在按值返回时复制构造函数的奇怪行为

Question

我测试了复制构造函数的行为，其中函数按值返回一个对象，我遇到了一个复制构造函数被调用的情况和一个不复制的情况。

请考虑以下代码：

class A {
public:
    A() {}
    A(const A& a) {
        cout << "Copy CTOR: " << "This address is " << this 
             << " input address is "<< &a <<       "\n";
    }
};

A returnMyself(A& a) {
    cout<<"Myself address is: "<< &a << "\n";
    return a;
}

A returnLocal(A& a) {
    A local;
    cout<<"local address in returnLocal is  "<< &local << "\n";
    return local;
}

int main () {
    A a;
    cout<<"Before returnMyself\n";
    returnMyself(a);
    cout<<"After returnMyself\n\n";

    cout<<"Before returnLocal\n";
    returnLocal(a);
    cout<<"After returnLocal\n";
}

main的输出是：

Before returnMyself.
Myself address is: 0x7fff6afd88f0.
Copy CTOR Invoked: This address is 0x7fff6afd88d8. Input address is 0x7fff6afd88f0.
After returnMyself.

Before returnLocal.
Local address in returnLocal is  0x7fff6afd88d0.
After returnLocal.

正如您所看到的，当我声明一个本地对象并返回它时，不会调用复制构造函数，而是返回一个调用复制构造函数的给定引用对象。

有没有人对此有解释？通常，从按函数返回的函数调用复制构造函数的情况是什么？

谢谢！

Answer 1

这是copy elision。编译器非常聪明，可以优化该副本。