在std :: forward如何接受右值？ [英] In std::forward how does it accept rvalue?

问题描述

查看Scott Meyer的有效现代C ++ 第200-201页，建议的 std :: forward 的简化实现可能是正确实现在别处）：

Looking at Scott Meyer's Effective Modern C++ pages 200-201, the suggested simplified implementation of std::forward could be (did see the proper implementation elsewhere):

template <typename T>
T&& forward(std::remove_reference_t<T>& param)
{ return static_cast<T&&>(param); }

当接受右值Widget时，它变成：

And when accepting an rvalue Widget, it becomes:

Widget&& forward(Widget& param)
{ return static_cast<Widget&&>(param); }

现在，如果你使用替换代码，并执行：

Now, if you take that substituted code, and do:

struct Widget { };

Widget&& forward(Widget& param)
{ return static_cast<Widget&&>(param); }

template <typename T>
void G(T&& uref)
{ }

template <typename T>
void F(T&& uref)
{ G(forward(uref)); }

int main()
{
  Widget x;
  F(std::move(x));
}

我无法包装我的头，直接回答SO仍然是：<$> forward 参数 Widget& param 管理从F（）接受 Widget&& ？通常gcc-5.0会抱怨非模板代码：

What I can't wrap my head around, and didn't see a direct answer on SO yet, is: in forward how does parameter Widget& param manage to accept Widget&& from F()? Normally gcc-5.0 would complain like so with non-template code:

错误：类型为'Widget&'的非const引用无效初始化类型为'std :: remove_reference ::类型{aka Widget}'

error: invalid initialization of non-const reference of type ‘Widget&’ from an rvalue of type ‘std::remove_reference::type {aka Widget}’

（问题＃27501400 几乎触及了这个主题，但不完全相同，它显示标准具有lvalue&和rvalue&& amp版本。）

(Question #27501400 nearly touches the topic, but not quite. It shows the standard as having both lvalue & and rvalue && versions.)

推荐答案

命名的右值引用是左值，

"Named rvalue references are lvalues",

$ b

so example works fine as noted in a comment.

但是，您的代码可以修改为

Nevertheless your code could be modified to

template <typename T>
void F(T && uref)
{
    G(forward(move(uref)));
}

$ b

which is accepted by another overload (compare):

template<typename T>
T && forward(typename std::remove_reference<T>::type & t)
{
    return static_cast<T &&>(t);
}


template<typename T>
T && forward(typename std::remove_reference<T>::type && t)
{
    static_assert(!std::is_lvalue_reference<T>::value, "T is an lvalue reference");
    return static_cast<T &&>(t);
}

第二个重载将用于右值。如果 T 是 Widget 或 Widget&&< / code >并且 Widget& 的assert失败。

The second overload will be used for rvalue. It works if T is Widget or Widget && and the assert fails for Widget &.

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