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。
最佳答案
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。
关于c++ - 为什么赋值运算符重载会创建一个对象的拷贝?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61947356/