Thread: Warning : returns address of a local variable YET correct output

This problem was suggested to me by a friend.
The code is -

Code:

/* Returning a pointer to an out of scope variable */

# include <stdio.h>

int* fun(){
  int k = 35;
  return &k;
}
int main(void){
  int *j;
  j = fun();
  printf("%d",*j);
  return 0;
}

Compiling this on GCC gives the above warning because k is a local variable and it will go out of scope once the function returns. Yet, when I run the program I get a perfectly correct output. That is 35. Why is this? Shouldn't I be getting garbage value?