c++ - 基类和派生类的复制构造函数

Question

我有这两个类

///////////BASE CLASS

    class Base{
    public:
     Base(int = 0);
     ~Base();
     Base(Base&);
     Base(Derived&);   ////<-may be problem here?, note: I tried without this function
     int valueOfBase();
    protected:
     int protectedData;
    private:
     int baseData;
    };

    /////////////DERIVED CLASS
    class Derived: public Base{
    public:
     Derived(int);
     //Derived(Derived&);
     ~Derived();
    private:
     int derivedData;
    };

and here my main

    int main(){
     Base base(1);
     Derived derived = base;
    return 0;
    }

我读到如果我的派生类没有复制 c'tor 将调用基的复制 c'tor 但每次我收到 从 Base 到非标量的转换type Derived requested 谁错了？我的编译器或我的书，还是我误解了？提前致谢

Answer 1

只是一个提示。

下面的代码是否给出同样的错误？

class base{};
class derived: public base{};

int main()
{
      derived d= base();
}

是吗？为什么？因为没有从基类到派生类的转换。如果您希望您的代码能够编译，您应该定义此转换。

如何将其添加到派生类？

derived(const base &b){} 

有道理吧？