了解右值引用 [英] Understanding rvalue references

问题描述

我认为有些东西我不太了解右值引用。为什么下面的编译失败（VS2012）时出现错误'foo'：无法将参数1从'int'转换为'int&&'？

I think there's something I'm not quite understanding about rvalue references. Why does the following fail to compile (VS2012) with the error 'foo' : cannot convert parameter 1 from 'int' to 'int &&'?

void foo(int &&) {}
void bar(int &&x) { foo(x); };

我会假设类型 int&&& c $ c>将在从bar传递到foo时被保留。为什么在函数体内部变成 int ？

我知道答案是使用 std :: forward ：

void bar(int &&x) { foo(std::forward<int>(x)); }

所以也许我没有清楚掌握 。 （另外，为什么 std :: move ？）

so maybe I just don't have a clear grasp on why. (Also, why not std::move?)

推荐答案

I始终将lvalue记住为具有名称或可以被寻址的值。由于x有一个名称，它作为一个左值传递。引用右值的目的是允许函数以任何它认为合适的方式完全删除值。如果我们在你的例子中通过引用传递x，那么我们没有办法知道是否可以这样做：

I always remember lvalue as a value that has a name or can be addressed. Since x has a name, it is passed as an lvalue. The purpose of reference to rvalue is to allow the function to completely clobber value in any way it sees fit. If we pass x by reference as in your example, then we have no way of knowing if is safe to do this:

void foo(int &&) {}
void bar(int &&x) { 
    foo(x); 
    x.DoSomething();   // what could x be?
};

执行 foo（std :: move（x））; 明确告诉编译器你完成了x，不再需要它。没有这一举动，现有代码可能会发生不良后果。 std :: move 是一个保护措施。

Doing foo(std::move(x)); is explicitly telling the compiler that you are done with x and no longer need it. Without that move, bad things could happen to existing code. The std::move is a safeguard.

std :: forward 用于模板中的完美转发。

std::forward is used for perfect forwarding in templates.

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