c++ - 为什么通过构造函数返回对象时没有临时对象？

Question

我想弄清楚通过构造函数(转换函数)返回对象时到底发生了什么。

Stonewt::Stonewt(const Stonewt & obj1) {
    cout << "Copy constructor shows up." << endl;
}

Stonewt::Stonewt(double lbs) {
    cout << "Construct object in lbs." << endl;
}

Stonewt::~Stonewt() {
    cout << "Deconstruct object." << endl;
}

Stonewt operator-(const Stonewt & obj1, const Stonewt & obj2) {
    double pounds_tmp = obj1.pounds - obj2.pounds;
    return Stonewt(pounds_tmp);
}

int main() {
    Stonewt incognito = 275;
    Stonewt wolfe(285.7);

    incognito = wolfe - incognito;
    cout << incognito << endl;
    return 0;
}

Output:
Construct object in lbs.
Construct object in lbs.

Construct object in lbs.
Deconstruct object.
10.7 pounds

Deconstruct object.
Deconstruct object.

所以我的问题是:

为什么通过构造函数返回对象时没有拷贝构造函数(没有临时对象)？

Answer 1

Stonewt operator-(const Stonewt & obj1, const Stonewt & obj2)
{
    ...
    return obj1;
}

   incognito = incognito - wolfe;

您的 operator - () 正在返回 incognito 的拷贝，然后您将其分配给 incognito。然后销毁该拷贝。