: 'uses', 'extends',
'includes'
What shall I go for Package
Diagram?
What is Polymorphism?
Is class an Object? Is object a
class?
Comment: C++ "includes"
behavior and java "imports"
What do you mean by
"Realization"?
What is a Presistent, Transient
Object?
What is the use of Operator
Overloading?
Does UML guarantee project
success?
Difference: Activity Diagram and
Sequence Diagram.
What is association?
How to resolve many to many
relationship?
How do you represent static
members and abstract classes in Class Diagram?
What does static variable mean?
What is a pointer?
What is a structure?
What are the differences between
structures and arrays?
In header files whether functions
are declared or defined?
What are the differences between
malloc() and calloc()?
What are macros? what are its
advantages and disadvantages?
Difference between pass by
reference and pass by value?
What is static identifier?
Where are the auto variables
stored?
Where does global, static, local,
register variables, free memory and C Program instructions get stored?
Difference between arrays and
linked list?
What are enumerations?
Describe about storage allocation
and scope of global, extern, static, local and register variables?
What are register variables? What
are the advantage of using register variables?
What is the use of typedef?
Can we specify variable field
width in a scanf() format string? If possible how?
Out of fgets() and gets() which
function is safe to use and why?
Difference between strdup and
strcpy?
What is recursion?
Differentiate between a for loop
and a while loop? What are it uses?
What are the different storage
classes in C?
Write down the equivalent pointer
_expression for referring the same element a[i][j][k][l]?
What is difference between
Structure and Unions?
What the advantages of using
Unions?
What are the advantages of using
pointers in a program?
What is the difference between
Strings and Arrays?
In a header file whether functions
are declared or defined?
What is a far pointer? where we
use it?
How will you declare an array of
three function pointers where each function receives two ints and returns a
float?
what is a NULL Pointer? Whether it
is same as an uninitialized pointer?
What is a NULL Macro? What is the
difference between a NULL Pointer and a NULL Macro?
What does the error 'Null Pointer
Assignment' mean and what causes this error?
What is near, far and huge
pointers? How many bytes are occupied by them?
How would you obtain segment and
offset addresses from a far address of a memory location?
Are the expressions arr and
&arr same for an array of integers?
Does mentioning the array name
gives the base address in all the contexts?
Explain one method to process an
entire string as one unit?
What is the similarity between a
Structure, Union and enumeration?
Can a Structure contain a Pointer
to itself?
How can we check whether the
contents of two structure variables are same or not?
How are Structure passing and
returning implemented by the complier?
How can we read/write Structures
from/to data files?
What is the difference between an
enumeration and a set of pre-processor # defines?
what do the 'c' and 'v' in argc
and argv stand for?
Are the variables argc and argv
are local to main?
What is the maximum combined
length of command line arguments including the space between adjacent
arguments?
If we want that any wildcard
characters in the command line arguments should be appropriately expanded, are
we required to make any special provision? If yes, which?
Does there exist any way to make
the command line arguments available to other functions without passing them as
arguments to the function?
What are bit fields? What is the
use of bit fields in a Structure declaration?
To which numbering system can the
binary number 1101100100111100 be easily converted to?
Which bit wise operator is
suitable for checking whether a particular bit is on or off?
Which bit wise operator is
suitable for turning off a particular bit in a number?
Which bit wise operator is
suitable for putting on a particular bit in a number?
Which bit wise operator is
suitable for checking whether a particular bit is on or off?
which one is equivalent to
multiplying by 2:Left shifting a number by 1 or Left shifting an unsigned int
or char by 1?
Write a program to compare two
strings without using the strcmp() function.
Write a program to concatenate two
strings.
Write a program to interchange 2
variables without using the third one.
Write programs for String Reversal
& Palindrome check
Write a program to find the
Factorial of a number
Write a program to generate the
Fibinocci Series
Write a program which employs
Recursion
Write a program which uses Command
Line Arguments
Write a program which uses
functions like strcmp(), strcpy()? etc
What are the advantages of using
typedef in a program?
How would you dynamically allocate
a one-dimensional and two-dimensional array of integers?
How can you increase the size of a
dynamically allocated array?
How can you increase the size of a
statically allocated array?
When reallocating memory if any
other pointers point into the same piece of memory do you have to readjust
these other pointers or do they get readjusted automatically?
Which function should be used to
free the memory allocated by calloc()?
How much maximum can you allocate
in a single call to malloc()?
Can you dynamically allocate
arrays in expanded memory?
What is object file? How can you
access object file?
Which header file should you
include if you are to develop a function which can accept variable number of
arguments?
Can you write a function similar
to printf()?
How can a called function
determine the number of arguments that have been passed to it?
Can there be at least some
solution to determine the number of arguments passed to a variable argument
list function?
How do you declare the following:
An array of three pointers to
chars
An array of three char pointers
A pointer to array of three chars
A pointer to function which
receives an int pointer and returns a float pointer
A pointer to a function which
receives nothing and returns nothing
What do the functions atoi(),
itoa() and gcvt() do?
Does there exist any other
function which can be used to convert an integer or a float to a string?
How would you use qsort() function
to sort an array of structures?
How would you use qsort() function
to sort the name stored in an array of pointers to string?
How would you use bsearch()
function to search a name stored in array of pointers to string?
How would you use the functions
sin(), pow(), sqrt()?
How would you use the functions
memcpy(), memset(), memmove()?
How would you use the functions
fseek(), freed(), fwrite() and ftell()?
How would you obtain the current
time and difference between two times?
How would you use the functions
randomize() and random()?
How would you implement a substr()
function that extracts a sub string from a given string?
What is the difference between the
functions rand(), random(), srand() and randomize()?
What is the difference between the
functions memmove() and memcpy()?
How do you print a string on the
printer?
Can you use the function fprintf()
to display the output on the screen?
What is an object?
What is the difference between an
object and a class?
What is the difference between
class and structure?
What is public, protected,
private?
What are virtual functions?
What is friend function?
What is a scope resolution
operator?
What do you mean by inheritance?
What is abstraction?
What is polymorphism? Explain with
an example.
What is encapsulation?
What do you mean by binding of
data and functions?
What is function overloading and
operator overloading?
What is virtual class and friend
class?
What do you mean by inline
function?
What do you mean by public,
private, protected and friendly?
When is an object created and what
is its lifetime?
What do you mean by multiple
inheritance and multilevel inheritance? Differentiate between them.
Difference between realloc() and
free?
What is a template?
What are the main differences
between procedure oriented languages and object oriented languages?
What is R T T I ?
What are generic functions and
generic classes?
What is namespace?
What is the difference between
pass by reference and pass by value?
Why do we use virtual functions?
What do you mean by pure virtual
functions?
What are virtual classes?
Does c++ support multilevel and
multiple inheritance?
What are the advantages of
inheritance?
When is a memory allocated to a
class?
What is the difference between
declaration and definition?
What is virtual
constructors/destructors?
In c++ there is only virtual destructors,
no constructors. Why?
What is late bound function call
and early bound function call? Differentiate.
How is exception handling carried
out in c++?
When will a constructor executed?
What is Dynamic Polymorphism?
Write a macro for swapping
integers.
Note : All the
programs are tested under Turbo C/C++ compilers.
It is assumed that,
Ø
Programs run under DOS
environment,
Ø
The underlying machine
is an x86 system,
Ø
Program is compiled
using Turbo C/C++ compiler.
The program output may depend on the information based on
this assumptions (for example sizeof(int) == 2 may be assumed).
Predict
the output or error(s) for the following:
1.
void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to
change the value of the "constant integer".
2.
main()
{
char
s[ ]="man";
int
i;
for(i=0;s[
i ];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing
the same idea. Generally array name is
the base address for that array. Here s
is the base address. i is the index
number/displacement from the base address. So, indirecting it with * is same as
s[i]. i[s] may be surprising. But in the
case of C it is same as s[i].
3.
main()
{
float
me = 1.1;
double
you = 1.1;
if(me==you)
printf("I
love U");
else
printf("I
hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float,
double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precession
with of the value represented varies.
Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with
less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point
numbers with relational operators (==
, >, <, <=, >=,!= )
.
4.
main()
{
static
int var = 5;
printf("%d
",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage
class is given, it is initialized once. The change in the value of a static variable is retained even between
the function calls. Main is also treated like
any other ordinary function, which can be called recursively.
5.
main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf("
%d ",*c);
++q; }
for(j=0;j<5;j++){
printf("
%d ",*p);
++p; }
}
Answer:
2
2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first
loop, since only q is incremented
and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values
2 3 4 6 5 will be printed.
6.
main()
{
extern
int i;
i=20;
printf("%d",i);
}
Answer:
Linker
Error : Undefined symbol '_i'
Explanation:
extern
storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given
to the current program at the time of linking. But linker finds that no other
variable of name i is available in
any other program with memory space allocated for it. Hence a linker error has
occurred .
7.
main()
{
int
i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d
%d %d %d %d",i,j,k,l,m);
}
Answer:
0
0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher
priority over the logical OR (||) operator. So the expression ‘i++
&& j++ && k++’ is executed first. The result of this
expression is 0 (-1 && -1
&& 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because
OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives
0). So the value of m is 1. The values of other variables are also incremented
by 1.
8.
main()
{
char
*p;
printf("%d
%d ",sizeof(*p),sizeof(p));
}
Answer:
1
2
Explanation:
The sizeof() operator gives the number of bytes taken by its
operand. P is a character pointer, which needs one byte for storing its value
(a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to
store the address of the character pointer sizeof(p) gives 2.
9.
main()
{
int
i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is
executed only when all other cases doesn't match.
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times
the least significant 4 bits are filled with 0's.The %x format specifier
specifies that the integer value be printed as a hexadecimal value.
11. main()
{
char string[]="Hello
World";
display(string);
}
void display(char
*string)
{
printf("%s",string);
}
Answer:
Compiler
Error : Type mismatch in
redeclaration of function display
Explanation
:
In third line, when the function display is encountered, the compiler doesn't know anything about
the function display. It assumes the arguments and return types to be integers,
(which is the default type). When it sees the actual function display, the arguments and type
contradicts with what it has assumed previously. Hence a compile time error
occurs.
12. main()
{
int
c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same
maths rules applies, ie. minus * minus=
plus.
