std :: forward的实现 [英] The implementation of std::forward

问题描述

我正在阅读新C ++概述（C ++ 11/14）（仅PDF） ，在幻灯片288上提供了 std :: forward ：

I'm reading Overview of the New C++ (C++11/14) (PDF only), at Slide 288 it gives an implementation of std::forward:

template<typename T>                // For lvalues (T is T&),
T&& std::forward(T&& param)         // take/return lvalue refs.
{                                   // For rvalues (T is T),
    return static_cast<T&&>(param); // take/return rvalue refs.
}

然后在文本中给出另一个实现：

And then gives another implemention in text:

通常的 std :: forward 实现是：

template<typename T>
struct identity {
    typedef T type;
};
template<typename T>
T&& forward(typename identity<T>::type&& param)
{
    return static_cast<identity<T>::type&&>(param);
}

有什么区别？为什么后者是通常的实现？

What is the difference? Why is latter the usual implementation?

推荐答案

第一个问题是您可以编写 std： ：forward（x），它不会做您想要的事情，因为它总是产生左值引用。

The problem with the first is that you can write std::forward(x), which doesn't do what you want, since it always produces lvalue references.

第二种情况下的参数是非推导上下文，防止自动推导模板参数。这迫使您编写 std :: forward< T（x），这是正确的选择。

The argument in the second case is a non-deduced context, preventing automatic deduction of the template argument. This forces you to write std::forward<T>(x), which is the right thing to do.

此外，第二次重载的参数类型应为 typename identity< T> :: type& ，因为习惯用法是 std的输入： ：forward 始终是左值。

Also, the argument type for the second overload should be typename identity<T>::type& because the input to idiomatic use of std::forward is always an lvalue.

编辑：该标准实际上要求的签名与此等效（ ，顺便说一句，正是libc ++所具有的）：

The standard actually mandates a signature equivalent to this one (which, incidentally, is exactly what libc++ has):

template <class T> T&& forward(typename remove_reference<T>::type& t) noexcept;
template <class T> T&& forward(typename remove_reference<T>::type&& t) noexcept;

这篇关于std :: forward的实现的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！