c++17 结构化绑定(bind)和 move 语义

Question

这个问题在这里已经有了答案:





What is std::move(), and when should it be used?

(8 个回答)


1年前关闭。




我在读这个https://leanpub.com/cppmove (C++17 - The Complete Guide First Edition Nicolai M. Josuttisand)关于 c++17 结构化绑定(bind)和 move 语义，它给出了以下示例，其中从对象 ms move 完好无损:

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"

我已经对此进行了测试，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(bind)的解释的方式，以上内容应该等同于

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，不会 auto&& e = std::move(ms)原因 ms立即将其内容“move ”/“被盗”/“无效”？这样在“Jim”上面的两个(等效？)变体中都不会(保证)打印......？
澄清:我知道 std::move 不会 move 任何东西，而是从左值转换为右值，但我认为这意味着至少你不能指望 std::move 对象(此处为 ms )内容在后续使用中被保留/可用
标记为 What is std::move(), and when should it be used? 的骗子.然而，对于这个问题，有大约 300 人赞成，这个答案是: https://stackoverflow.com/a/27026280/11608725

The first thing to note is that std::move() doesn't actually moveanything. It changes an expression from being an lvalue (such as anamed 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 thecompiler to cannibalize x

所以我想困惑是允许的:-P
请为 c++17 的超傻瓜推荐一本好书并 move 语义——这本书的请求肯定是正确的；-)

Answer 1

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

std::move()只准备 ms要被 move ，它不会完全从它 move 。这就是为什么 e是对 MyStruct 的引用，而不是实际的 MyStruct目的。