c ++ 11的举动没有生效吗? [英] c++11 move not take effective？

问题描述

我测试了c ++ 11 move函数，但是没有生效.谁能告诉我为什么?谢谢.代码如下:

I tested c++11 move function, but not become effective. Who can tell me why ? Thanks. The code is as follows：

class Base {
  public:
   Base() { cout << "Base" << endl;}
   ~Base() { cout << "~Base" << endl;}

   Base(const Base& base) { cout << "Copy" << endl; }
   Base& operator=(const Base& base) {cout << "operator=" << endl;}

   Base(Base&& base) { cout << "move" << endl;}
   Base& operator=(Base&& base) { cout << "move=" << endl;}
};

Base b;

Base&& GetResult() {
  return std::move(b);
} 

int main() {
Base&& tmp = GetResult();

cout << &b << endl;
cout << &tmp << endl;

}

输出:

 Base
 0x6013a0
 0x6013a0
 ~Base

为什么不调用move copy和move operator=?为什么地址相同?

Why move copy and move operator= not be called ? And why address is the same ？

推荐答案

要添加到现有的出色答案中，我相信这里的主要困惑是std::move的作用.

To add to the excellent existing answers, I believe the main point of confusion here is what std::move does.

std::move不会移动.

std::move does not move.

它是 abysmally 的名字.

它只为您提供一个xvalue，该xvalue指向您所提供的任何内容；此xvalue将绑定到右值引用，而左值则不会.这意味着可以将std::move 的结果提供给move构造函数或move赋值运算符.但是，它不会为您做到这一点，您在这里也不会这样做.

It only gives you an xvalue referring to whatever you gave it; this xvalue will bind to a rvalue reference where an lvalue won't. This means the result of std::move can be given to a move constructor or move assignment operator. However, it does not do that for you, and you do not do it here.

鉴于对命名的强烈批评，您肯定有其他建议
– user2079303

Given such strong criticism for the naming, surely you have an alternative suggestion
– user2079303

这是一个很容易理解的话题，但是 C ++的创建者建议std::rval 和

This is a well-trodden topic, but the creator of C++ suggests std::rval, and one of the architects of modern C++ suggests std::rvalue_cast (even though you actually get an xvalue).

我个人认为std::moveable应该是一个很好的中间立场.

Personally, I think std::moveable would have been a nice middle ground.

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