为什么在c ++中创建复制构造函数之前需要显式构造函数？

Question

我试图在我的例子中定义一个拷贝构造函数。但是，我发现如果必须使用复制构造函数，默认/隐式构造函数不会使编译器满意。为什么会这样？背后有什么理由吗？

class DemoCpyConstructor
{

private:

    int priv_var1;
    int priv_var2;

public:

    void setDemoCpyConstructor(int b1, int b2)
    {
        std::cout<<"The Demo Cpy Constructor Invoked"<<std::endl;
        priv_var1 = b1;
        priv_var2 = b2;

    }

    void showDemoCpyConstructor()
    {
        std::cout<<"The priv_var1 = "<<priv_var1<<std::endl;
        std::cout<<"The priv_var2 = "<<priv_var2<<std::endl;
    }

    DemoCpyConstructor(const DemoCpyConstructor &oldObj)
    {
        std::cout<<"Copy Constructor Invoked.."<<std::endl;
        priv_var1 = oldObj.priv_var1;
        std::cout<<"Tweaking the copy constructor"<<std::endl;
        priv_var2 = 400;
    }


};


int main(int argc, char *argv[])
{
    DemoCpyConstructor oldObj;
    oldObj.setDemoCpyConstructor(120,200);
    oldObj.showDemoCpyConstructor();

    DemoCpyConstructor newObj = oldObj;

    newObj.showDemoCpyConstructor();

    return 0;
}

这就是错误，我得到了 -

error: no matching function for call to ‘DemoCpyConstructor::DemoCpyConstructor()’

Answer 1

每当你定义任何类型的构造函数（意味着转换，复制）并且你想要使用默认构造函数时，你必须明确地提供它。它的规则