Question

我有以下类，我在类向量上重载+运算符。我也定义了自己的拷贝构造函数。这是代码::

class vector {

public:
    int size;

    vector() {
        cout << "In empty constructor\n";
        this->size = 5;
        cout << "size = " << size << endl;
    }

    vector(int size) {
        cout << "In size constructor\n";
        this->size = size;
        cout << "size = " << size << endl;
    }

    vector(const vector &v) {
        cout << "inside copy constructor\n";
        this->size = v.size;
        cout << "size = " << this->size << endl;
    }

    vector operator ++() {
        vector v = *this;
        (this->size)++;
        return v;
    }

    vector operator ++(int a) {
        cout << "a = " << a << endl;
        vector v = *this;
        (this->size)++;
        return v;
    }

    vector operator+(vector &a) {
        vector  v;
        v.size = this->size + a.size;
        return v;
    }

    vector& operator=(vector &a) {
        cout << "Inside = assignment operator\n";
        this->size = a.size;
        return *this;
    }

    ~vector() {
        cout << "In destructor for vector of size = " << this->size << endl;
    }

};

这是主要的::

int main(int argc, char** argv) {
    vector v;
    vector v2(27);
    vector v3 = v + v2;
    return 0;
}

我在构造函数中插入了cout语句，以查看何时调用哪个构造函数。根据我的理解，在我的行vector v3 = v + v2中，应该调用复制构造函数，因为我们在调用期间初始化对象，因此在计算v + v2复制构造函数之后应该被调用。但它并没有。我还尝试重载=运算符以检查它是否正在分配值，但这也没有发生。

about程序的输出是这个::

In empty constructor
size = 5
In size constructor
size = 27
In empty constructor
size = 5
In destructor for vector of size = 32
In destructor for vector of size = 27
In destructor for vector of size = 5

在+的重载函数中，我按值返回结果向量，所以我应该复制它。

为什么复制构造函数不会被调用？

感谢您提前提供任何帮助。

Ideone链接（如果有帮助）:: http://ideone.com/S9EtjR

复制构造函数和运算符重载

0 个答案: