bind first argument of function without knowing its arity

问题

I'd like to have a function BindFirst that binds the first argument of a function without me having to explicitly know/state the arity of the function by using std::placeholders. I'd like the client code to look something like that.

#include <functional>
#include <iostream>

void print2(int a, int b)
{
    std::cout << a << std::endl;
    std::cout << b << std::endl;
}

void print3(int a, int b, int c)
{
    std::cout << a << std::endl;
    std::cout << b << std::endl;
    std::cout << c << std::endl;
}

int main()
{ 
    auto f = BindFirst(print2, 1); // std::bind(print2, 1, std::placeholders::_1);
    auto g = BindFirst(print3, 1); // std::bind(print3, 1, std::placeholders::_1, std::placeholders::_2);
    f(2);
    g(2,3);
}

Any ideas how BindFirst could be implemented?

回答1:

In C++11:

#include <type_traits>
#include <utility>

template <typename F, typename T>
struct binder
{
    F f; T t;
    template <typename... Args>
    auto operator()(Args&&... args) const
        -> decltype(f(t, std::forward<Args>(args)...))
    {
        return f(t, std::forward<Args>(args)...);
    }
};

template <typename F, typename T>
binder<typename std::decay<F>::type
     , typename std::decay<T>::type> BindFirst(F&& f, T&& t)
{
    return { std::forward<F>(f), std::forward<T>(t) };
}

DEMO 1

In C++14:

#include <utility>

template <typename F, typename T>
auto BindFirst(F&& f, T&& t)
{
    return [f = std::forward<F>(f), t = std::forward<T>(t)]
           (auto&&... args)
           { return f(t, std::forward<decltype(args)>(args)...); };
}

DEMO 2

来源：https://stackoverflow.com/questions/33724042/bind-first-argument-of-function-without-knowing-its-arity