c++ - 不调用赋值运算符

Question

#include <iostream>

class MemoryBlock {
private:
    int length = 0;
    int* m_arr = nullptr;

public:
    // Constructor
    explicit MemoryBlock(int p_length) : length(p_length), m_arr(new int[length]) {
        std::cout << "In MemoryBlock(int), length is " << length << ".\n";
    }

    // Destructor
    ~MemoryBlock() {
        std::cout << "~MemoryBlock() is called. Deleting resources.\n";

        delete[] m_arr;
    }

    // Copy constructor
    MemoryBlock(const MemoryBlock& other) : length(other.length), m_arr(new int[other.length]) {
        std::cout << "In MemoryBlock(const MemoryBlock&), length is " << length << ".\n";

        for (int i = 0; i < length; ++i) {
            m_arr[i] = other.m_arr[i];
        }
    }

    // Move constructor
    MemoryBlock(MemoryBlock&& other) noexcept : length(0), m_arr(nullptr) {
        std::cout << "In MemoryBlock(MemoryBlock&&), length is " << other.length << ".\n";

        length = other.length;
        m_arr = other.m_arr;

        other.length = 0;
        other.m_arr = nullptr;
    }

    // Addition operator overloading
    MemoryBlock operator+(const MemoryBlock& other) {
        std::cout << "In operator+(const MemoryBlock&), adds length and array space. New length: " << length + other.length << '\n';

        int added_length = length + other.length;
        MemoryBlock sum(added_length);

        int idx = 0;
        int other_idx = 0;

        for (int i = 0; i < added_length; ++i) {
            if (i < length) {
                sum.m_arr[i] = m_arr[idx];
                ++idx;
            } else {
                sum.m_arr[i] = other.m_arr[other_idx];
                ++other_idx;
            }
        }

        return sum;
    }

    // Copy assignment operator
    MemoryBlock& operator=(const MemoryBlock& other) {
        std::cout << "In operator=(const MemoryBlock&), length is " << other.length << ".\n";

        if (this != &other) {
            delete[] m_arr;

            length = other.length;
            m_arr = new int[other.length];

            for (int i = 0; i < length; ++i) {
                m_arr[i] = other.m_arr[i];
            }
        }

        return *this;
    }

    // Move assignment operator
    MemoryBlock& operator=(MemoryBlock&& other) noexcept {
        std::cout << "In operator=(MemoryBlock&&), length is " << other.length << ".\n";

        if (this != &other) {
            delete[] m_arr;

            length = other.length;
            m_arr = other.m_arr;

            other.length = 0;
            other.m_arr = nullptr;
        }

        return *this;
    }
};

int main() {
    MemoryBlock mb1(11);
    MemoryBlock mb2(mb1);
    MemoryBlock mb3 = mb1 + mb2;

    std::cout << "----------------- End of main() -----------------\n";

    return 0;
}

Run code

我从 Microsoft developer docs 中获取了这个示例代码.我生成了对象 mb1 并将其复制到 mb2。然后将mb1 + mb2赋值给mb3。但是正如您在结果中看到的那样，没有调用赋值运算符。

另外，似乎临时对象是在调用加法运算符后生成的，但并未调用其析构函数。被调用的析构函数是mb1、mb2、mb3的析构函数。

为什么有一句MemoryBlock mb3 = mb1 + mb2; 赋值运算符没有被调用？为什么没有调用临时对象的析构函数？

结果:

In MemoryBlock(int), length is 11.
In MemoryBlock(const MemoryBlock&), length is 11.
In operator+(const MemoryBlock&), adds length and array space. New length: 22
In MemoryBlock(int), length is 22.  // Maybe temporary object?
----------------- End of main() -----------------
~MemoryBlock() is called. Deleting resources.
~MemoryBlock() is called. Deleting resources.
~MemoryBlock() is called. Deleting resources.

Answer 1

MemoryBlock mb3 = mb1 + mb2; 调用复制构造函数。

内存块 mb3； mb3 = mb1 + mb2; 调用赋值运算符。