C++: operator overloading: in-class and out-class. Ambiguity with preincrement operator

Question

Check it out this code:

struct A
{
    A operator+(A const& a) { cout << 1 << endl; return A(); }
    A& operator++() { cout << 2 << endl; return *this; }
    A operator++(int) { cout << 3 << endl; return *this; }
    bool operator!() { cout << 4 << endl; return true; }

};

A operator+(A const& a, A const& b)
{ cout << 5 << endl; return A(); }

A& operator++(A& a) { cout << 6 << endl; return a; }
A operator++(A const& a, int) { cout << 7 << endl; return A(); }
bool operator!(A const& a) { cout << 8 << endl; return false; }

int main()
{
    A a, b;

    a + b; // Prints 1 instead 5
    ++a;   // Ambiguity
    a++;   // Prints 3 instead 7
    !a;    // Prints 4 instead 8

    return 0;
}

In every case, the in-class overload of an operator is choosen against any other out-class overload of the same operator, but the preincrement operator is different: it causes an ambiguity between the in-class and out-class overload.

Why?

Answer 1

Your post-increment member operator is not a const member, whereas the non-member one const A& parameter. There is no ambiguity, and the non-const version chosen. The same does not apply for the pre-increment version. Both member and non-member bind to a non-const A, hence the ambiguity.

Look:

// non-const member. Does not require A operand to be const
A operator++(int)

// non-member. Requires A operand to be const
A operator++(A const& a, int)