Question

我正在查看以下有关移动构造函数/赋值的示例： https://msdn.microsoft.com/en-us/library/dd293665.aspx

我通过添加交换函数来修改它，以简化移动构造函数/赋值和复制赋值：

#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通过移动构造函数调用复制构造函数？

如果我删除了移动构造函数定义，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

许多向量操作要求在抛出异常时没有效果（强异常保证）。如果移动构造函数可以抛出强大的异常保证中断：

push_back的简介：

如果新大小大于旧容量，则会导致重新分配。如果没有重新分配，则之前的所有迭代器和引用插入点仍然有效。如果抛出异常而不是通过复制构造函数，移动构造函数，赋值运算符或移动T的赋值运算符或任何InputIterator操作没有任何影响。如果在插入时抛出异常最后的单个元素T是CopyInsertable或is_nothrow_move_constructible<T>::value是真的，没有效果。 否则，如果移动引发异常非CopyInsertable T的构造函数，效果是未指定的。

由于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

1 个答案: