C# syntax operator-overloading equals-operator

当使用接口而不是实际类型时，如何覆盖运算符 == ？

发布于 2024-09-06 08:53:10 字数 827 浏览 5 评论 0原文

我有一些这样的代码：

我应该如何实现运算符 == 以便当变量属于接口 IMyClass 时调用它？

public class MyClass : IMyClass
{
    public static bool operator ==(MyClass a, MyClass b)
    {
        if (ReferenceEquals(a, b))
            return true;

        if ((Object)a == null || (Object)b == null)
            return false;

        return false;
    }

    public static bool operator !=(MyClass a, MyClass b)
    {
        return !(a == b);
    }
}

class Program
{
    static void Main(string[] args)
    {
        IMyClass m1 = new MyClass();
        IMyClass m2 = new MyClass();

        MyClass m3 = new MyClass();
        MyClass m4 = new MyClass();

        Console.WriteLine(m1 == m2); // does not go into custom == function. why not?
        Console.WriteLine(m3 == m4); // DOES go into custom == function
    } 
}

原文

I have some code like this:

How should I implement the operator == so that it will be called when the variables are of interface IMyClass?

public class MyClass : IMyClass
{
    public static bool operator ==(MyClass a, MyClass b)
    {
        if (ReferenceEquals(a, b))
            return true;

        if ((Object)a == null || (Object)b == null)
            return false;

        return false;
    }

    public static bool operator !=(MyClass a, MyClass b)
    {
        return !(a == b);
    }
}

class Program
{
    static void Main(string[] args)
    {
        IMyClass m1 = new MyClass();
        IMyClass m2 = new MyClass();

        MyClass m3 = new MyClass();
        MyClass m4 = new MyClass();

        Console.WriteLine(m1 == m2); // does not go into custom == function. why not?
        Console.WriteLine(m3 == m4); // DOES go into custom == function
    } 
}

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樱桃奶球 2024-09-13 08:53:10

关键是您没有重写某个运算符，而是重载它。

没有定义任何运算符，

operator ==(IMyClass x, IMyClass y)

因此编译器无法调用任何内容。它无法调用

operator ==(MyClass x, MyClass y)

，因为它不知道 m1 和 m2 实际上会引用 MyClass 的实例。

据我所知，没有办法实现用于接口的运算符 - 毕竟，多个实现都可以提供自己的一个，只是为了可能存在歧义。

就我个人而言，我对一开始就尝试谈论非密封类型的平等有点谨慎——平等和继承不能很好地混合。当然，这对于接口来说是双重的：）您最好实现一个适当的 IEqualityComparer并使用它。

The key is that you're not overriding an operator - you're overloading it.

There's no operator defined for

operator ==(IMyClass x, IMyClass y)

so the compiler has nothing it could call. It can't call

operator ==(MyClass x, MyClass y)

as it doesn't know that m1 and m2 will actually refer to instance of MyClass.

As far as I know there's no way of implementing an operator to be used for interfaces - after all, multiple implementations could all provide their own one, just for one point of possible ambiguity.

Personally I'm somewhat wary of trying to talk about equality over non-sealed types to start with - equality and inheritance don't mix terribly nicely. That goes doubly for interfaces, of course :) You might be best implementing an appropriate IEqualityComparer<IMyClass> and using that instead.

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