Note:
However you cannot give like --2. Because -- operator can only be
applied to variables as a decrement operator
(eg., i--). 2 is a constant and not a variable.
13. #define int char
main()
{
int
i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int
by the macro char
14. main()
{
int i=10;
i=! i>14;
Printf
("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14
, NOT (!) operator has more precedence than ‘ >’ symbol. !
is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
15. #include<stdio.h>
main()
{
char
s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p
+ ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a'
++*p. "p is pointing to '\n' and that is incremented by one." the
ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p
is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes
'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 –
32), we get 77("M");
So we get the output 77 ::
"M" (Ascii is 77).
16. #include<stdio.h>
main()
{
int a[2][2][2] = {
{10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you
declare only two 2D arrays, but you are trying to access the third 2D(which you
are not declared) it will print garbage values. *q=***a starting address of a
is assigned integer pointer. Now q is pointing to starting address of a. If you
print *q, it will print first element of 3D array.
17. #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18. #include<stdio.h>
main()
{
struct xx
{
int
x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the
elements are of yy are to be accessed through the instance of structure xx,
which needs an instance of yy to be known. If the instance is created after
defining the structure the compiler will not know about the instance relative
to xx. Hence for nested structure yy you have to declare member.
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from
left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the
result.
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression
becomes i = 64/4*4 . Since / and * has equal priority the expression will be
evaluated as (64/4)*4 i.e. 16*4 = 64
22. main()
{
char *p="hai
friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
Ø
*p that is value at
the location currently pointed by p will be taken
Ø
++*p the retrieved
value will be incremented
Ø
when ; is encountered
the location will be incremented that is p++ will be executed
Hence,
in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by
executing ++*p and pointer moves to point, ‘a’ which is similarly changed to
‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value
in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus
p1doesnot print anything.
23. #include
<stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the
program. So the most recently assigned value will be taken.
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of
the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't
give any problem
25. main()
{
printf("%p",main);
}
Answer:
Some address will be
printed.
Explanation:
Function names are just addresses (just like array names
are addresses).
main()
is also a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal
numbers.
27) main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a
function call. In the second clrscr(); is a function declaration (because it is
not inside any function).
28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
char far
*farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the
program. Any number of printf's may be given. All of them take only the first
two values. If more number of assignments given in the program,then printf will
take garbage values.
31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They
can be applied any number of times
provided it is meaningful. Here p points
to the first character in the string
"Hello". *p dereferences it and so its value is H. Again & references it to an address and *
dereferences it to the value H.
32)
main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the
labels is limited to functions . The label 'here' is available in function
fun() Hence it is not visible in function main.
33) main()
{
static char
names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
34) void main()
{
int i=5;
printf("%d",i++
+ ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an
illegal combination of operators.
36) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this
implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have
been scanned successfully. So number of items read is 1.
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
39) main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is
"evaluated". Here it evaluates to 0 (false) and comes out of the
loop, and i is incremented (note the semicolon after the for loop).
40) #include<stdio.h>
main()
{
char
s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p
+ ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a'
++*p meAnswer:"p is pointing to '\n' and that is incremented by one."
the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p
is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1
and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and
result is subtracted from 32.
i.e. (11+98-32)=77("M");
41) #include<stdio.h>
main()
{
struct xx
{
int x=3;
char
name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the
structure declaration
42) #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined
somewhere else. The compiler passes the external variable to be resolved by the
linker. So compiler doesn't find an error. During linking the linker searches
for the definition of i. Since it is not found the linker flags an error.
44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of
declaration. Even though a is a global variable, it is not available for main.
Hence an error.
45) main()
{
extern out;
printf("%d",
out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
46) main()
{
show();
}
void show()
{
printf("I'm the
greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything
about it. So the default return type (ie, int) is assumed. But when compiler
sees the actual definition of show mismatch occurs since it is declared as
void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
47) main( )
{
int a[2][3][2] =
{{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d
\n”,a,*a,**a,***a);
printf(“%u %u %u %d
\n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The
given array is a 3-D one. It can also be viewed as a 1-D array.
2
|
4
|
7
|
8
|
3
|
4
|
2
|
2
|
2
|
3
|
3
|
4
|
100 102
104 106 108 110
112 114 116
118 120 122
thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a
gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension
thus points to value at 114, *a+1 increments in second dimension thus points to
104, **a +1 increments the first dimension thus points to 102 and ***a+1 first
gets the value at first location and then increments it by 1. Hence, the
output.
48) main( )
{
int a[ ] =
{10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5;
j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an
lvalue and may be of any of scalar type for the any operator, array name only
when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
49) main( )
{
static int a[ ]
= {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr =
p;
ptr++;
printf(“\n %d %d
%d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d
%d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d
%d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d
%d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some
address
a
0
|
1
|
2
|
3
|
4
|
100 102 104
106 108
p
100
|
102
|
104
|
106
|
108
|
1000
1002 1004 1006
1008
ptr
1000
|
2000
After execution of the
instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2
bytes. Now ptr – p is value in ptr – starting location of array p, (1002 –
1000) / (scaling factor) = 1, *ptr – a =
value at address pointed by ptr – starting value of array a, 1002 has a value
102 so the value is (102 – 100)/(scaling
factor) = 1, **ptr is the value stored
in the location pointed by the pointer
of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1.
