viceroy: Feb 25, 2012

1.What will be output if you will compile and execute the following c code?

void main(){

int i=10;

static int x=i;

if(x==i)

printf("Equal");

else if(x>i)

printf("Greater than");

else

printf("Less than");

}

(a) Equal

(b) Greater than

(d) Compiler error

(e) None of above

Answer: (d)

Explanation:

static variables are load time entity while auto variables are run time entity. We can not initialize any load time variable by the run time variable.

In this example i is run time variable while x is load time variable.

2.What will be output if you will compile and execute the following c code?

void main(){

int i;

float a=5.2;

char *ptr;

ptr=(char *)&a;

for(i=0;i<=3;i++)

printf("%d ",*ptr++);

}

(a)0 0 0 0

(b)Garbage Garbage Garbage Garbage

(d)102 102 -90 64

(e)Compiler error

Answer: (d)

Explanation:

In c float data type is four byte data type while char pointer ptr can point one byte of memory at a time.

Memory representation of float a=5.2

ptr pointer will point first fourth byte then third byte then second byte then first byte.

Content of fourth byte:

Binary value=01100110

Decimal value= 64+32+4+2=102

Content of third byte:

Binary value=01100110

Decimal value=64+32+4+2=102

Content of second byte:

Binary value=10100110

Decimal value=-128+32+4+2=-90

Content of first byte:

Binary value=01000000

Decimal value=64

Note: Character pointer treats MSB bit of each byte i.e. left most bit of above figure as sign bit.

3.What will be output if you will compile and execute the following c code?

void main(){

printf("%s","c" "question" "bank");

}

(a) c question bank

(b) c

(d) cquestionbank

(e) Compiler error

Answer: (d)

Explanation:

In c string constant “xy” is same as “x” “y”

4.What will be output if you will compile and execute the following c code?

void main(){

printf("%s",__DATE__);

}

(a) Current system date

(b) Current system date with time

(d) Compiler error

(e) None of these

Answer: (a)

Explanation:

__DATE__ is global identifier which returns current system date.

5.What will be output if you will compile and execute the following c code?

void main(){

char *str="c-pointer";

printf("%*.*s",10,7,str);

}

(a) c-pointer

(b) c-pointer

(d) cpointer null null

(e) c-point

Answer: (e)

Explanation:

Meaning of %*.*s in the printf function:

First * indicates the width i.e. how many spaces will take to print the string and second * indicates how many characters will print of any string.

Following figure illustrates output of above code:

6.What will be output if you will compile and execute the following c code?

void main(){

int a=-12;

a=a>>3;

printf("%d",a);

}

(a) -4

(b) -3

(d) -96

(e) Compiler error

Answer :( c)

Explanation:

Binary value of 12 is: 00000000 00001100

Binary value of -12 wills 2’s complement of 12 i.e.

So binary value of -12 is: 11111111 11110100

Right shifting rule:

Rule 1: If number is positive the fill vacant spaces in the left side by 0.

Rule 2: If number is negative the fill vacant spaces in the left side by 1.

In this case number is negative. So right shift all the binary digits by three space and fill vacant space by 1 as shown following figure:

Since it is negative number so output will also a negative number but its 2’s complement.

Hence final out put will be:

And its decimal value is: 2

7.What will be output if you will compile and execute the following c code?

#include "string.h"

void main(){

clrscr();

printf("%d%d",sizeof("string"),strlen("string"));

getch();

}

(a) 6 6

(b) 7 7

(d) 7 6

(e) None of these

Answer: (d)

Explanation:

Sizeof operator returns the size of string including null character while strlen function returns length of a string excluding null character.

8.What will be output if you will compile and execute the following c code?

void main(){

static main;

int x;

x=call(main);

clrscr();

printf("%d ",x);

getch();

}

int call(int address){

address++;

return address;

}

(a) 0

(b) 1

(d) Compiler error

(e) None of these

Answer: (b)

Explanation:

As we know main is not keyword of c but is special type of function. Word main can be name variable in the main and other functions.

