在子字段上使用std :: forward [英] Using std::forward on sub fields

问题描述

我正在尝试 std :: move 和 std :: forward 有何不同，我发现我无法在类字段上使用std :: forward：

I was experimenting with how std::move and std::forward differs, and I have found that I am not able to use std::forward on class field:

name = std::forward<T>(rhs.name);

是完整示例。我在gcc 6.3下遇到的错误是：

below is full example. The error I am getting under gcc 6.3 is:

C:/PROGRA~1/MINGW-~1/X86_64~3.0-P/mingw64/lib/gcc/x86_64-w64-mingw32/6.3.0/include/c++/bits/move.h:89:7: error: static assertion failed: template argument substituting _Tp is an lvalue reference type
       static_assert(!std::is_lvalue_reference<_Tp>::value, "template argument"
       ^~~~~~~~~~~~~

我了解原因可能是因为 T 类型为 WrongUseOfMove 。但是我不知道是否只能转发一个子变量，例如，我可以使用传入的 rhs 参数并将其字段转发给不同的类变量。

I understand that the cause is probably because T is of type WrongUseOfMove. But I wonder if forwarding only a sub variable is possible. For example I could use passed in rhs parameter and forward its fields to different class variables.

#include <iostream>
#include <string>
#include <vector>

class WrongUseOfMove {
 public:
  template<typename T>
  WrongUseOfMove(T&& rhs)
  {
    //name = std::move(rhs.name); // Very wrong!!
    //name = std::forward<T>(rhs.name); // Does not compile, T is WrongUseOfMove instead decltype(rhs.name);
    name = std::forward<decltype(rhs.name)>(rhs.name); // compiles - but is it correct? 
    std::cout << __PRETTY_FUNCTION__ << "\n";
  }

  WrongUseOfMove(){}

  std::string name;
};

int main()
{
  WrongUseOfMove wm;
  WrongUseOfMove wm2 = wm;
}

http://coliru.stacked-crooked.com/a/88d8591ee1478a3f

推荐答案

您可以使用直观的方式：

You may use the intuitive way:

name = std::forward<T>(rhs).name;

演示

您的尝试：

name = std::forward<decltype(rhs.name)>(rhs.name);

进行无条件移动：

< a href = http://coliru.stacked-crooked.com/a/7dfe6e0d9a66f2f6 rel = nofollow noreferrer>演示

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