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NOW SOME QUESTIONS ON POINTERS

1)What will be output of following program?

#include
#include
int main(){
char *ptr1 = NULL;
char *ptr2 = 0;
strcpy(ptr1,"jay");
strcpy(ptr2,"sanghvi");
printf("\n%s %s",ptr1,ptr2);
return 0;
}

Explanation:

Turbo C++ 3.0: (null) (null)

Turbo C ++4.5: Run time error

Linux GCC: Run time error

Visual C++: Run time error



We cannot assign any string constant in null pointer by strcpy function.

2)What will be output of following program?

#include
int main(){
int a = 10;
void *p = &a;
int *ptr = p;
printf("%u",*ptr);
return 0;
}
ANS:-10

Void pointer can hold address of any data type without type casting. Any pointer can hold void pointer without type casting.

Excellent questions with answers

1)

#include<stdio.h>

void main()

{

char arr[6]="jaynam" ;

printf("%s",arr);

}

output:-  Garbage Value

Explanation:-

Size of  a character array should one greater than total number of characters in any string which it stores.In c every string has one terminating null character.this represents end of string.

so in the string "jaynam" there are 6 characters and they are 'j','a','y','n','a','m' and '/o'.size of array six.so array will only store first 6 characters and it will not store null character.

As we know %s in printf statement prints stream of characters until it doesn't get first null characters.since array arr has not store any null character so it will print garbage value...

2)

#include<stdio.h>

void main()

{

int goto=4;

printf("%d",goto);

}

output-Compilation error

Explanation:-

goto is a keyword in c.variable name cannot be any keyword of c language..

3).

#include<stdio.h>

int main(){

     int i=1;

     i=2+2*i++;

     printf("%d",i);

     return 0;

}

output-5

Explanation:-

i++ i.e. when postfix increment operator is used any expression the it first assign the value in the expression the it increments the value of variable by one. So,

i = 2 + 2 * 1

i = 4

Now i will be incremented by one so i = 4 + 1 = 5

4)

#include<stdio.h>

int main(){

     int a=1,b=2,c=3;

     c=a==b;

     printf("%d",c);

     return 0;

}

output-0

Explanation:-

 == is relational operator which returns only two values.

0: If a == b is false

1: If a == b is true

Since

a=2

b=7

So, a == b is false hence c=0

5)

#include<stdio.h>

void main(){

      int a=5;

      a=a>=4;

      switch(2){

          case 0:int a=8;

          case 1:int a=10;

          case 2:++a;

          case 3:printf("%d",a);

      }

}

output-error

Explanation:-

We can not declare any variable in any case of switch case statement.

6)

#include<stdio.h>

#include<string.h>

void main(){

  int i=0;

  for(;i<=2;)

    printf(" %d",++i);

}

output-1 2 3

Explanation:-

In for loop each part is optional.

7)Program for Subtract two integer numbers without using subtraction operator

#include<stdio.h>

void main(){

     int a,b,d;

     printf("enter two numbers u want to subtract");

     scanf("%d%d",&a,&b);

     d=a+~b+1;

     printf("ans=%d",d);

}

Output:-enter two numbers u want to subtract

5

4

ans=1

Explanation:-

 ~ means ones complement

ones complement of 4=1011

add 1 to this that is 1100

binary number for 5=0101

nw add 0101+1100=0001

that is equal to 1 

8)

Write a c program without using any semicolon whose output is: jaynam.

#include<stdio.h>

void main(){

     switch(printf("jaynam")){

     }

}

Output:-jaynam

9)what is the output of the following program??

#include<stdio.h>

void main(){

char arr[10];

arr = "jaynam";

printf("%s",arr);

}

Output:-Compilation error

Explanation:-

Compilation error Lvalue required

Array name is constant pointer and we cannot assign any value in constant data type after declaration.

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nice concept behind this

what will be the output of this program?

#include<stdio.h>
int main()
{
int x=011,i;
for(i=0;i<x;i+=3)
{
printf("start");
continue;
printf("end");
}
return 0;
}

output:-start start start 

Explanation:-
011 is octal number. Its equivalent decimal value is 9.
So, x = 9