emplace_back调用移动构造函数和析构函数

Question

我尝试将类cBar的两个实例置于具有emplace_back函数的向量中.

根据引用调用,emplace_back仅保留向量中的位置,然后"就地"创建新实例.

现在,我试着尝试一下:

#include <vector>
#include <iostream>
#include <memory>
#include <string>


class cBar
{
public:

    cBar(const int index);
    cBar(cBar&& other); //needed for emplace_back?
    ~cBar();

private:
    cBar(const cBar& other) = delete;
    cBar& operator=(const cBar& other) = delete;

public:
    int                         mIndex;

};




cBar::cBar(const int index) : mIndex(index)
{
    std::cout << "cBar being created ..." << std::endl;
}
cBar::cBar(cBar&& other) : mIndex(other.mIndex)
{
    std::cout << "cBar being moved ..." << std::endl;
}
cBar::~cBar()
{
    std::cout << "cBar being destroyed ..." << std::endl;
}

int main()
{
    std::vector<cBar> mBars;

    std::cout << "Begin to create 2 cBar instance, I am expecting 2 \"cBar being created ...\" messages here" << std::endl;
    mBars.emplace_back(0);//this supposed to call only cBar(const int index) constructor, and nothing more
    mBars.emplace_back(1);//this works as expected, only one constructor call

    //destroy all
    std::cout << "Destroy the 2 isntances (this works, I see the two expected \"cBar being destroyed ...\" messages here" << std::endl;
    mBars.clear();

    std::cin.get();
    return 0;
}

输出:

Begin to create 2 cBar instance, I am expecting 2 "cBar being created ..." messages here
cBar being created ...
cBar being moved ...
cBar being destroyed ...
cBar being created ...
Destroy the 2 isntances (this works, I see the two expected "cBar being destroyed ..." messages here
cBar being destroyed ...
cBar being destroyed ...

如果你运行上面的那个,你将看到第一个emplace_back创建实例"就地",但然后立即调用移动构造函数,然后调用析构函数.

更奇怪的是,在第二个安慰的情况下,我看到了预期的行为:只有一个构造函数调用.

我有两个问题:

1. 如果我只想要emplace_back项目,并且从不使用push_back,为什么我需要在我的类中定义移动构造函数.

2. 在第一个实例创建的情况下,为什么移动构造函数,然后调用析构函数？如果我访问第一个实例的数据似乎都是正确的,所以我不知道为什么移动构造函数和析构函数被调用.

我使用Visual Studio 2015.

每步输出矢量大小:

Begin to create 2 cBar instance, I am expecting 2 "cBar being created ..." messages here
Vector size:0
cBar being created ...
Vector size:1
cBar being moved ...
cBar being destroyed ...
cBar being created ...
Vector size:2
Destroy the 2 isntances (this works, I see the two expected "cBar being destroyed ..." messages here
cBar being destroyed ...
cBar being destroyed ...

Answer 1

2.在第一个实例创建的情况下,为什么移动构造函数,然后调用析构函数？

因为emplace_back插入第二个元素会导致重新分配; vector需要扩展的内部存储,旧存储中的元素必须被复制/移动到新存储,然后销毁.

1.如果我只想要emplace_back项目,并且从不使用push_back,我为什么需要在我的类中定义移动构造函数.

如上所述,vector需要通过复制/移动操作来移动元素.因此,您必须为类定义复制或移动构造函数.这是双方真实emplace_back和push_back,因为它们都添加元素vector,并可能导致重新分配.