9. What will be output if you will compile and execute the following c code?

void main(){

int a,b;

a=1,3,15;

b=(2,4,6);

clrscr();

printf("%d ",a+b);

getch();

}

(a) 3

(b) 21

(d) 7

(e) Compiler error

Answer: (d)

Explanation:

In c comma behaves as separator as well as operator.

a=1, 3, 15;

b= (2, 4, 6);

In the above two statements comma is working as operator. Comma enjoys least precedence and associative is left to right.

Assigning the priority of each operator in the first statement:

Hence 1 will assign to a.

Assigning the priority of each operator in the second statement:

10. What will be output if you will compile and execute the following c code?

int extern x;

void main()

printf("%d",x);
x=2;

getch();

}

int x=23;

(a) 0

(b) 2

(d) Compiler error

(e) None of these

Answer: (c)

Explanation:

extern variables can search the declaration of variable any where in the program.

11. What will be output if you will compile and execute the following c code?

int extern x;

void main()

printf("%d",x);
x=2;

getch();

}

int x=23;

(a) 0

(b) 2

(d) Compiler error

(e) None of these

Answer: (c)

Explanation:

extern variables can search the declaration of variable any where in the program.

12.What will be output if you will compile and execute the following c code?

void main(){

int a=25;

clrscr();

printf("%o %x",a,a);

getch();

}

(a) 25 25

(b) 025 0x25

(d) 31 19

(e) None of these

Answer: (d)

Explanation:

%o is used to print the number in octal number format.

%x is used to print the number in hexadecimal number format.

Note: In c octal number starts with 0 and hexadecimal number starts with 0x.

13. What will be output if you will compile and execute the following c code?

#define message "union is\

power of c"

void main(){

clrscr();

printf("%s",message);

getch();

}

(a) union is power of c

(b) union ispower of c

Power of c

(d) Compiler error

(e) None of these

Answer: (b)

Explanation:

If you want to write macro constant in new line the end with the character \.

14. What will be output if you will compile and execute the following c code?

#define call(x) #x

void main(){

printf("%s",call(c/c++));

}

(a)c

(b)c++

(c)#c/c++

(d)c/c++

(e)Compiler error

Answer: (d)

Explanation:

# is string operator. It converts the macro function call argument in the string. First see the intermediate file:

test.c 1:

test.c 2: void main(){

test.c 3: printf("%s","c/c++");

test.c 4: }

test.c 5:

It is clear macro call is replaced by its argument in the string format.

II.DOES NOT contain any options

(15) What will be output of following code?

#include<stdio.h>

#define max 10

int main(){

int i;

i=++max;

printf("%d",i);

return 0;

}

(1)
output: compiler error.
Explanation:

Here max is preprocessor macro symbol which process first before the actual compilation. First preprocessor replace the symbol to its value in entire the program before the compilation. So in this program max will be replaced by 10 before compilation. Thus program will be converted like this:

int main(){

int i;

i=++10;

printf("%d",i);

return 0;

}

In this program we are trying to increment a constant symbol.

Meaning of ++10 is:
10=10+1
or 10=11

Which is error because we cannot assign constant value to another constant value .Hence compiler will give error.

(16) What will be output of following code?

#include<stdio.h>

#define max 10+2

int main(){

int i;

i=max*max;

printf("%d",i);

return 0;

}
Output: 32

Explanation:
Here max is preprocessor macro symbol which process first before the actual compilation start. Preprocessor replace the symbol to its value in entire the program before the compilation. So in this program max will be replaced by 10+2 before compilation. Thus program will be converted as:

int main(){

int i;

i=10+2*10+2;

printf("%d",i);

return 0;

}

now i=10+2*10+2
i=10+20+2
i=32

(17) What will be output of following code?

