带有调用运算符的std :: bind on member

Question

This might be a silly and stupid thing to do - however I would like to understand what happens here.

I have the following code:

#include <iostream>
#include <functional>

namespace
{
    struct call
    {
        void operator()() const
        {
            std::cout << "call::operator()" << std::endl;
        }
    };

    struct dummy
    {
        dummy() = default;
        dummy(const dummy&) = delete;

        call member;
    }; 
}

So member essentially would work like any other object method, allowing it to be invoked as:

dummy d;
d.member()

Which would print call::operator() .

Now I would like to use bind to do that, the initial implementation looked like this:

int main() 
{
    dummy d;

    auto b = std::bind(&dummy::member, &d);
    b();
    return 0;
}

This compiles, but nothing is printed. I don't really understand what is happening - the fact that it compiles, but produces no output puzzles me :) surely some magic is going on inside the belly of std::bind , but what?

Here is a link to play with the code: https://ideone.com/P81PND

Answer 1

Currently, your bind return a member, so b() is d.member . You would have to call operator () on that:

b()(); // call::operator()

As alternative, you may use any of:

b = std::bind(&call::operator(), &d.member);
b = [&]() {d.member();};

Answer 2

You can also call through a std::reference_wrapper . No need for bind at all.

int main() 
{
    dummy d;

    auto b= std::cref(d.member);  // create reference wrapper
    b();
    return 0;
}