std::bind of class member function

Question

I have this code:

#include <iostream>
#include <functional>

struct Foo
{
        int get(int n) { return 5+n; }
};

int main()
{
        Foo foo;
        auto L = std::bind(&Foo::get, &foo, 3);

        std::cout << L() << std::endl;

        return 0;
}

Seems that this:

auto L = std::bind(&Foo::get, &foo, 3);

is equivalento to:

auto L = std::bind(&Foo::get, foo, 3);

Why?

Answer 1

std::bind() accepts its arguments by value. This means that in the first case you are passing a pointer by value, resulting in the copy of a pointer. In the second case, you are passing an object of type foo by value, resulting in a copy of an object of type Foo.

As a consequence, in the second case the evaluation of the expression L() causes the member function get() to be invoked on a copy of the original object foo, which may or may not be what you want.

This example illustrates the difference (forget the violation of the Rule of Three/Rule of Five, this is just for illustration purposes):

#include <iostream>
#include <functional>

struct Foo
{
    int _x;

    Foo(int x) : _x(x) { }

    Foo(Foo const& f) : _x(f._x)
    {
        std::cout << "Foo(Foo const&)" << std::endl;
    }

    int get(int n) { return _x + n; }
};

int main()
{
   Foo foo1(42);

   std::cout << "=== FIRST CALL ===" << std::endl;
   auto L1 = std::bind(&Foo::get, foo1, 3);
   foo1._x = 1729;
   std::cout << L1() << std::endl; // Prints 45

   Foo foo2(42);

   std::cout << "=== SECOND CALL ===" << std::endl;
   auto L2 = std::bind(&Foo::get, &foo2, 3);
   foo2._x = 1729;
   std::cout << L2() << std::endl; // Prints 1732
}

Live example.

If, for any reason, you don't want to use the pointer form, you can use std::ref() to prevent a copy of the argument from being created:

auto L = std::bind(&Foo::get, std::ref(foo), 3);