Question

Below is the code

The Code:

#include <stdio.h>

int * num(void);

int main(void)
{
    int * num2;
    num2 =num();
    printf("%d\n" , *num2);

    return 0;
}


int * num(void)
{
    int num = 20;

    return &num;
}

The Question :

1. As we know , the function num is local to its function num() , so in this code I try to return the address of the variable num in the function to the function that calls it , which is main().

2. After that I just use the dereferencing operator to extract the value of the specific num variable and print it out in main() function.

3. There’s one thing i’m confused . I remember i read a book about javascript that mention a variable lifetime is within the function , which mean after the function finish performing its instructions and pass the control back to the function that calls it , the value of each variable in the function will be clean out(garbage collector).But why in this code my main() function still can point to the value of that specific memory address??

Answer 1

The reason you can see the value of the variable is because of how the stack works. Effectively as you enter the function num, a pointer (the stack pointer) is moved to add space in for the local storage of the function. When you exit the function the stack pointer is moved back effectively meaning that the next function call will overwrite the stack storage used in the previous function call. Until it is overwritten, however, the value exists in a sort of limbo. It’s still actually there in memory but may get overwritten at any moment. The existence of the actual value there may or may not be the case. That is why doing as you do above is known as undefined behaviour.

Basically … don’t do it.