c++ - 派生类中virtual operator==的重新定义

Question

我将要在派生类上重新定义运算符==。基类本身重新定义了operator==，即:

virtual bool operator==(const Dress &c) const {
    return brand==c.brand && size==c.size
}

在这种情况下，我认为 brand==c.brand 与 this.brand==c.brand 相同。在我的派生类上，我使用相同的函数签名。

virtual bool operator==(const Dress &c) const

我的问题是:

使用检查参数类型是否匹配是否正确

typeid(this) == typeid(c)

因为这个重新定义的函数只会在派生类的元素上调用？我不明白的是，如何重新定义函数以在调用 derived_class_object == base_class_object 等对象时返回 false？

基类

class Dress {
private:
    string brand;
    int size;
public:
Dress(string b="unknown", int s=40): brand(b), size(s) {}
virtual bool operator==(const Dress &c) const {
        return brand==c.brand && size==c.size
    }

派生类

class Tshirt: public Dress {
private:
bool is_shortsleeve;
public:
Tshirt(string b="unknown", int s=40, bool t=true):Capo(b,s),is_shortsleeve(t) {}
virtual bool operator==(const Dress &c) const {

...

}

显然，除了其他控件之外，我还需要检查这两个元素是否属于派生类。像我之前写的那样做可以接受吗？

Answer 1

这就是您要找的吗？

#include <iostream>
#include <string>

class base
{
protected:
    int a;
public:
    base(int a) : a(a) {}
    virtual bool operator == (base & other) const
    {
        return (typeid(*this) == typeid(other)) && this->a == other.a;
    }
};

class derived : public base
{
public:
    derived(int a) : base(a)
    {}
    virtual bool operator == (base & other) const
    {
        derived* otherDerived = dynamic_cast<derived*>(&other);
        bool canBeCastToDerived = (0 != otherDerived);
        bool isExactlyTheSameType = (typeid(*this) == typeid(other));
        bool hasOtherProperties = false;

        if (otherDerived)
        {
            hasOtherProperties = (this->a == otherDerived->a);
        }

        return  canBeCastToDerived && isExactlyTheSameType && hasOtherProperties;
    }
};

int main()
{
    using std::cout;
    base one(1);
    derived two(2);

    cout << (one == one) << "\n";
    cout << (one == two) << "\n";
    cout << (two == one) << "\n";
    cout << (two == two) << "\n";
}

编辑 1:在评论中包含所有有效的批评。

编辑 2:更改示例以明确包含其他变体。