Why doesn't explicit bool() conversion happen in contextual conversion

c++, c++11, operator-overloading

Solution

When performing overload resolution on a reference binding, the less cv-qualified type is preferred. This is discussed in 13.3.3.2p3, with the example given:

struct X {
  void f() const;
  void f();
};
void g(const X& a, X b) {
  a.f(); // calls X::f() const
  b.f(); // calls X::f()
}

Note that binding an object to the implicit object parameter of a member function (13.3.1.1.1p2) is a reference binding (13.3.3.1.4).

Conversion operators are treated as member functions (13.3.1.5) for the purposes of overload resolution (13.3p2). Contextual conversion to `bool` has the semantics of initialization (4p4).

Importantly, any conversion required on the return type of the conversion operator is considered only after considering overload resolution between the conversion operators themselves (13.3.3p1).

The solution is to ensure that all conversion operators have the same `const`-qualification, especially to scalar type.

Problem

If the following test-programm ``` #include <iostream> class A { public: A() {} explicit operator bool() const { std::cout << __PRETTY_FUNCTION__ << std::endl; return true; } // explicit operator bool() { // std::cout << __PRETTY_FUNCTION__ << std::endl; // return true; // } const operator int() const { std::cout << __PRETTY_FUNCTION__ << std::endl; return 1; } operator int() { std::cout << __PRETTY_FUNCTION__ << std::endl; return 1; } }; int main() { A a; if (a) { std::cout << "bool()" << std::endl; } if (a + 0) { std::cout << "int()" << std::endl; } } ``` is run, the output is ``` int A::operator int() bool() int A::operator int() int() ``` and not ``` bool A::operator _Bool() bool() int A::operator int() int() ``` what I expected (and what you get if you uncomment the commented parts). So the question is what are the rules giving the conversion to non-const-int precedence over converting to const-bool?

Original source