Hence the output of the firs printf is
1, 1, 1.
After execution of *ptr++
increments value of the value in ptr by scaling factor, so it becomes1004.
Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr =
2.
After execution of *++ptr
increments value of the value in ptr by scaling factor, so it becomes1004.
Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr =
3.
After execution of ++*ptr value
in ptr remains the same, the value pointed by the value is incremented by the
scaling factor. So the value in array p at location 1006 changes from 106 10
108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3,
*ptr – a = 108 – 100 = 4, **ptr = 4.
50) main( )
{
char *q;
int j;
for (j=0; j<3;
j++) scanf(“%s” ,(q+j));
for (j=0; j<3;
j++) printf(“%c” ,*(q+j));
for (j=0; j<3;
j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we
take input in the same pointer thus we keep writing over in the same location,
each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input
suppose the pointer starts at location 100 then the input one is stored as
M
|
O
|
U
|
S
|
E
|
\0
|
When the second input is given the pointer is incremented
as j value becomes 1, so the input is filled in memory starting from 101.
M
|
T
|
R
|
A
|
C
|
K
|
\0
|
The third input
starts filling from the location 102
M
|
T
|
V
|
I
|
R
|
T
|
U
|
A
|
L
|
\0
|
This is the final value stored .
The first printf prints the values at the position q, q+1
and q+2 = M T V
The second printf prints three strings starting from
locations q, q+1, q+2
i.e
MTVIRTUAL, TVIRTUAL and VIRTUAL.
51) main( )
{
void *vp;
char ch = ‘g’, *cp =
“goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char
*)vp);
vp = &j;
printf(“%d”,*(int
*)vp);
vp = cp;
printf(“%s”,(char
*)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to
any other type pointer. vp =
&ch stores address of char ch and
the next statement prints the value stored in vp after type casting it to the
proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The
third printf statement type casts it to print the string from the 4th
value hence the output is ‘fy’.
52) main ( )
{
static char *s[
] = {“black”, “white”, “yellow”,
“violet”};
char **ptr[ ] =
{s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p
+ 3);
}
Answer:
ck
Explanation:
In this problem we have an array
of char pointers pointing to start of 4 strings. Then we have ptr which is a
pointer to a pointer of type char and a variable p which is a pointer to a
pointer to a pointer of type char. p hold the initial value of ptr, i.e. p =
s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is executed and
we get s+1 – 1 = s . the indirection operator now gets the value from the array
of s and adds 3 to the starting address. The string is printed starting from
this position. Thus, the output is ‘ck’.
53) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n;
++i)
{
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is
initialized with a value “girl”. The
strlen function returns the length of the string, thus n has a value 4. The
next statement assigns value at the nth location (‘\0’) to the first location.
Now the string becomes “\0irl” . Now the printf statement prints the string
after each iteration it increments it starting position. Loop starts
from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer
value is incremented. The second time it prints from x[1] i.e “irl” and the
third time it prints “rl” and the last time it prints “l” and the loop
terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert
failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain
conditions are satisfied. If assertion fails, the program will terminate
reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code.
Assertion
is a good debugging tool to make use of.
55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d
\n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C.
Where-ever it comes you can just ignore it just because it has no effect in the
expressions (hence the name dummy operator).
56) What are the files which are
automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard
error).
57)
what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET
sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the
file position marker to the current position
of the file.
58) main()
{
char name[10],s[12];
scanf("
\"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then
it matches with a quotation mark and then it
reads all character upto another quotation mark.
59) What is the problem with the following
code segment?
while ((fgets(receiving
array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is
checking for != NULL.
60) main()
{
main();
}
Answer:
Runtime error : Stack
overflow.
Explanation:
main function calls itself again and again. Each time the function
is called its return address is stored in the call stack. Since there is no
condition to terminate the function call, the call stack overflows at runtime.
So it terminates the program and results in an error.
61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void,
since void is an empty type. In the second line you are creating variable vptr
of type void * and v of type void hence an error.
62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d
%d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the
size of the pointer variable. In second sizeof the name str2 indicates the name
of the array whose size is 5 (including the '\0' termination character). The
third sizeof is similar to the second one.
63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the
boolean value FALSE, and any non-zero value is considered to be the boolean
value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints
0.
64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the
preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double
quotes. The check by if condition is boolean value false so it goes to else. In
second if -1 is boolean value true hence "TRUE" is printed.
65) main()
{
int k=1;
printf("%d==1 is
""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space)
they are concatenated (this is called as "stringization" operation).
So the string is as if it is given as "%d==1 is %s". The conditional
operator( ?: ) evaluates to "TRUE".
66) main()
{
int y;
scanf("%d",&y); //
input given is 2000
if( (y%4==0 && y%100 != 0)
|| y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
67) #define
max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can
be used to declare the variable name of the type arr2. But it is not the case
of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are
used for declaring new types.
