Thread: Doubt in C Code (printed return value of local variable's address is 0)

Hello Cboard members,

I have written the following code. I am unable to understand why the returned value is 0 while in the function hi() it is non-zero.

Code:

#include<stdio.h>
int* hi(void);
int main() {
  int *Dptr;
  int n=5;
  Dptr = &n;
  printf("\n %u", Dptr);
  Dptr = hi();
  printf("\n %u", Dptr); // Why is this printed 0 on execution ?
}
int* hi(void)
{
  int *Rptr;
  int n=5;
  Rptr = &n;
  printf("\n %u", Rptr);
  return &n;    


}

Regards,
HitX11

The "address of" a local (non-static) variable is undefined after returning from the function; the fact that it's 0 in this case is immaterial (it could be anything it wants to be simply because it's undefined).

Just so you know, use "%p" when printing a pointer.

Code:

#include <stdio.h>

int main(void)
{
    int n = 5;
    int *p = &n;

    printf("%p -> %d\n", (void *)p, *p);

    return 0;
}