重写运算符以在C中使用派生类

我试图覆盖子类中的虚拟比较运算符,但我得到一个编译器错误,说派生类没有实现基类的虚拟运算符.

我感觉这与我的派生运算符不使用基类’参数类型这一事实有关.

简化版如下：

struct Base {
  virtual bool operator ==(const Base) const;
};

struct Derived : Base {
  bool operator ==(const Derived) const {
    // implementation goes here
  }
};

有没有办法让我这样做,或者我是否必须在Derived实现中进行类型检查以查看它是否是正确的类型？

解决方法:

I have a feeling it’s to do with the fact that my derived operator
doesn’t use the base class’ argument type.

正是如此.基类必须采用const引用(以便它可以具有动态类型Derived,然后将覆盖声明为：

bool operator ==(const Base& rhs) const {
    const auto pRhs = dynamic_cast<const Derived*>(&rhs);
    if (pRhs == nullptr)
    {
        return false;  // Not a derived.  Cannot be equal.
    }
    // Derived/Derived implementation goes here
}

请注意：像这样的虚拟比较运算符很容易出错.你需要一个很好的激励范例来做到这一点.特别是,如果你写：

Derived d;
Base b;
if (d == b)  // All is well - derived override called, and returns false.

if (b == d) // Uh-oh!  This will call the *base* version.  Is that what you want?

也：

Derived d;
DerivedDerived dd;

if (d == dd) // Did you want to use the DerivedDerived comparison?

标签：method-overriding,subtyping,c,operator-overloading,inheritance
来源： https://codeday.me/bug/20190828/1755398.html