68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost
block i is declared as,
const volatile
unsigned
which is a valid declaration. i is assumed of type int. So printf
prints 30. In the next block, i has value 20 and so printf prints 20. In the
outermost block, i is declared as extern, so no storage space is allocated for
it. After compilation is over the linker resolves it to global variable i
(since it is the only variable visible there). So it prints i's value as 10.
69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is
inside that block only. But the lifetime of i is lifetime of the function so it
lives upto the exit of main function. Since the i is still allocated space, *j
prints the value stored in i since j points i.
70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d
\n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i.
In printf first you just print the value of i. After that the value of the
expression -i = -(-1) is printed.
71) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
72) #include<stdio.h>
main()
{
int a[2][2][2] = {
{10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you
declare only two 2D arrays. but you are trying to access the third 2D(which you
are not declared) it will print garbage values. *q=***a starting address of a
is assigned integer pointer. now q is pointing to starting address of a.if you
print *q meAnswer:it will print first element of 3D array.
73) #include<stdio.h>
main()
{
register i=5;
char j[]=
"hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register
compiler will treat it as ordinary integer and it will take integer
value. i value may be stored either in register or in memory.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements
form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular
node is 2.
77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the
structure either with arrow mark or with indirection operator.
Note:
Since structure point is
globally declared x & y are initialized as zeroes
78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10
will be returned.
79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to
pointers increments address according to their corresponding data-types.
80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') &&
(c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
printf("enter the
character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function
sum without converting it to integer values.
82) # include <stdio.h>
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
83) # include<stdio.h>
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is
assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb
and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2]
points to ccc.
85) #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against
NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine
the truth value of the boolean expression. So the statement following the if
statement is not executed. The values of
i and j remain unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the
information from the accumulator. Here _AH is the pseudo global variable
denoting the accumulator. Hence, the value of the accumulator is set 1000 so the
function returns value 1000.
88) int i;
main(){
int t;
for (
t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find
the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values
during execution
will
be,
t i
x
4 0
-4
3 1
-2
2 2
0
89) main(){
int a= 0;int b = 20;char x
=1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the
rightmost value is returned and the other values are evaluated and ignored.
Thus the value of last variable y is returned to check in if. Since it is a non
zero value if becomes true so, "hello" will be printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c
aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value.
Since the both types doesn't match, signed is promoted to unsigned value. The
unsigned equivalent of -2 is a huge value so condition becomes false and
control comes out of the loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
92) What are the following notations of
defining functions known as?
i. int abc(int a,float
b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/*
some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again
decremented by 2, it points to '%d\n' and 300 is printed.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a
points to 'b' then after incrementing to 'c' so bc will be printed.
95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n
",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another
function 2 and 3, integers. When this
function is invoked from main, the following substitutions for formal
parameters take place: func for pf, 3 for val1 and 6 for val2. This function
returns the result of the operation performed by the function 'func'. The
function func has two integer parameters. The formal parameters are substituted
as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore
the function returns 0 which in turn is returned by the function 'process'.
96) void main()
{
static int
i=5;
if(--i){
main();
printf("%d
",i);
}
}
Answer:
0 0 0 0
Explanation:
The
variable "I" is declared as static, hence memory for I will be
allocated for only once, as it encounters the statement. The function main()
will be called recursively unless I becomes equal to 0, and since main() is
recursively called, so the value of static I ie., 0 will be printed every time
the control is returned.
97) void main()
{
int
k=ret(sizeof(float));
printf("\n
here value is %d",++k);
}
int ret(int ret)
{
ret +=
2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name
can be the same.
Firstly,
the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret
will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed
after and preincrement.
98) void main()
{
char
a[]="12345\0";
int
i=strlen(a);
printf("here
in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char
array 'a' will hold the initialized string, whose length will be counted from 0
till the null character. Hence the 'I' will hold the value equal to 5, after
the pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned
giveit=-1;
int gotit;
printf("%u
",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char
a[]="\0";
if(printf("%s\n",a))
printf("Ok
here \n");
else
printf("Forget
it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence
printing a null character returns 1 which makes the if statement true, thus
"Ok here" is printed.
101) void main()
{
void *v;
int
integer=2;
int
*i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer
arithmetic can be done on it. Void pointers are normally used for,
1.
Passing generic
pointers to functions and returning such pointers.
2.
As a intermediate
pointer type.
3.
Used when the exact
pointer type will be known at a later point of time.
102) void main()
{
int
i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is
available to use in program code from the point of its declaration.
So expressions such as
i = i++ are valid statements. The i, j and k are automatic variables and
so they contain some garbage value.
Garbage in is garbage out (GIGO).
103) void main()
{
static int
i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The
printf returns no of characters printed and this value also cannot be
predicted. Still the outer printf prints
something and so returns a non-zero value. So it encounters the break statement
and comes out of the while statement.
104) main()
{
unsigned
int i=10;
while(i-->=0)
printf("%u
",i);
}
Answer:
10
9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become
negative. So the expression i-- >=0
will always be true, leading to an infinite loop.
105) #include<conio.h>
main()
{
int
x,y=2,z,a;
if(x=y%2)
z=2;
a=2;
printf("%d
%d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition
reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control
paths to write bug free code.
106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a
cancels out. The result is as simple as 1 + 3 = 4 !
