用多态在 C++ 中重载 == 和 != 运算符

Question

Here is the thing. I have one base class and 4 child classes.

class Base{
  public:
    virtual int getType() const = 0;
};

class Type1 : public Base{
  public:
    virtual int getType() const {
        return 1;
    };
};

class Type2 : public Base{
  public:
    virtual int getType() const {
        return 2;
    };
};

class Type3 : public Base{
  public:
    virtual int getType() const {
        return 3;
    };
};

class Type4 : public Base{
  public:
    virtual int getType() const {
        return 4;
    };
};

I need to overload the == and != operators which do the same thing for all child classes, just retrieve the type values and compare them. So I would naturally implement the operator in the Base class with references to Base as both operands but when I do that, my IDE starts screaming when I use the operators on child views, that it cannot compare structs.

So the question is. Is there a way I can implement the operators just once without having to specify them for each combination of child classes ?

Thanks!

Answer 1

I do not have any problem with this operator:

bool operator==(Base const& x, Base const& y)
{
    return x.getType() == y.getType();
}

unless with an accessibility issue: Your getType function is private. If you do not provide any access modifier with classes, all members, variables and functions, are implicitly private.

So you either need a friend declaration or make the getType function public.

Answer 2

Yes, you can do it in your Base class. There will be no error for doing this.

class Base{
    public:
        virtual int getType() const = 0;

        bool operator==(const Base& rhs) const{
                return getType() == rhs.getType();
        }

        bool operator!=(const Base& rhs) const{
                return !(*this == rhs);
        }
};

Answer 3

class Base {
public:
    virtual int getType() const = 0;
    bool operator ==(const Base &b) const {
        return getType() == b.getType();
    }
};

class Type1 : public Base {
public:
    virtual int getType() const {
        cout << "Type1.getType()" << endl;
        return 1;
    };
};

class Type2 : public Base {
public:
    virtual int getType() const {
        cout << "Type2.getType()" << endl;
        return 2;
    };
};

Usage :

Base *t1 = new Type1(), *t2 = new Type2();
bool res1 = *t1 == *t1; // true, calls Type1.getType() twice
bool res2 = *t1 == *t2; // false, calls Type1.getType() and Type2.getType()