#include<stdio.h>

#define A 4-2

#define B 3-1

int main(){

int ratio=A/B;

printf("%d ",ratio);

return 0;

}

Output:3 3
Explanation:

A and B are preprocessor macro symbol which process first before the actual compilation start. Preprocessor replace the symbol to its value in entire the program before the compilation. So in this program A and B will be replaced by 4-2 and 3-1 respectively before compilation. Thus program will be converted as:

int main(){

int ratio=4-2/3-1;

printf("%d ",ratio);

return 0;

}

Here ratio=4-2/3-1
ratio=4-0-1
ratio=3

(18) What will be output of following code?

#include<stdio.h>

#define MAN(x,y) (x)>(y)?(x):(y)

int main(){

int i=10,j=9,k=0;

k=MAN(i++,++j);

printf("%d %d %d",i,j,k);

return 0;

}

Output: 11 11 11

Explanation:

Preprocessor’s macro which process first before the actual compilation. Thus program will be converted as:

int main(){

int i=10,j=9,k=0;

k=(i++)>(++j)?(i++):(++j);

printf("%d %d %d",i,j,k);

return 0;

}
now k=(i++)>(++j)?(i++):(++j);
first it will check the condition
(i++)>(++j)
i++ i.e. when postfix is used with variable in expression then expression is evaluated first with original value then variable is incremented

Or 10>10
This condition is false.
Now i = 10+1 = 11
There is rule, only false part will execute after? i.e. ++j, i++ will be not execute.
So after ++j
j=10+1=11;

And k will assign value of j .so k=11;

(19) What will be output of following code?

#include<stdio.h>

#define START main() {

#define PRINT printf("*******");

#define END }

START

END

Output: *******
Explanation:

This program will be converted as:

main(){

printf("*******");

}

(20) What will be output of following code?

#define CUBE(x) (x*x*x)

#define M 5

#define N M+1

#define PRINT printf("RITESH");

int main(){

int volume =CUBE(3+2);

printf("%d %d ",volume,N);

return 0;

}

Output: 17 6
Explanation: This program will be converted as:

int main(){

int volume =(3+2*3+2*3+2);

printf("%d %d ",volume,5+1);

return 0;

}

(21)

#include<stdio.h>

#define ABC 25

#define PQR "Exact Help"

int main(){

int num = 3;

#ifdef ABC

printf("%d",ABC * ABC);

#else

printf("%s",PQR);

#endif

return 0;

}

Output: 625

Explanatiopn: Since macro constant ABC has defined so #ifdef condition is true.

Pragma is implementation specific directive i.e each pragma directive has different implementation rule and use . There are many type of pragma directive and varies from one compiler to another compiler .If compiler does not recognize particular pragma the it simply ignore that pragma statement without showing any error or warning message and execute the whole program assuming this pragma statement is not present. For example suppose there is any pragma directive is #pragma world .

(22).#include<stdio.h>

#pragma world

int main(){

printf("C is powerful language ");

return 0;

}

Output : C is powerful language

Explanation:

Since #pragma world is unknown for Turbo c 3.0 compiler so it will ignore this directive without showing any error or warning message and execute the whole program assuming #pragma world statement is not present.

List of pragma directives in turbo c 3.0:

1. #pragma startup

2. #pragma exit

3. #pragma warn

4. #pragma option

5. #pragma inline

6. #pragma argsused

7. #pragma hdrfile

8. #pragma hdrstop

9. #pragma saveregs

What is dangling pointer in c?

Explanation:

Dangling pointer:

If any pointer is pointing the memory address of any variable but after some variable has deleted from that memory location while pointer is still pointing such memory location. Such pointer is known as dangling pointer and this problem is known as dangling pointer problem.

Initially:

Later:

(23).What will be output of following c program?

#include<stdio.h>

int *call();

int main(){

int *ptr;

ptr=call();

fflush(stdin);

printf("%d",*ptr);

return 0;

}

int * call(){

int x=25;

++x;

return &x;

}

Output: Garbage value

Note: In some compiler you may get warning message returning address of local variable or temporary

Explanation: variable x is local variable. Its scope and lifetime is within the function call hence after returning address of x variable x became dead and pointer is still pointing ptr is still pointing to that location.

