为什么在传递给另一个对象时调用const对象上的std :: move调用复制构造函数？ [英] Why does calling std::move on a const object call the copy constructor when passed to another object?

问题描述

为什么在传递给另一个对象时，调用std :: move在 const 对象上调用复制构造函数？具体来说，代码

  #include< iostream> 
 
 struct Foo {
 Foo（）= default; 
 Foo（Foo&& x）{std :: cout<< 移动<< std :: endl; } 
 Foo（Foo const& x）= delete; 
}; 
 
 int main（）{
 Foo const x; Foo y（std :: move（x））; 
} 
  

无法使用以下消息进行编译：

  g ++ -std = c ++ 14 test07.cpp -o test07 
 test07.cpp：在函数'int main（）'：
 test07.cpp：10：3​​6：error：使用删除的函数'Foo :: Foo（const Foo&）'
 Foo const x; Foo y（std :: move（x））; 
 ^ 
 test07.cpp：6：5：注意：这里声明了
 Foo（Foo const& x）= delete; 
 ^ 
 Makefile：2：目标'all'失败的食谱
 make：*** [all]错误1 
  

当然，我期望它失败，因为我们不能移动 const 值。同时，我不明白代码在尝试调用复制构造函数之前所采用的路由。意思是，我知道 std :: move 将元素转换为x值，但我不知道如何处理之后相对于

调用 std的结果类型： ：move 与 T const 参数是 T const&& 到 T&&& 参数。下一个最佳匹配是您的副本构造函数，它被删除，因此会出错。

显式地 delete 并不意味着它不能用于重载解析，但是如果它确实是通过重载解析选择的最可行的候选者，则它是编译器错误。

结果是有意义的，因为移动构造是一种操作，从源对象窃取资源，从而改变它，所以你不能这样做到一个 const 对象简单地调用 std :: move 。

Why does calling std::move on a const object call the copy constructor when passed to another object? Specifically, the code

#include <iostream>

struct Foo {
    Foo() = default;
    Foo(Foo && x) { std::cout << "Move" << std::endl; }
    Foo(Foo const & x) = delete;
};

int main() {
    Foo const x; Foo y(std::move(x)); 
}

fails to compile with the message:

g++ -std=c++14 test07.cpp -o test07
test07.cpp: In function 'int main()':
test07.cpp:10:36: error: use of deleted function 'Foo::Foo(const Foo&)'
     Foo const x; Foo y(std::move(x)); 
                                    ^
test07.cpp:6:5: note: declared here
     Foo(Foo const & x) = delete;
     ^
Makefile:2: recipe for target 'all' failed
make: *** [all] Error 1

Certainly, I expect it to fail because we can't move a const value. At the same time, I don't understand the route that the code takes before it tries to call the copy constructor. Meaning, I know that std::move converts the element to an x-value, but I don't know how things proceed after that with respect to const.

解决方案

The type of the result of calling std::move with a T const argument is T const&&, which cannot bind to a T&& parameter. The next best match is your copy constructor, which is deleted, hence the error.

Explicitly deleteing a function doesn't mean it is not available for overload resolution, but that if it is indeed the most viable candidate selected by overload resolution, then it's a compiler error.

The result makes sense because a move construction is an operation that steals resources from the source object, thus mutating it, so you shouldn't be able to do that to a const object simply by calling std::move.

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