107) #define prod(a,b) a*b
main()
{
int
x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro
expands and evaluates to as:
x+2*y-1
=> x+(2*y)-1 => 10
108) main()
{
unsigned
int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the
value of i becomes 0 it comes out of while loop. Due to post-increment on i the
value of i while printing is 1.
109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the
only dummy operator in C. So it has no
effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while
loop. The value –1 is printed due to the post-decrement operator.
113) main()
{
float
f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants
cannot be modified, so you cannot apply ++.
Bit-wise operators and %
operators cannot be applied on float values.
fmod() is to find the
modulus values for floats as % operator is for ints.
110) main()
{
int i=10;
f(i++,i++,i++);
printf("
%d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown
type integer
Compiler error:
undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used.
It means that the function follows Pascal argument passing mechanism in calling
the functions.
111) void pascal
f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
printf("
%d\n",i);
i=10;
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to
be called from left to right. cdecl is the normal C argument passing mechanism
where the arguments are passed from right to left.
112). What is the output of the program given below
main()
{
signed char
i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe
initial value of the i is set to 0. The inner loop executes to increment the
value from 0 to 127 (the positive range of char) and then it rotates to the
negative value of -128. The condition in the for loop fails and so comes out of
the for loop. It prints the current value of i that is -128.
113) main()
{
unsigned char
i=0;
for(;i>=0;i++)
;
printf("%d\n",i);
}
Answer
infinite
loop
Explanation
The difference between the previous question and this one
is that the char is declared to be unsigned. So the i++ can never yield
negative value and i>=0 never becomes false so that it can come out of the
for loop.
114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior
is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by
default is implementation dependent. If the implementation treats the char to
be signed by default the program will print –128 and terminate. On the other
hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation
dependent behavior. But dont write programs that depend on such behavior.
115) Is the following statement a declaration/definition.
Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a
pointer to array of(size 10) integers.
Apply clock-wise rule to find the
meaning of this definition.
116). What is the output for the program given below
typedef enum
errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler
error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One
means that it is a enumerator constant with value 1. The another use is that it
is a type name (due to typedef) for enum errorType. Given a situation the
compiler cannot distinguish the meaning of error to know in what sense the
error is used:
error
g1;
g1=error;
// which
error it refers in each case?
When the compiler can distinguish between
usages then it will not issue error (in pure technical terms, names can only be
overloaded in different namespaces).
Note:
the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided
just for programmer’s convenience.
117) typedef struct error{int warning, error,
exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The three usages of name errors can be distinguishable by the
compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error,
exception;}error;
This error can be used only by preceding the error by
struct kayword as in:
struct error someError;
typedef struct error{int warning, error,
exception;}error;
This can be used only after . (dot) or -> (arrow)
operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int
warning, error, exception;}error;
This can be used to define variables without using the
preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these
three usages, it is perfectly legal and valid.
Note
This code is given here to just explain the concept behind.
In real programming don’t use such overloading of names. It reduces the
readability of the code. Possible doesn’t mean that we should use it!
118) #ifdef
something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler
error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation.
The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
119) #if something
== 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
0
0
Explanation
This code is to show that preprocessor expressions are not
the same as the ordinary expressions. If a name is not known the preprocessor
treats it to be equal to zero.
120). What is the output for the following program
main()
{
int
arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D ==
arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and
pointers. N dimensional arrays are made up of (N-1) dimensional arrays.
arr2D is
made up of a 3 single arrays that contains 3 integers each .
|
|
The name arr2D refers to the beginning of all the 3 arrays.
*arr2D refers to the start of the first 1D array (of 3 integers) that is the
same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a
zero doesn’t change the value/meaning. Again arr2D[0] is the another way of
telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the
result is true(1) and the same is printed.
121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are
represented in memory”);
}
Answer
You can answer this if you know how values are
represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates
on 0 to produce all ones to fill the space for an integer. –1 is represented in
unsigned value as all 1’s and so both are equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int
x=10,y=20;
swap(&x,&y);
printf("x=
%d y = %d\n",x,y);
}
Answer
x = 20 y =
10
Explanation
This is one way of swapping two values. Simple checking
will help understand this.
123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler
error: Lvalue required in function main
Explanation:
++i
yields an rvalue. For postfix ++ to
operate an lvalue is required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and
++*p++. Parenthesis just works as a visual clue for the reader to see which
expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to
functions that takes no arguments and returns the type int. By the assignment
ptr[0] = aaa; it means that the first function pointer in the array is
initialized with the address of the function aaa. Similarly, the other two
array elements also get initialized with the addresses of the functions bbb and
ccc. Since ptr[2] contains the address of the function ccc, the call to the
function ptr[2]() is same as calling ccc(). So it results in printing "bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because ==
is of higher precedence than = operator. In the inner expression, ++i is equal
to 6 yielding true(1). Hence the result.
128) main()
{
char
p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes,
“%c\n”. Since this string becomes the format string for printf and ASCII value
of 65 is ‘A’, the same gets printed.
129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a ptr to a
function which takes 2 parameters .(a). an integer variable.(b). a
ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise
rule to find the result.