Solution of this problem:

Make the variable x is as static variable. In other word we can say a pointer whose pointing object has been deleted is called dangling pointer.

#include<stdio.h>

int *call();

int main(){

int *ptr;

ptr=call();

fflush(stdin);

printf("%d",*ptr);

return 0;

}

int * call(){

static int x=25;

++x;

return &x;

}

Output: 26

What is pointer to a function?

Explanation:

(24) What will be output if you will execute following code?

int * function();

int main(){

auto int *x;

int *(*ptr)();

ptr=&function;

x=(*ptr)();

printf("%d",*x);

}

int *function(){

static int a=10;

return &a;

}

Output: 10

Explanation: Here function is function whose parameter is void data type and return type is pointer to int data type.

x=(*ptr)()

=> x=(*&functyion)() //ptr=&function

=> x=function() //From rule *&p=p

=> x=&a

So, *x = *&a = a =10

(25) What will be output if you will execute following code?

int find(char);

int(*function())(char);

int main(){

int x;

int(*ptr)(char);

ptr=function();

x=(*ptr)('A');

printf("%d",x);

return 0;

}

int find(char c){

return c;

}

int(*function())(char){

return find;

}

Output: 65

Explanation: Here function whose name is function which passing void data type and returning another function whose parameter is char data type and return type is int data type.

x=(*ptr)(‘A’)

=> x= (*function ()) (‘A’) //ptr=function ()

//&find=function () i.e. return type of function ()

=> x= (* &find) (‘A’)

=> x= find (‘A’) //From rule*&p=p

=> x= 65

(26) What will be output if you will execute following code?

char * call(int *,float *);

int main(){

char *string;

int a=2;

float b=2.0l;

char *(*ptr)(int*,float *);

ptr=&call;

string=(*ptr)(&a,&b);

printf("%s",string);

return 0;

}

char *call(int *i,float *j){

static char *str="c-pointers”;

str=str+*i+(int)(*j);

return str;

}

Output: inters

Explanation: Here call is function whose return type is pointer to character and one parameter is pointer to int data type and second parameter is pointer to float data type and ptr is pointer to such function.

str= str+*i+ (int) (*j)

=”c-pointers” + *&a+ (int) (*&b)

//i=&a, j=&b

=”c-pointers” + a+ (int) (b)

=”c-pointers” +2 + (int) (2.0)

=”c-pointers” +4

=”inters”

(27) What will be output if you will execute following code?

char far * display(char far*);

int main(){

char far* string="cquestionbank 1 ";

char far *(*ptr)(char far *);

ptr=&display;

string=(*ptr)(string);

printf("%s",string);

}

char far *display(char far * str){

char far * temp=str;

temp=temp+13;

*temp='\0';

return str;

}

Output: cquestionbank

Explanation: Here display is function whose parameter is pointer to character and return type is also pointer to character and ptr is its pointer.

temp is char pointer

temp=temp+13

temp=’\0’

Above two lines replaces first dot character by null character of string of variable string i.e.

"cquestionbank\0 1 "

As we know %s print the character of stream up to null character.

28. Which of the following statements should be used to obtain the reminder after dividing 56.4/6.3?

a. rem = 56.4%6.3;

b. rem = fmod(56.4,6.3);

c. rem = modf(56.4,6.3);

d. Remainder cannot be obtained for floating point division

(b) rem = fmod( 56.4,6.3)

29. How will you round off the value 7.66 to 8.0 ?

a. roundto(7.66);

b. ceil(7.66);

c. floor(7.66);

d. roundup(7.66);

(b) ceil(7.66);

30. Which of the following function sets first n characters of a string to a given character?

a. strinit()

b. strset()

c. strnset()

d. strcset()

viceroy

Saturday, February 25, 2012

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