为什么选择复制构造函数而不是移动构造函数

Question

我正在查看以下关于移动 constructor/assignment 的示例： https://msdn.microsoft.com/en-us/library/dd293665.aspx

我对它进行了一些修改，添加了交换功能以简化移动 constructor/assignment 和复制赋值：

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class MemoryBlock
{
public:

    // Simple constructor that initializes the resource.
    explicit MemoryBlock(size_t length)
        : _length(length)
        , _data(new int[length])
    {
        std::cout << "In MemoryBlock(size_t). length = "
                  << _length << "." << std::endl;
    }

    // Destructor.
    ~MemoryBlock()
    {
        std::cout << "In ~MemoryBlock(). length = "
                  << _length << ".";

        if (_data != nullptr)
        {
            std::cout << " Deleting resource.";
            // Delete the resource.
            delete[] _data;
        }

        std::cout << std::endl;
    }

    // Copy constructor.
    MemoryBlock(const MemoryBlock& other)
        : _length(other._length)
        , _data(new int[other._length])
    {
        std::cout << "In MemoryBlock(const MemoryBlock&). length = "
                  << other._length << ". Copying resource." << std::endl;

        std::copy(other._data, other._data + _length, _data);
    }

    // Copy assignment operator.
    MemoryBlock& operator=(MemoryBlock& other)
    {
        std::cout << "In operator=(const MemoryBlock&). length = "
                  << other._length << ". Copying resource." << std::endl;

        swap(*this, other);
        return *this;
    }

    // Retrieves the length of the data resource.
    size_t Length() const
    {
        return _length;
    }

    // Move constructor.
    MemoryBlock(MemoryBlock&& other)
        : _data(nullptr)
        , _length(0)
    {
        std::cout << "In MemoryBlock(MemoryBlock&&). length = "
                  << other._length << ". Moving resource." << std::endl;


        *this = std::move(other);
    }

    // Move assignment operator.
    MemoryBlock& operator=(MemoryBlock&& other)
    {
        std::cout << "In operator=(MemoryBlock&&). length = "
                  << other._length << "." << std::endl;

        swap(*this, other);
        return *this;
    }

    void swap(MemoryBlock& first, MemoryBlock& second)
    {
        using std::swap;
        swap(first._length, second._length);
        swap(first._data, second._data);
    }

private:
    size_t _length; // The length of the resource.
    int* _data; // The resource.
};

int main()
{
   // Create a vector object and add a few elements to it.
   vector<MemoryBlock> v;
   v.push_back(MemoryBlock(25));
   v.push_back(MemoryBlock(75));

   // Insert a new element into the second position of the vector.

   v.insert(v.begin() + 1, MemoryBlock(50));
}

现在，当我运行代码时，我有以下输出：

In MemoryBlock(size_t). length = 25.
In MemoryBlock(MemoryBlock&&). length = 25. Moving resource.
In operator=(MemoryBlock&&). length = 25.
In ~MemoryBlock(). length = 0.
In MemoryBlock(size_t). length = 75.
In MemoryBlock(MemoryBlock&&). length = 75. Moving resource.
In operator=(MemoryBlock&&). length = 75.
In MemoryBlock(const MemoryBlock&). length = 25. Copying resource.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 0.
In MemoryBlock(size_t). length = 50.
In MemoryBlock(MemoryBlock&&). length = 50. Moving resource.
In operator=(MemoryBlock&&). length = 50.
In MemoryBlock(const MemoryBlock&). length = 25. Copying resource.
In MemoryBlock(const MemoryBlock&). length = 75. Copying resource.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 75. Deleting resource.
In ~MemoryBlock(). length = 0.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 50. Deleting resource.
In ~MemoryBlock(). length = 75. Deleting resource.

我不明白为什么有时会在移动构造函数上调用复制构造函数？

如果我删除移动构造函数定义，并将其声明为默认值：

// Move constructor.
MemoryBlock(MemoryBlock&& other) = default;

然后我得到正确的输出：

In MemoryBlock(size_t). length = 25.
In ~MemoryBlock(). length = 25. Deleting resource.
In MemoryBlock(size_t). length = 75.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 75. Deleting resource.
In MemoryBlock(size_t). length = 50.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 75. Deleting resource.
In ~MemoryBlock(). length = 50. Deleting resource.
In ~MemoryBlock(). length = 25. Deleting resource.
In ~MemoryBlock(). length = 50. Deleting resource.
In ~MemoryBlock(). length = 75. Deleting resource.

（输出中缺少构造函数调用，这意味着使用了移动构造函数）

Answer 1

许多矢量操作要求在抛出异常时没有任何影响（强异常保证）。如果移动构造函数可以抛出强异常保证会中断：

push_back 的概要：

Causes reallocation if the new size is greater than the old capacity. If no reallocation happens, all the iterators and references before the insertion point remain valid. If an exception is thrown other than by the copy constructor, move constructor, assignment operator, or move assignment operator of T or by any InputIterator operation there are no effects. If an exception is thrown while inserting a single element at the end and T is CopyInsertable or is_nothrow_move_constructible<T>::value is true, there are no effects. Otherwise, if an exception is thrown by the move constructor of a non-CopyInsertable T, the effects are unspecified.

因为T是CopyInsertible，它使用复制构造函数而不是移动构造函数。

为什么选择复制构造函数而不是移动构造函数

Why is copy constructor chosen over move constructor

c++

copy-constructor

move-constructor

move-semantics

c++11