130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no
difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false)
hence breaking out of the while loop.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with
characters, ‘\0’ is not appended automatically to the string. Since str1
doesn’t have null termination, it treats whatever the values that are in the
following positions as part of the string until it randomly reaches a ‘\0’. So
str1 and str2 are not the same, hence the result.
132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
Compiler Error: Lvalue
required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side
of an assignment operation.
133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized,
whereas calloc returns the allocated memory space initialized to zeros.
134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”,
i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while
loop the conditional operator evaluates to false, executing i--. This continues
till the integer value rotates to positive value (32767). The while condition
becomes false and hence, comes out of the while loop, printing the i value.
135) main()
{
int i=10,j=20;
j = i,
j?(i,j)?i:j:j;
printf("%d
%d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ?
: ) is equivalent for if-then-else statement. So the question can be written
as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a
constant char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a'
rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
3. Same as 1.
137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d
%d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as
i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1
because j==10. i is 5. i = 5&1 is 1. Hence the result.
138) main()
{
int i=4,j=7;
j = j ||
i++ && printf("YOU CAN");
printf("%d
%d", i, j);
}
Answer:
4 1
Explanation:
The boolean
expression needs to be evaluated only till the truth value of the expression is
not known. j is not equal to zero itself
means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where
(anything) will not be evaluated. So the remaining expression is not evaluated
and so the value of i remains the same.
Similarly when && operator is involved in an
expression, when any of the operands become false, the whole expression’s truth
value becomes false and hence the remaining expression will not be
evaluated.
false && (anything) => false
where (anything) will not be evaluated.
139) main()
{
register int a=2;
printf("Address
of a = %d",&a);
printf("Value of
a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied
on register variables.
140) main()
{
float i=1.5;
switch(i)
{
case
1: printf("1");
case 2:
printf("2");
default :
printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.
141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so
using extern has no use in resolving it.
142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d
%d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So
changes through f1 and f2 ultimately affects only the value of a.
143) main()
{
char
*p="GOOD";
char a[
]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p)
= %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n
sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a)
= 5, strlen(a) = 4
Explanation:
sizeof(p) =>
sizeof(char*) => 2
sizeof(*p)
=> sizeof(char) => 1
Similarly,
sizeof(a)
=> size of the character array => 5
When sizeof operator is applied to an array it returns the
sizeof the array and it is not the same as the sizeof the pointer variable.
Here the sizeof(a) where a is the character array and the size of the array is
5 because the space necessary for the terminating NULL character should also be
taken into account.
144) #define
DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int
arr[10];
printf(“The
dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer
array of 10 elements is 10 * sizeof(int). The macro expands to
sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
145) int
DIM(int array[])
{
return
sizeof(array)/sizeof(int );
}
main()
{
int
arr[10];
printf(“The
dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be
passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one
of the very few places where [] and * usage are equivalent). The return
statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this
case.
146) main()
{
static int
a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static
*p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1
1 1
2 4
2 4
3 7
3 7
4 2
4 2
5 5
5 5
6 8
6 8
7 3
7 3
8 6
8 6
9 9
9 9
Explanation:
*(*(p+i)+j) is
equivalent to p[i][j].
147) main()
{
void swap();
int
x=10,y=8;
swap(&x,&y);
printf("x=%d
y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without
using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function
that may take any number of arguments (not no arguments) and returns nothing.
So this doesn’t issue a compiler error by the call swap(&x,&y); that
has two arguments.
This convention is historically due to pre-ANSI style
(referred to as Kernighan and Ritchie style) style of function declaration. In
that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^=
*b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the
declaration for swap will look like, void swap() which means the swap can take
any number of arguments.
148) main()
{
int i = 257;
int *iPtr
= &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001
00000001, so the individual bytes are taken by casting it to char * and get
printed.
149) main()
{
int i = 258;
int *iPtr
= &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as,
00000001 00000001. Remember that the INTEL machines are ‘small-endian’
machines. Small-endian means that the
lower order bytes are stored in the higher memory addresses and the higher
order bytes are stored in lower addresses. The integer value 258 is stored
in memory as: 00000001 00000010.
150) main()
{
int i=300;
char *ptr
= &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300
in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100
00000001. Result of the expression *++ptr = 2 makes the memory representation
as: 00101100 00000010. So the integer corresponding to it is
00000010 00101100 => 556.
151) #include <stdio.h>
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr<least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed
by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the
value of ‘least’ finally is 0.
152) Declare an
array of N pointers to functions returning pointers to functions returning
pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])(
);
153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int
day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day,
&student.dob.month,
&student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of
type struct date is given. The compiler doesn’t have the definition of date
structure (forward reference is not
allowed in C in this case) so it issues an error.
154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int
day,month,year;
};
scanf("%s%d%d%d", stud.rollno,
&student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the
structure definition of ‘student’ but to have a variable of type struct date
the definition of the structure is required.
155) There were 10
records stored in “somefile.dat” but the following program printed 11 names.
What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud,
sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully.
It will return EOF only when fread tries to read another record and fails
reading EOF (and returning EOF). So it prints the last record again. After this
only the condition feof(fp) becomes false, hence comes out of the while loop.
