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Explain this code.

#include <stdio.h>

void f1(int *k)
{
*k = *k + 10;
}

main ( ){
int i;
i = 0;
printf (" The value of i before call %d \n", i);
f1 (&i);
printf (" The value of i after call %d \n", i);
}

Answers were Sorted based on User's Feedback



Explain this code. #include <stdio.h> void f1(int *k) { *k = *k + 10; } main (..

Answer / pooja agarwal

10

Is This Answer Correct ?    11 Yes 0 No

Explain this code. #include <stdio.h> void f1(int *k) { *k = *k + 10; } main (..

Answer / ankita sharma

answer will be 10. as k has the address of i so when we write *k=*k+10; *k meand that value to which k is pointing so it is pointing to i and i has the value 0 as intial value so 10 would be added to the value of i. so output  will be 10.

Is This Answer Correct ?    0 Yes 0 No

Explain this code. #include <stdio.h> void f1(int *k) { *k = *k + 10; } main (..

Answer / vivek

it will produce an error as constant cant be added to a pointer

Is This Answer Correct ?    1 Yes 4 No

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