Is it possible to check if a member function is defined for a class even if the member is inherited from an unknown base class

Question

I found similar questions and answers like this one. However, as I tried out, this SFINAE tests only succeeded if the tested member is directly defined in the class being te

Answer 1

In C++03, this is unfortunately not possible, sorry.

In C++11, things get much easier thanks to the magic of decltype. decltype lets you write expressions to deduce the type of their result, so you can perfectly name a member of a base class. And if the method is template, then SFINAE applies to the decltype expression.

#include 

template 
auto has_foo(T& t) -> decltype(t.foo(), bool()) { return true; }

bool has_foo(...) { return false; }


struct Base {
    void foo() {}
};

struct Derived1: Base {
    void foo() {}
};

struct Derived2: Base {
    using Base::foo;
};

struct Derived3: Base {
};

int main() {
    Base b; Derived1 d1; Derived2 d2; Derived3 d3;

    std::cout << has_foo(b) << " "
              << has_foo(d1) << " "
              << has_foo(d2) << " "
              << has_foo(d3) << "\n";
}

Unfortunately ideone has a version of gcc that's too old for this and clang 3.0 is no better.