关于函数返回到底是值还是引用 - 编程论坛

注意重载的+运算符，这里我返回的是引用（按道理这里应该是不能返回引用的），但是为什么s3的结果却能正确打印？
在完成+运算之后，函数内部的变量不是会被释放吗？为什么s3的取值正确？

程序代码：

#include <iostream>
#include <string>

using namespace std;

class Stone {
    friend Stone& operator + (const Stone& s1, const Stone& s2);
public:
    Stone(int weight = 0) { this->weight = weight; }
    int getWeight() const { return weight; }
private:
    int weight;
};

Stone& operator + (const Stone& s1, const Stone& s2) {
    // Stone temp;
    // temp.weight = s1.getWeight() + s2.getWeight();
    return Stone(s1.getWeight() + s2.getWeight());
}

int main() {
    Stone s1(1);
    Stone s2(2);
    Stone s3;

    s3 = s1 + s2;
    cout << s3.getWeight() << endl;
    cout << s3.getWeight() << endl;
    cout << s3.getWeight() << endl;
}

[此贴子已经被作者于2021-4-15 23:13编辑过]