为什么这个C ++ 0x代码调用move构造函数？ [英] Why doesn't this C++0x code call the move constructor?

问题描述

由于某些原因，以下代码不会调用 Event :: Event（Event& e）

For some reason, the following code never calls Event::Event(Event&& e)

Event a;
Event b;
Event temp;
temp = move(a);
a = move(b);
b = move(temp);

为什么不？

使用 std :: swap 调用一次。

class Event {
public:
    Event(): myTime(0.0), myNode(NULL) {}
    Event(fpreal t, Node* n);
    Event(Event&& other);
    Event(Event const& other) = delete;
    ~Event();

    bool				operator<(Event const& other) const { return myTime < other.myTime; }
    bool				operator>(Event const& other) const { return myTime > other.myTime; }
    fpreal				getTime() const { return myTime; }
    void				setTime(fpreal time) { myTime = time; }
    Node*				getNode() const { return myNode; }

private:
    fpreal				myTime;
    Node*				myNode;
};

推荐答案

您的代码有两个潜在位置移动构造函数被调用（但不是）：

Your code has two potential locations for where one may expect the move constructor to get called (but it doesn't):

1）调用std :: move

2）。

1) calling std::move
2) during assignment.

关于1），std :: move会执行一个简单的转换 - 它不会从副本创建一个对象移动构造函数可能会被它调用，但由于它做了一个简单的右值转换，它不会被调用。 std :: move的定义类似于 static_cast< Event&&>（temp）。

Regarding 1), std::move does a simple cast - it does not create an object from a copy - if it did then the move constructor might get invoked by it, but since it does a simple rvalue cast it doesn't get invoked. The definition of std::move is similar to static_cast<Event&&>(temp).

关于2），初始化和赋值是两个完全不同的操作（即使一些形式的初始化使用'='符号）。您的代码分配，因此使用默认的赋值操作符，它被声明为接受一个const lvalue引用。因为你从来没有用另一个事件对象初始化一个事件对象，你不会看到你的移动构造函数被调用。如果你声明了一个移动赋值运算符： Event& operator =（Event&& other），那么你当前的代码将调用它，或者如果你写： Event a;事件tmp = move（a）; 你的移动构造函数，如写的，将被调用。

Regarding 2), Initialization and assignment are two entirely different operations (even though some forms of initialization use the '=' symbol). Your code does assignment and therefore uses the default assignment operator which is declared to accept a const lvalue reference. Since you never initialize one event object with another, you won't see your move constructor get invoked. If you declared a move assignment operator: Event& operator=(Event&& other), then your current code would invoke it or if you wrote: Event a; Event tmp = move(a); your move constructor, as written, would get invoked.

这篇关于为什么这个C ++ 0x代码调用move构造函数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！