前向T（a）和（T& amp;）（a）之间的区别是什么， [英] what is the difference between forward<T>(a) and (T&&)(a)

问题描述

template<typename T>  
void outer(T&& t) {inner(forward<T>(t));}  

template<typename T>  
void outer(T&& t) {inner((T&&)(t));}  

有什么区别？

推荐答案

std :: forward< T>（v）被指定为 static_cast< T&& 。

There is no practical difference. std::forward<T>(v) is specified as static_cast<T&&>(v).

§20.2.3[forward]

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




2 返回： c> static_cast< T&&>（t）。

2 Returns: static_cast<T&&>(t).

通过大多数C ++风格的cast来确定第一个工作。其中之一是 static_cast ，这也是在这种情况下工作的第一个。

A C-style cast goes through most C++-style casts to determine the first working one. One of them is static_cast, which also is the first one that works in this case.

§5.4[expr.cast] p4

由

The conversions performed by

• a const_cast （5.2.11），


• a static_cast （5.2.9），


• a static_cast 后跟< $ c> const_cast ，


• a reinterpret_cast （5.2.10），或


• a reinterpret_cast 后跟 const_cast ，

• a const_cast (5.2.11),
• a static_cast (5.2.9),
• a static_cast followed by a const_cast,
• a reinterpret_cast (5.2.10), or
• a reinterpret_cast followed by a const_cast,

可以使用显式类型转换的cast符号执行。 [...]如果一个转换可以用上面列出的多种方式中的一种来解释，则使用列表中首先出现的解释，即使由该解释产生的转换是错误的。

can be performed using the cast notation of explicit type conversion. [...] If a conversion can be interpreted in more than one of the ways listed above, the interpretation that appears first in the list is used, even if a cast resulting from that interpretation is ill-formed.

但我建议坚持使用 std :: forward 。这个意图从名字上是清楚的，人们会知道它做什么比知道什么奇怪的 static_cast< T&&> （甚至（T&&））。

I'd advise to stick with std::forward, though. The intent is clear from the name and and people will know what it does much more likely than knowing what a weird static_cast<T&&> (or even (T&&)) does.

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