156) Is there any difference between the two declarations,
1.
int foo(int *arr[])
and
2.
int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The
numbers that are allowed inside the [] is just for more readability. So there
is no difference between the two declarations.
157) What is the subtle error in the following
code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++<n)
p
= &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no
address. So p remains NULL where *p =0 may result in problem (may rise to
runtime error “NULL pointer assignment” and terminate the program).
158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out
bugs. The check s != NULL is for error/exception handling and for that assert
shouldn’t be used. A plain if and the corresponding remedy statement has to be
given.
159) What is the hidden bug with the
following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release
version. In assert, the experssion involves side-effects. So the behavior of
the code becomes different in case of debug version and the release version
thus leading to a subtle bug.
Rule to Remember:
Don’t use
expressions that have side-effects in assert statements.
160) void main()
{
int *i = 0x400; // i
points to the address 400
*i = 0; //
set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because
it points to some location whose value may not be available for
modification. This type of pointer in which the non-availability of the
implementation of the referenced location is known as 'incomplete type'.
161) #define assert(cond) if(!(cond)) \
(fprintf(stderr,
"assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__),
abort())
void main()
{
int i = 10;
if(i==0)
assert(i <
100);
else
printf("This
statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else
for if in the assert macro. Hence nothing is printed.
The solution is to use conditional operator instead of if
statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr,
"assertion failed: \ %s, file %s, line %d \n",#cond,
__FILE__,__LINE__), abort()))
Note:
However this problem of “matching with nearest else” cannot
be solved by the usual method of placing the if statement inside a block like
this,
#define assert(cond) { \
if(!(cond)) \
(fprintf(stderr,
"assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__),
abort()) \
}
162) Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is
of the same
type as in this case. Because this will cause the structure
declaration to be recursive without end.
163) Is the following code legal?
struct a
{
int x;
struct a
*b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The
compiler knows, the size of the pointer to a structure even before the size of
the structure
is determined(as you know the pointer to any type is of
same size). This type of structures is known as ‘self-referencing’ structure.
164) Is the following code legal?
typedef struct a
{
int x;
aType *b;
}aType
Answer:
No
Explanation:
The typename aType is not known at the point of declaring
the structure (forward references are not made for typedefs).
165) Is the following code legal?
typedef struct a aType;
struct a
{
int x;
aType *b;
};
Answer:
Yes
Explanation:
The typename aType is known at the point of declaring the
structure, because it is already typedefined.
166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
aType *b;
};
}
Answer:
No
Explanation:
When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This is known
as ‘incomplete types’.
167) void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
printf(“sizeof (int *) = %d \n”, sizeof(int *));
printf(“sizeof (double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct unknown *) =
%d \n”, sizeof(struct unknown *));
}
Answer :
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *)
= 2
sizeof(struct unknown *) =
2
Explanation:
The pointer to any type is of same size.
168) char inputString[100] = {0};
To get string input from the keyboard which one of the
following is better?
1)
gets(inputString)
2)
fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString) doesn't
know the size of the string passed and so, if a very big input (here, more than
100 chars) the charactes will be written past the input string. When fgets is
used with stdin performs the same operation as gets but is safe.
169) Which version do you prefer of the
following two,
1) printf(“%s”,str); //
or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any format characters like %d then it will result
in a subtle bug.
170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started
in line 5”.
Explanation:
The programmer intended to divide two integers, but by the
“maximum munch” rule, the compiler treats the operator sequence / and * as /*
which happens to be the starting of comment. To force what is intended by the
programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“,
ch, ch);
}
Answer:
Implementaion
dependent
Explanation:
The char type may be signed or unsigned by default. If it
is signed then ch++ is executed after ch reaches 127 and rotates back to -128.
Thus ch is always smaller than 127.
172) Is this code legal?
int *ptr;
ptr = (int *) 0x400;
Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location
0x400.
173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many
initializers
Explanation:
The array a is of size 4 but the string constant requires 6
bytes to get stored.
174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the
string “HELL” and doesnt have enough space to store the terminating null
character. So it prints the HELL correctly and continues to print garbage
values till it accidentally comes
across a NULL character.
175) main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u %u ",j,k);
}
Answer:
Compiler error: Cannot
increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used
only when the type is not known and as an intermediate address storage type. No
pointer arithmetic can be done on it and you cannot apply indirection operator
(*) on void pointers.
176) main()
{
extern int
i;
{ int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int i;
177) Printf can be implemented by using __________ list.
Answer:
Variable length argument
lists
178) char *someFun()
{
char *temp
= “string constant";
return
temp;
}
int main()
{
puts(someFun());
}
Answer:
string
constant
Explanation:
The
program suffers no problem and gives the output correctly because the character
constants are stored in code/data area and not allocated in stack, so this
doesn’t lead to dangling pointers.
179) char
*someFun1()
{
char temp[
] = “string";
return
temp;
}
char
*someFun2()
{
char temp[
] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return
temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the
problem of dangling pointers. In someFun1() temp is a character array and so
the space for it is allocated in heap and is initialized with character string
“string”. This is created dynamically as the function is called, so is also
deleted dynamically on exiting the function so the string data is not available
in the calling function main() leading to print some garbage values. The
function someFun2() also suffers from the same problem but the problem can be
easily identified in this case.
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