c++17 结构化绑定和移动语义

Question

我正在阅读此 https://leanpub.com/cppmove（C++17 - 完整指南第一版 Nicolai M. Josuttisand）并且关于 c++17 结构化绑定和移动语义，它给出了以下示例，其中移动自对象 ms 保持原样：

MyStruct ms = { 42, "Jim" };   // struct has fields {int i, std::string s}
auto&& [v,n] = std::move(ms); 

// author says: the structured bindings v and n refer to an anonymous
// entity being an rvalue reference to ms, while ms still holds its value:

std::cout << "ms.s: " << ms.s << '\n'; // prints "Jim"

我已经测试过了，至少在实际运行中，这是真的。但这通常也是它应该如何工作的吗？按照作者之前对结构化绑定的解释，上面应该等同于

MyStruct ms = { 42, "Jim" };
auto&& e = std::move(ms); // anonymous compiler constructed entity "e"
/*aliasname*/ v = e.i; // v and n act as if e was constructed as
                       // above and v/n assigned to e.i/e.s as here
/*aliasname*/ n = e.s;

std::cout << "ms.s: " << ms.s << '\n'; // prints "Jim"

问题：

所以如果这是一个有效的解释，会不会 auto&& e = std::move(ms) 导致 ms 立即使其内容“移动”/“被盗”/“无效”？因此，在“Jim”上方的两个（等效？）变体中都没有（保证）打印...？

澄清：我知道 std::move 不会移动任何东西，而是从左值到右值的转换，但我认为这暗示至少你不能指望 std::move-ed 对象（此处 ms）的内容在后续使用中 retained/available

标记为 What is std::move(), and when should it be used?. however, with ~300 upvotes, to that question, this answer:

的欺骗

The first thing to note is that std::move() doesn't actually move anything. It changes an expression from being an lvalue (such as a named variable) to being an xvalue. An xvalue tells the compiler:

You can plunder me, move anything I'm holding and use it elsewhere (since I'm going to be destroyed soon anyway)".

in other words, when you use std::move(x), you're allowing the compiler to cannibalize x

所以我想混淆是允许的:-P

请为超级傻瓜推荐一本关于 c++17 和移动语义的好书 - 这本书的请求肯定是切合主题的；-)

Answer 1

would not auto&& e = std::move(ms) cause ms to immediately have its contents "moved"/"stolen"/"invalidated"?

std::move() 只是准备 ms 被移动，它并没有完全离开它。这就是为什么 e 是对 MyStruct 的引用，而不是实际的 MyStruct 对象。

c++17 结构化绑定和移动语义

c++17 structured bindings and move semantics

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structured-bindings