在移动赋值和移动构造函数方面实现 std::swap

问题描述

这是std::swap的可能定义:

template<class T>
void swap(T& a, T& b) {
  T tmp(std::move(a));
  a = std::move(b);
  b = std::move(tmp);
}

我相信

1. std::swap(v,v) 保证没有影响，并且
2. std::swap 可以如上实现.

1. std::swap(v,v) is guaranteed to have no effects and
2. std::swap can be implemented as above.

以下引述在我看来暗示这些信念是相互矛盾的.

The following quote seems to me to imply that these beliefs are contradictory.

17.6.4.9 函数参数 [res.on.arguments]

1 以下每一项都适用于定义的函数的所有参数在 C++ 标准库中，除非另有明确说明.

1 Each of the following applies to all arguments to functions defined in the C++ standard library, unless explicitly stated otherwise.

...

• 如果函数参数绑定到右值引用参数，则实现可能会假定此参数是对这个论点.[注意:如果参数是泛型参数表格 T&&并且绑定了类型 A 的左值，参数绑定到左值引用 (14.8.2.1) 因此不包括在前面的句子.— end note ] [ 注意:如果程序将左值转换为xvalue 同时将该左值传递给库函数(例如通过使用参数 move(x)) 调用函数，程序是有效地要求该函数将该左值视为临时值.该实现可以自由地优化掉别名检查如果参数是左值，则可能需要.——尾注]

(感谢 Howard Hinnant 提供报价)

让 v 成为从标准模板库中提取的某种可移动类型的对象，并考虑调用 std::swap(v, v).在上面的 a = std::move(b); 行中，T::operator=(T&& t) 的情况是 this == &b，所以参数不是唯一的引用.这违反了上述要求，因此当从 std::swap(v, v)a = std::move(b) 调用未定义的行为>.

Let v be an object of some movable type taken from the Standard Template Library and consider the call std::swap(v, v). In the line a = std::move(b); above, it is the case inside T::operator=(T&& t) that this == &b, so the parameter is not a unique reference. That is a violation of the requirement made above, so the line a = std::move(b) invokes undefined behavior when called from std::swap(v, v).

这里的解释是什么?

推荐答案

[res.on.arguments] 是关于客户端应该如何使用 std::lib 的声明.当客户端向 std::lib 函数发送 xvalue 时，客户端必须愿意假装 xvalue 确实是一个纯右值，并期望 std::lib 能够利用这一点.

[res.on.arguments] is a statement about how the client should use the std::lib. When the client sends an xvalue to a std::lib function, the client has to be willing to pretend that the xvalue is really a prvalue, and expect the std::lib to take advantage of that.

然而，当客户端调用 std::swap(x, x) 时，客户端不会向 std::lib 函数发送 xvalue.相反，它是这样做的实现.因此，责任在于使 std::swap(x, x) 工作的实现.

However when the client calls std::swap(x, x), the client isn't sending an xvalue to a std::lib function. It is the implementation that is doing so instead. And so the onus is on the implementation to make std::swap(x, x) work.

话虽如此，标准已经给了实现者一个保证:X 应该满足 MoveAssignable.即使处于移动状态，客户端也必须确保 X 是 MoveAssignable.此外，std::swap 的实现并不真正关心自移动赋值的作用，只要它不是 X 的未定义行为.即只要不崩溃.

That being said, the std has given the implementor a guarantee: X shall satisfy MoveAssignable. Even if in a moved-from state, the client must ensure that X is MoveAssignable. Furthermore, the implementation of std::swap doesn't really care what self-move-assignment does, as long as it is not undefined behavior for X. I.e. as long as it doesn't crash.

a = std::move(b);

当 &a == &b 时，此赋值的源和目标都具有未指定(移动自)的值.这可以是空操作，也可以执行其他操作.只要它不崩溃，std::swap 就会正常工作.这是因为在下一行:

When &a == &b, both the source and target of this assignment have an unspecified (moved-from) value. This can be a no-op, or it can do something else. As long as it doesn't crash, std::swap will work correctly. This is because in the next line:

b = std::move(tmp);

从前一行进入 a 的任何值都将从 tmp 获得一个新值.而tmp的原始值是a.所以除了消耗大量的 CPU 周期之外，swap(a, a) 是无操作的.

Whatever value went into a from the previous line is going to be given a new value from tmp. And tmp has the original value of a. So besides burning up a lot of cpu cycles, swap(a, a) is a no-op.

更新

最新工作草案，N4618 已被修改以明确说明在 MoveAssignable 要求中的表达式:

The latest working draft, N4618 has been modified to clearly state that in the MoveAssignable requirements the expression:

t = rv

(其中rv 是一个右值)，如果t，t 只需在赋值前等于rv 和 rv 不引用同一个对象.无论如何，在赋值之后 rv 的状态是未指定的.还有一个额外的说明需要进一步说明: