编译器遇到return语句怎么办？

Question

I have this code:-

int i;
class A
{
public:
    ~A()
    {
        i = 10;
    }
};

int& foo()
{
    i = 3;
    A ob;
    return i;
}

int main()
{
    cout << "i = " << foo() << endl; //call to foo
    return 0;
}

I'm confused about the order in which the compiler executes this code after encountering the return i statement. My guess: A variable of the caller(main) is made(say K) and a reference to it is passed to foo(), when the return statement is encountered, the compiler copies the value of the return statement into K, in this case the address of i, as a reference is being returned. Then it jumps to the end of foo() and the destructor for class object ob is called which changes the value of i to 10. This is followed by the removal of foo() and all its variables(including the reference to K passed to foo) from the call stack and the control is returned to main where K replaces the call to foo() and the cout is implemented. Is this the correct order of actions taken by the compiler ?

I searched extensively for what happens in memory when a return is called and this was the best result, though it is for C#.

Answer 1

The function foo is in scope any place inside this source file, so there's no problem calling it.

As for the reference it's returning, i is a file scope variable and thus has static storage duration, which means its lifetime is the life of the program. This means you're allowed to return a reference to it from foo .

Had you attempted to return a reference to a local variable like this:

int& foo()
{
    int x = 3;
    return x;
}

That would invokes undefined behavior since the variable no longer exists after the function exits.