C ++：std :: bind（）如何“知道”模板args是引用静态函数还是成员函数？

Question

Using bind() I can do both of these:

void f(int a) { }

class C {
public:
    void f (int a) { }
};

int main()
{
    auto f1 = std::bind (f,3);
    f1();
    C myC;
    auto f2 = std::bind (&C::f, &myC, 3);
    f2();
}

Presumably underneath, f1() is in someway translated to f(3), and f2() is translated to (&myC)->f (3). I'm not too concerned with the binding of the '3' argument, but I would like to understand how bind() automagically knows that f1 should be just a straight function call whereas f2 should be a member call of an object. What is the technique used to check what 'flavour' the first arg is at compile time? I'd like to leverage this kind of technique for my own program.

Answer 1

A simple, packed approach is to use

std::is_member_function_pointer<F>::value

where F is the type of the first argument, probably with potential reference and cv qualification removed.