转发引用是否仍然是右值引用? [英] Is a forwarding reference still an rvalue reference or not?

问题描述

我仍然对为支持移动和转发而发明的规则感到困惑.我仍然不确定的一件事是:

是转发参考，只是右值参考(带有是否应用了参考折叠规则)?

如果它是右值引用，那么为什么要执行该功能:

  template< typename T>void func(T&); 

不仅接受右值，还接受左值吗?

解决方案

(显然)在替换 T 之前， T&&是右值引用./p>

在替换 T 之后(并且折叠引用之后)， T& 要么仍然是右值引用(如果 T 为而不是引用)，或成为左值引用(如果 T 是左值引用).

I'm still confused by the rules invented to support moving and forwarding. One thing I'm still not sure about is:

Is a forwarding reference just an rvalue reference (with reference collapsing rules applied)?

If it is an rvalue reference, then why does the function:

template<typename T>
void func(T&&);

accept not only rvalues, but also lvalues?

解决方案

Before T is substituted, T && is an rvalue reference (obviously).

After T is substituted (and after references are collapsed), T && either remains an rvalue reference (if T is not a reference), or becomes an lvalue reference (if T is an lvalue reference).

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