为什么我必须在右值引用上调用move? [英] Why do I have to call move on an rvalue reference?

问题描述

在下面的代码中，为什么第一次调用mkme = mvme_rv不会分派给T& operator=(const T&&)?

In the code below, why doesn't the first call mkme = mvme_rv dispatch to T& operator=(const T&&)?

#include <iostream>
#include <string>
#include <vector>

using namespace std;
using T = vector<int>;

int main()
{
  T mvme(10, 1), mkme;
  T&& mvme_rv = move(mvme); // rvalue ref?
  mkme = mvme_rv;           // calls T& operator=(const T&)?
  cout << mvme.empty();     // 0
  mkme = move(mvme_rv);     // calls T& operator=(const T&&)?
  cout << mvme.empty();     // 1
}

推荐答案

当skypjack正确注释时，通过对象名称访问对象始终会产生左值引用.

As skypjack correctly comments, accessing an object through its name always results in an lvalue reference.

这是一项安全功能，如果您认为它是正确的，那么您将意识到您对此感到高兴.

This is a safety feature and if you think it through you will realise that you are glad of it.

如您所知，std::move只是将l值引用转换为r值引用.如果我们立即使用返回的r值引用(即未命名)，则它仍然是r值引用.

As you know, std::move simply casts an l-value reference to an r-value reference. If we use the returned r-value reference immediately (i.e. un-named) then it remains an r-value reference.

这意味着只能在代码中提到move(x)的位置使用r值.从代码阅读器的角度来看，现在很容易看到x的状态变为未定义状态.

This means that the use of the r-value can only be at the point in the code where move(x) is mentioned. From a code-reader's perspective, it's now easy to see where x's state became undefined.

如此:

 1: auto x = make_x();
 2: auto&& r = std::move(x);
 3: // lots of other stuff
35: // ...
54: // ...
55: take_my_x(r);

不起作用.如果确实如此，则维护代码的人将很难看到(并记住)x(在第1行定义)通过第2行上的引用进入第55行的未定义状态.

does not work. If it did, someone maintaining the code would have a hard time seeing (and remembering) that x (defined on line 1) enters an undefined state on line 55 through a reference taken on line 2.

这很明确:

 1: auto x = make_x();
 2: //
 3: // lots of other stuff
35: // ...
54: // ...
55: take_my_x(std::move(x));

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