c++ - Why is the address of a local variable identical for different executions?

Question

int fun(int x);

int main()
{
    fun(10);
    fun(11);   
    return 0;
}

int fun(int x)
{
    int loc;//local variable
    cout<<&loc;
    return 0;
}

Output is

0xbfb8e610 
0xbfb8e610

Here loc is a local variable, which goes out of scope after the 1st execution of the function f(10), then get allocated again for the next execution of fun(11). So the address of loc variable has to be different as per my understanding. Then why is the address &loc same for both the execution?

Answer 1

Each invocation of fun needs its own place to store the variable. But as soon as the function returns, the variable no longer exists. There's no reason the address can't be re-used. It doesn't have to be, but there's no reason it can't be.

In a typical implementation, stack space is used to hold the information needed to return from a function and their local variables when a function is invoked. When the function returns, the local variables are removed from the stack and the return information popped off it, leaving the stack back where it was when the function was called. Since the two function invocations are the same, they wind up with the stack the same in both cases, making the local variable have the same address. This is what an experienced programmer would would expect, but not rely on.