C++ nonconst-const reference function overloading

Question

In the following code:

int foo(const int& f) //version 1
{
    int g = f;
    return int(foo(g)); // calls itself, turning into SO
}

int& foo(int& f) //version 2
{
    f *= -1;
    return f;
}

int main()
{
    int f = 11;
    cout << foo(f) << endl;
    cout << foo(22) << endl;
}

The first cout prints -11 as expected; f is a lvalue, so it binds to the second version of foo (although it could bind to 1st version as well, the 2nd version it's a better match).

The second call of foo is using a rvalue as parameter, so the only viable version of foo is the first one. So far, so good. Inside the first version of foo, I made a copy of the parameter so I could call the second version (with a lvalue) and return a copy of it after the call of the second version of foo. The thing is this will turn into a stack overflow; still the first version of foo will be called.

Could someone please explain to me why this happens? I would expect that g inside the first version of foo to bind to the second version of foo when passed as parameter.

Answer 1

It's simple really - foo at that point only means foo(const int& f). There's no second choice. Not yet. Switch up the definitions. Or separate them:

int foo(const int& f);
int& foo(int& f);

int main()
{
    int f = 11;
    cout << foo(f) << endl;
    cout << foo(22) << endl;
}


int foo(const int& f) //version 1
{
    int g = f;
    return int(foo(g)); // calls itself, turning into SO
}

int& foo(int& f) //version 2
{
    f *= -1;
    return f;
}

Answer 2

The first declaration of foo has no idea about the existence of the second one. Try this:

int foo(int& f);
int foo(const int& f) //version 1
{
    int g = f;
    return int(foo(g)); // calls itself, turning into SO
}

Answer 3

When the compiler reaches the line:

return int(foo(g))

It doesn't know about your overloaded version 2. Add prototype declarations to the top of the file:

int foo(const int& f);
int& foo(int& f);

This way the compiler will know about the existence of version 2 and can consider it when working out whith foo to call.