c++ - 为什么要在这里调用复制构造函数？

Question

#include <iostream>
using namespace std;
class A
{
    int x;
public:

    A(int a)
    {   
        x = a;
        cout << "CTOR CALLED";
    }
    A(A &t)
    {
        cout << "COPY CTOR CALLED";
    }
    void display()
    {
        cout << "Random stuff";
    }
    A operator = (A &d)
    {   
        d.x = x;
        cout << "Assignment operator called";
        return *this;
    }
};

int main()
{
    A a(3), b(4);
    a = b;
    return 0;
}

这段代码的输出是:

CTOR CALLED
CTOR CALLED
Assignment operator called
COPY CTOR CALLED

当我在 visual studio 中使用 watch 时，它显示 a 中 x 的值甚至在调用重载赋值运算符之前就已更改。

那么为什么还要在这里调用复制构造函数呢？

Answer 1

因为您从赋值运算符返回按值。它应该返回一个引用:

A& operator = (A &d) { ... }