Question

It recently came to my attention that member functions completely shadow free functions with the same name when inside the class. And by completely I mean that every free function with the same name is not considered for overload resolution at all. I can understand why it’s done with something like this:

void f();

struct S
{
    void f();

    void g()
    {
        f(); // calls S::f instead of ::f
    }
};

where the functions have identical signatures, its only natural as variable scoping works the same way. But why prohibit unambigious calls where free function has different signature like this:

void f();

struct S
{
    void f(int x);

    void g()
    {
        f(); // fails to compile attempting to call S::f, which has wrong signature
    }
};

I am not asking how to call a shadowed free function from inside the class. What i want to know is the rationale behind this design.

Answer 1

For unqualified name lookup, only one scope at a time is considered, and if the search in that scope doesn’t yield any results, the next higher scope is searched. In your case, only S‘s scope is searched.

But why prohibit unambigious calls where free function has different signature like this:

The problem is that name lookup doesn’t concern itself with anything but the name, the identifier. It is completely oblivious to the fact that you want to call a function, it just sees an identifier. The same name lookup happens if you just use auto x = f;, and if you think of it that way, there are very good reasons you only want a very limited scope to search. Anything else would just surprise the user.