为什么在使用 std::bind 时复制构造函数被调用两次？

Question

我正在研究 std::function 和 std::bind 以了解如何复制参数以及是否可以保存一些复制操作。

我知道当使用 std::bind 时，参数是按值而不是引用传递的（除非指定了 std::ref）。但是，当我运行下面的代码片段时，复制构造函数被调用了两次。有人可以解释为什么吗？

struct token
{
    static int i;
    int code;

    token()
        : code(i++)
    {
        cout << __FUNCTION__ << ": " << code << endl;
    }
    virtual ~token()
    {
        cout << __FUNCTION__ << endl;
    }

    token (token const & other)
        : code (other.code)
    {
        cout << "copy ctor: " << code << endl;
    }

    // update -- adding a move ctor
    token (token const && other)
        : code (std::move(other.code))
    {
        cout << "move ctor: " << code << endl;
    }
    // update -- end

    void boo() const
    {
            cout << __FUNCTION__ << ": " << code << endl;
    }


};

void call_boo(token const & t)
{
    t.boo();
}


int main()
{
    token t2;

    cout << "default" << endl;
    std::function< void () >(std::bind(&call_boo, t2));

    cout << "ref" << endl;
    std::function< void () >(std::bind(&call_boo, std::ref(t2)));

    cout << "move" << endl;
    std::function< void () >(std::bind(&call_boo, std::move(t2)));


    cout << "end" << endl;
    return 0;
}

当运行时，会产生以下输出：

token: 1
default
// Without move ctor
// copy ctor: 1 // Makes sense. This is the passing by value.
// copy ctor: 1 // Why does this happen?
// With move ctor    
copy ctor: 1
move ctor: 1
~token
~token
ref // No copies. Once again, makes sense.
move
// Without move ctor
// copy ctor: 1
// copy ctor: 1
// With move ctor
move ctor: 1
move ctor: 1
~token
~token
end
~token

Answer 1

std::function 的这个构造函数的参数总是被复制，因此在 std::function.

http://en.cppreference.com/w/cpp/utility/functional/function/function

template< class F > 
function( F f );

5) Initializes the target with a copy of f. If f is a null pointer to function or null pointer to member, *this will be empty after the call. This constructor does not participate in overload resolution unless f is Callable for argument types Args... and return type R. (since C++14)

为什么在使用 std::bind 时复制构造函数被调用两次？

Why is the copy ctor invoked twice when using std::bind?

c++

stdbind

std-function