What is the output!.

Question

#include <stdio.h>
void main()
{
int k = 5;
int *p = &k;
int **m = &p;
printf("%d%d%d\n", k, *p, **p);
}

Answer 1

athik rehman First of all you need to know this
(*) - Dereference Operator
(&) - Reference/Address Operator

int k = 5
// 'k' is assigned the value 5

int *p = &k
// a pointer variable 'p' is created which holds the address of variable 'k'
//p = &k, k = 5
//*p = 5

int **m = &p
// a double pointer variable is created which holds the address of pointer variable 'p'. The pointer variable 'p' holds the address of 'k'
//m = &p, p = &k, k = 5
//**m = 5

Answer 2

athik rehman Corrected the code, guess the output and then check if your guess is correct or not.

#include <stdio.h>
int main()
{
int k = 5;
int *p = &k;
int **m = &p;
printf("%d%d%d\n", k, *p, **m);
}

Answer 3

Why don't you run it?

Answer 4

k, the address of k, and the address where the address of k is stored.

Answer 5

but i got compile time error cant understand how?

Answer 6

nAutAxH AhmAd can u pls explain once! thanks