How a function can access value of local variable of another function whose stack is already gone?

Question

Even I have similar problem:

int (*pfun)(void);
int *pInt = 0;
void fun1()
{
    int i = 5; /* Local to fun1*/

    printf("Outer function");
    pInt = &i; /* As I know address of local variable is valid till function execution */

    int fun2()
    {
      printf("innerfunction");
      printf("%d", *pInt);
    }
    /* fun2 address assign to pfun so that It can be called even after completion of fun1 */
    pfun = fun2;
}

int main()
{
    fun1();
    pfun();
    return 0;
}

Can someone please explain ? I am getting *pInt value 5.

Comments

related to thread:
http://queryhome.com/16927/can-we-define-function-with-in-function?show=16990#c16990

Answer 1

1) Because address space for stack is not overwritten. So you are getting the same same value "5".
2) Stack got memory form heap, heap exist in address space of that process.

If you want to access the local variable.
Declare the local variable as static. Static variables goes in data section.
Now you can store the address of these static variables in global pointers
OR you can return address of variable.

Please correct if I am wrong.

Comments

Is this behavior consistence for all the system ?