转发返回值。是否需要std :: forward？ [英] Forwarding of return values. Is std::forward is needed?
问题描述
我正在编写一个库，其中包含了其他库中的许多函数和方法。为了避免处理返回值，我在应用 std :: forward 像这样：
I am writing library which wraps a lot of functions and methods from other library. To avoid coping of return values I am applying std::forward like so:
template<class T>
T&& wrapper(T&& t) {
f(t); // t passed as lvalue
return std::forward<T>(t);
}
f 返回 void 并使用 T& （或超载）。包装器始终返回包装器的参数，返回值应保留参数的价值。我是否真的需要在返回中使用 std :: forward ？ RVO是否多余？它是参考（R或L）的事实是否使它多余？如果return不是最后一个函数语句（是否在if中），是否需要？
f returns void and takes T&& (or overloaded on valueness). Wrapper always returns wrappers's param and on returned value should preserve valuness of argument. Do I actually need to use std::forward in return? Does RVO makes it superfluous? Does the fact that it is a reference (R or L) makes it superfluous? Is it needed if return is not last function statement (inside some if)?
如果 wrapper（）应该返回 void 或 T& ，因为调用者可以通过arg（即引用，R或L）访问评估值。但就我而言，我需要返回值，以便 wrapper（）可以在表达式中使用。
It is debatable if wrapper() should return void or T&&, because caller have access to evaluated value via arg (which is reference, R or L). But in my case I need to return value so that wrapper() can be used in expressions.
这可能与问题无关，但是众所周知，函数 f 不会从<$ c中窃取$ c> t ，因此 f（std :: forward< T>（）中第一次使用是多余的，被我删除了。 std :: forward t））
It might be irrelevant to the question, but it is known that functions f does not steal from t, so 1st use of std::forward in f(std::forward<T>(t)) is superfluous and it was removed by me.
我写了一个小测试：https://gist.github.com/3910503
测试显示，返回的未转发 T -确实在带有-O3的gcc48和clang32中创建了额外的副本（RVO不起作用）。
Test shows, that returning unforwarded T- does creates extra copy in gcc48 and clang32 with -O3 (RVO does not kicks in).
此外，我无法从UB中获得不良行为：
Also, I was not able to get bad behavior from UB in:
auto&& tmp = wrapper(42);
它不能证明任何原因，因为它是 undefined 行为（如果它是UB）。
It does not prove anything of cause because it is undefined behavior (if it is UB).
推荐答案
如果您确实知道 t 不会在调用 f 后处于移出状态，您的两个比较明智的选择是：
In the case that you do know that t will not be in a moved-from state after the call to f, your two somewhat sensible options are:
-
返回类型为
T&的std :: forward< T（t），它避免了任何构造，但允许编写例如自动& ref = wrapper（42）;，这会给ref一个悬空的引用
return
std::forward<T>(t)with typeT&&, which avoids any construction but allows for writing e.g.auto&& ref = wrapper(42);, which leavesrefa dangling reference
返回类型为 T 的 std :: forward< T（t），在最坏的情况下要求移动构造当参数为右值时–可以避免上述prvalue的问题，但有可能从xvalues中窃取
return std::forward<T>(t) with type T, which at worst requests a move construction when the parameter is an rvalue -- this avoids the above problem for prvalues but potentially steals from xvalues
您需要 std :: forward 。不考虑使用复制省略符，因为 t 始终是引用。
In all cases you need std::forward. Copy elision is not considered because t is always a reference.
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