运算符+在基类中重载并在派生类中使用它

Question

我有2个类,base(带有复制构造函数)和派生,在baseΙ有重载operator+:

class Base {

  public:

     Base(const Base& x) {
         // some code for copying
     }

     Base operator+(const Base &bignum) const {
         Base x;
         /* ... */
         return x;
     }
};

class Derived : public Base {
};

当我尝试做那样的事情

Derived x;
Derived y;
Derived c=x+y;

我得到错误:conversion from "Base" to non-scalar type "Derived" derived 问题可能在于该运算符+返回Base类型的对象,我想将它分配给Derived类型对象？

Answer 1

事实上,你不需要重新定义operator+(除非你的设计需要它,正如Ajay的例子所指出的那样).

它比你想象的更好

采取以下简约示例:

struct Base {
    Base operator+ (Base a) const
        { cout <<"Base+Base\n"; }
    Base& operator= (Base a)  
        { cout<<"Base=Base\n"; }
};
struct Derived : public Base { };

int main() {
    Base a,b,c;  
    c=a+b;     // prints out "Base+Base" and "Base=Base"
    Derived e,f,g; 
    e+f;       // prints out "Base+Base" (implicit conversion); 
}  

这非常有效,因为当遇到e+f编译器找到operator+基类时,他会隐式转换Derived为Base,并计算类型的结果Base.你可以轻松写c=e+f.

少了什么东西 ？

问题始于仅对Derived的分配.一旦你尝试,g=e+f;你会收到一个错误.编译器不确定如何将A放入B中.这种谨慎是通过普遍的智慧证明的:所有类人猿都是动物,但所有动物都不一定是猿.

如果Derived有更多的字段,那就更明显了Base:应该如何编译它们？基本上,如何告诉编译器他将如何制作Derived出其他东西？用构造函数!

struct Derived : public Base {
    Derived()=default; 
    Derived(const Base& a) : Base(a) { cout<<"construct B from A\n"; }
};

一旦你定义了这个,一切都按照你的期望工作:

 g=e+f;   // will automatically construct a Derived from the result
          // and then execute `Derived`'s default `operator=` which 
          // will call `Base`'s `operator=`  

这是一个现场演示.