当您进行完美转发时，类型名称T变为T& amp;.或T& amp; amp; amp; amp; amp; amp; amp; amp;但当您不这样做时，T根本不是参考.如何?

问题描述

我正在阅读有关完美转发的信息，这使我感到困惑:
当您尝试实现完美转发时，您将执行以下操作:

I'm reading about perfect forwarding, and this is something I've learnt that has confused me:
When you're trying to achieve perfect forwarding, you'll do something like this:

template<class T> // If I were to call foo with an l-value int, foo would look
void foo(T &&x);  // like this: void foo(int& &&x)

所以我想，等等，这是否意味着如果我这样做了

So then I thought, wait, does that mean that if I did this:

template<class T> // If I were to call foo with an l-value int, foo would look
void foo(T x);    // like this: void foo(int& x);

但事实并非如此. foo 看起来像这样: void foo(int x);

But that's not what happens. foo instead looks like this: void foo(int x);

我的问题:完美的转发功能如何将T变成T&或T&& amp;但在另一种情况下，T不是参考吗?有人可以告诉我确切的规则吗?我需要澄清一下！

推荐答案

如果模板参数 T 出现在函数参数中，则只能将其推断为引用类型 形式为 T&&

A template parameter T can be deduced as a reference type only if it appears in a function parameter of the form T&&

具有以下形式的功能模板:

A function template of the form:

• template< class T>void f(T x)
  将推导出 T 作为对象类型(而 x 是对象类型，因此按值传递)

• template<class T> void f(T x)
  will deduce T as an object type (and x is an object type so is passed by value)

template< class T>void f(T& x)

将推导 T 作为对象类型(然后 x 具有左值引用类型)

template<class T> void f(T& x)

will deduce T as an object type (and then x has lvalue reference type)

template< class T>void f(T& x)

将推导 T 为

• 两个一个左值引用(由于引用折叠规则，因此 x 具有左值引用类型)
• 或作为对象类型(因此 x 具有右值引用类型)
  

• either an lvalue reference (so x has lvalue reference type due to reference collapsing rules)
• or as an object type (so x has rvalue reference type)
  

这是错误的. T 成为引用类型 L& 或对象类型 R ，不是参考 R&& .
形式为 T&& 的功能参数因此变为

This is wrong. T becomes a reference type L& or an object type R, not a reference R&&.
The function parameter of the form T&& thus becomes

• L& (因为将左值引用添加到左值引用仍然是左值引用，就像 add_rvalue_reference< L& :: type 仍 L& )
• 或变成 R& (因为 add_rvalue_reference< R> :: type 是 R&&))

• either L& (because adding an rvalue reference to an lvalue reference is still an lvalue reference, just like add_rvalue_reference<L&>::type is still L&)
• or it becomes R&& (because add_rvalue_reference<R>::type is R&&)

这篇关于当您进行完美转发时，类型名称T变为T& amp;.或T& amp; amp; amp; amp; amp; amp; amp; amp;但当您不这样做时，T根本不是参考.如何?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！