为什么你使用std :: move当你有&&在C ++ 11？ [英] Why do you use std::move when you have && in C++11?

问题描述

可能重复：

我最近参加了一个C ++ 11研讨会，并给出了以下建议。

I recently attended a C++11 seminar and the following tidbit of advice was given.

when you have && and you are unsure, you will almost always use std::move

您应该使用 std :: move 而不是某些替代方案，以及在您不应使用 std :: move ？

Could any one explain to me why you should use std::move as opposed to some alternatives and some cases when you should not use std::move?

推荐答案

首先，我会解决的问题可能有一个误解：

每当你看到 T&& t 在代码中（And T是一个实际类型，而不是模板类型），记住 t 的值类别是一个左值），而不是右值（临时）。这很混乱。 T&&& 仅表示 t 可以 $ rvalue ¹ ，但 t 本身是一个左值，而不是右值。如果它有一个名称（在这种情况下， t ），那么它是一个左值，不会自动移动，但如果没有名称（结果 3 + 4 ），那么它是一个右值，并且会自动移动到它的结果中。 （在这种情况下 T&& ）几乎与变量的值类别无关（在这种情况下， ）。

First, there's probably a misconception in the question I'll address:
Whenever you see T&& t in code (And T is an actual type, not a template type), keep in mind the value category of t is an lvalue(reference), not an rvalue(temporary) anymore. It's very confusing. The T&& merely means that t can be constructed from an object that was an rvalue ¹, but t itself is an lvalue, not an rvalue. If it has a name (in this case, t) then it's an lvalue and won't automatically move, but if it has no name (the result of 3+4) then it is an rvalue and will automatically move into it's result if it can. The type (in this case T&&) has almost nothing to do with the value category of the variable (in this case, an lvalue).

话虽如此，如果你有 T&& t 写入您的代码，这意味着您有一个变量的引用， 是一个临时的，如果你愿意，可以销毁。如果您需要多次访问该变量，您可以不要> cd> cd>，否则会丢失它的值。但是，最后一次你接受 t 它是安全的 std :: move 它的值到另一个 T 。 （95％的时间，这是你想做的）

That being said, if you have T&& t written in your code, that means you have a reference to a variable that was a temporary, and it is ok to destroy if you want to. If you need to access the variable multiple times, you do not want to std::move from it, or else it would lose it's value. But the last time you acccess t it is safe to std::move it's value to another T if you wish. (And 95% of the time, that's what you want to do)

^{1。如果 T 是模板类型，则 T&& 是一个通用引用，在这种情况下，您使用 std :: forward< T>（t）而不是 std :: move（t）请参见}此问题

^{1. if T is a template type, T&& is a universal reference instead, in which case you use std::forward<T>(t) instead of std::move(t) the last time. See this question}

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