为什么C ++ 11的移动构造函数/赋值操作符不如预期 [英] Why do not C++11's move constructor/assignment operator act as expected

问题描述

#include <iostream>

using namespace std;

struct A
{
    A()
    {
        cout << "A()" << endl;
    }

    ~A()
    {
        cout << "~A()" << endl;
    }

    A(A&&)
    {
        cout << "A(A&&)" << endl;
    }

    A& operator =(A&&)
    {
        cout << "A& operator =(A&&)" << endl;
        return *this;
    }
};

struct B
{
    // According to the C++11, the move ctor/assignment operator
    // should be implicitly declared and defined. The move ctor
    // /assignment operator should implicitly call class A's move
    // ctor/assignment operator to move member a.
    A a;
};

B f()
{
    B b;

    // The compiler knows b is a temporary object, so implicitly 
    // defined move ctor/assignment operator of class B should be
    // called here. Which will cause A's move ctor is called.
    return b; 
}

int main()
{
    f();
    return 0;
}

我的预期输出应为：

A()
A(A&&)
~A()
~A()

However, the actual output is: (The C++ compiler is: Visual Studio 2012)

A()
~A()
~A()

这是VC ++的错误吗？

Is this a bug of VC++? or just my misunderstanding?

推荐答案

根据这篇博文，VC ++ 2012目前实现了 N2844 + DR1138 ，但不能 N3053 。因此，编译器不会为您隐式生成移动构造函数或赋值运算符。如果你添加显式默认和移动构造函数到 B ，那么你会得到你期望的输出。

According to this blog post, VC++ 2012 currently implements N2844 + DR1138, but not N3053. As a result, the compiler is not implicitly generating move constructors or assignment operators for you. If you add explicit default and move constructors to B then you will get the output you expect.

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