没有调用复制构造函数？

Question

据我所知,在下列情况下调用复制构造函数:
1实例化一个对象并使用另一个对象的值初始化时
2按值传递对象时.
3按值从函数返回对象时.
我决定把它用于测试,我做了这个小程序测试这个(每次调用构造函数时都会显示消息.它似乎适用于前两种情况,但不适用于第三种情况.我想找出我的错误.欢迎您的想法.

#include <iostream>

using namespace std;

class Circle{
private:
    double* data;
public:
    Circle();
    Circle(double* set);
    Circle(const Circle& tt1);
    ~Circle();
    Circle& operator=(const Circle& tt1);
};



Circle :: Circle()
{
    cout << "Default constructor called" << endl;
    data = NULL;
}

Circle :: Circle(double* set)
{
    cout << "Set up constructor called" << endl;
    data = new double[3];
    copy(set, set+3, data);
}

Circle :: Circle(const Circle& tt1)
{
    cout << "Copy constructor called" << endl;
    data = new double[3];
    copy(tt1.data, tt1.data+3, this->data);
}
Circle :: ~Circle()
{
    cout << "Destructor called!" << endl;
    delete[] data;
}

Circle& Circle :: operator=(const Circle& tt1)
{
    cout << "Overloaded = called" << endl;
    if(this != &tt1)
    {
        delete[] this->data;
        this->data  = new double[3];
        copy(tt1.data, tt1.data+3, this->data);
    }
    return *this;
}

void test2(Circle a)
{

}
Circle test3()
{
    double arr [] = { 3, 5, 8, 2};
    Circle asd(arr);
    cout<< "end of test 3 function" << endl;
    return asd;
}

int main()
{
    cout <<"-------------Test for initialization" << endl;
    double arr [] = { 16, 2, 7};
   Circle z(arr);
    Circle y = z;
   cout << "-------------Test for pass by value" << endl;
   test2(z);
   cout <<"------------- Test for return value-------"<<endl;
   Circle work = test3();

    cout<< "-----------Relese allocated data" << endl;
    return 0;
}

Answer 1

由于返回值优化,第三个测试用例的复制构造函数调用由编译器优化.