Rvalue引用：为什么不会隐式移动右值？

问题描述：

关于C ++右值参考的Artima文章（ http://www.artima.com/cppsource/ rvalue.html ）有词语：这就是为什么当向下传递给基类时，需要说move（x）而不是x。这是移动语义的一个关键安全特性，旨在防止意外地从某个命名变量移动两次。

On Artima article about C++ rvalue reference (http://www.artima.com/cppsource/rvalue.html) there is words: That's why it is necessary to say move(x) instead of just x when passing down to the base class. This is a key safety feature of move semantics designed to prevent accidently moving twice from some named variable.

执行。你能举个例子吗？换句话说，如果 T&& 的所有成员都是右值引用而不仅仅是引用，那么会出错？

I can't think situation when such double move can perform. Can you give an example of this? In other words, what will go wrong if all members of T&& would be rvalue references and not just references?

答

请考虑这种情况：

void foo(std::string x) {}
void bar(std::string y) {}

void test(std::string&& str)
{
    // to be determined
}

我们要调用 foo str ， bar 两者具有相同的值。最好的方法是：

We want to call foo with str, then bar with str, both with the same value. The best way to do this is:

foo(str); // copy str to x
bar(std::move(str)); // move str to y; we move it because we're done with it

这样做是错误的： p>

It would be a mistake to do this:

foo(std::move(str)); // move str to x
bar(std::move(str)); // move str to y...er, except now it's empty

str 的值未指定。

因此在右值引用的设计中，隐式移动不存在。如果是，我们最好的方法不会工作，因为 str 的第一次提及将是 std :: move（str）。

So in the design of rvalue references, this implicit move is not there. If it were, our best way above would not work because the first mention of str would be std::move(str) instead.

Rvalue引用：为什么不会隐式移动右值？

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