c++ - 为什么赋值运算符重载会创建一个对象的拷贝？

Question

在下面给出的代码中，我在所有的类构造函数、析构函数和重载赋值运算符中编写了cout语句。

#include <iostream>
using namespace std;

class person {
    string name;
    int age ;
    int id ;
    static int num;
public :
    person (string name , int age) : name(name) , age(age) {
        id = num++;
        cout << "creating person : " << id << "(" << name <<")"<< endl;
    }
    person (const person &other) : name(other.name) , age(other.age) {
            id = num++;
            cout << "CREATING PERSON  : " << id << "(" << name <<")" << " from : " << other.id << endl;
    }
    ~person () {
        cout << "killing person : " << id << "(" << name <<")" << endl;
    }
    const person operator= (const person &other) {
        name = other.name ;
        age = other.age;
        //id = num++;
        cout << "copying in : " << id << "(" << name <<")" << " from : " << other.id << endl;
        return *this;
    }
    void print () {
        cout << "name : " << name << ", age : " << age << ", id : " << id << endl;
    }

int person::num = 1;

int main() {
    person per1 ("p1" , 20);
    person per2 ("p2" , 30);
    person per3 ("p3" , 40);
    cout << "see the strange object creation here: " << endl << endl;
    per3 = per2 = per1;
    return 0;
}

给定代码的输出结果是:

creating person : 1(p1)creating person : 2(p2)creating person : 3(p3)see the strange object creation here:copying in : 2(p1) from : 1*CREATING PERSON  : 4(p1) from : 2*copying in : 3(p1) from : 4*CREATING PERSON  : 5(p1) from : 3*killing person : 5(p1)killing person : 4(p1)killing person : 3(p1)killing person : 2(p1)killing person : 1(p1)

我的问题是，是什么导致使用复制构造函数创建两个对象(4 和 5)？分配中使用的对象已经存在。有没有办法在不创建虚拟对象的情况下重载赋值运算符？这个方法看起来不是很优化。

Answer 1

那是因为你的 operator=() 看起来像这样:

const person operator= (const person &other)

按值返回。 const 在此上下文中没有真正意义。

您实际上想做的是通过常量引用返回:

const person& operator= (const person &other)

那个 & 会让一切变得不同。