转发转发参考的单个成员 [英] Forward individual members of a Forward reference

问题描述

我有一个函数可以通过其成员移动通用参数.这些选项中哪个更正确:

I have a function that potentially moves a generic argument but through their members. What of these options is more correct:

1. 这似乎更自然，但很奇怪，因为该参数可能会移动两次[a]，这很奇怪，因为对象可能变得无效.

1. This seems the more natural but it is strange because the argument is potentially moved twice [a], which is odd because the object can become invalid.

template<class T> 
void fun(T&& t){
    myhead_ = std::forward<T>(t).head_;
    myrest_ = std::forward<T>(t).rest_;
}

这可能是不正确的，但可能不会移动任何东西.

This can't be incorrect but it may not be moving anything.

template<class T> void fun(T&& t){
    myhead_ = std::forward<decltype(t.head_)>(t.head_);
    myrest_ = std::forward<decltype(t.rest_)>(t.rest_);
}

这似乎是正确的，但是代码太多.

This seems correct but too much code.

template<class T> void fun(T& t){
    myhead_ = t.head_;
    myrest_ = t.rest_;
}
template<class T> void fun(T&& t){
    myhead_ = std::move(t.head_);
    myrest_ = std::move(t.rest_);
}

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[a]正如@Angew所指出的那样，该语句是不正确的，它仅看起来 好像被移动了两次. std::forward(如std::move)实际上不移动任何内容.最多移动成员(通过后续操作decltype(myhead)::operator=，但这恰恰是目标.)

---

[a] This statement is incorrect as @Angew pointed out, it only looks as if it is moved twice. std::forward (like std::move) doesn't actually move anything. At most the member is moved (by the subsequent operation decltype(myhead)::operator= but that is precisely the objective.)

推荐答案

您的第一个代码非常好:

Your first code is perfectly fine:

template<class T> 
void fun(T&& t){
    myhead_ = std::forward<T>(t).head_;
    myrest_ = std::forward<T>(t).rest_;
}

这是因为该标准保证当执行a.b和a时是x值(例如转发的rvalue引用)，a.b的结果也是一个exvalue(即可以从中移出).另请注意，std::forward和std::move本身并不进行任何实际的 moving 动作，它们只是强制转换.因此，在代码中两次从t移出没有任何风险.

That is because the standard guarantees that when doing a.b and a is an xvalue (such as a forwarded rvalue reference), the result of a.b is also an exvalue (i.e. can be moved from). Also note that std::forward and std::move do not do any actual moving themselves, they're just casts. So there is no risk in moving from t twice in your code.

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