為什么賦值運算符重載會創建 object 的副本？

Question

在下面給出的代碼中，我在所有 class 構造函數、析構函數和重載賦值運算符中編寫了cout語句。

#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;
}

給定代碼的 output 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)

我的問題是，是什么導致使用復制構造函數創建兩個對象（4 和 5）？ 分配中使用的對象已經存在。 有沒有辦法在不創建虛擬對象的情況下重載賦值運算符？ 這種方法似乎不是很優化。

Answer 1

那是因為您的operator=()看起來像這樣：

const person operator= (const person &other)

按價值返回。 const在這種情況下並沒有真正的意義。

您實際上要做的是通過常量引用返回：

const person& operator= (const person &other)

那一個&將使一切變得不同。