在 C++ 中覆盖运算符以使用派生的 class

Question

我试图覆盖子 class 中的虚拟比较运算符，但我收到一个编译器错误，指出派生的 class 没有实现基础 class' 虚拟运算符。

我感觉这与我的派生运算符不使用基本 class' 参数类型有关。

简化版如下：

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

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

我有办法做到这一点，还是我必须在 Derived 实现中进行类型检查以查看它是否是正确的类型？

Answer 1

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

确实如此。基础 class 必须采用 const reference（以便它可以具有动态类型 Derived，然后您将覆盖声明为：

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
}

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例子来做到这一点。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?

还有：

Derived d;
DerivedDerived dd;

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

Answer 2

您必须在 Derived 实现中键入检查参数是否具有预期类型。

对于运算符，您可能更愿意定义一个虚拟标准方法，然后通过调用该方法来实现您的运算符。这样可以避免操作员出现意外或过大的签名。

struct Base {
  virtual int compare(const Base& source) const { return 0; }

  bool operator ==(const Base& source) const
    { return compare(source) == 0; }
};

struct Derived : Base {
  int compare(const Base& asource) const override
    { const Derived* source = dynamic_cast<const Derived*>(&asource);
      int result = -2;
      if (source) { ... result = ...; }
      return result;
    }

  // redefinition to force the expected/right signature at this level
  bool operator==(const Derived& source) const
    { return compare(source) == 0; }
};

在 C++ 中覆盖运算符以使用派生的 class

Overriding an operator to use the derived class in C++

c++

inheritance

overriding

operator-overloading

subtyping