将继承接口的 class 与接口引用进行比较

Question

In my class ClassA: ISomeInterface i have:

public class ClassA : ISomeInterface {
    Entity entity;
    public void Test(){
       if(entity.Target == this){}
    }    
}

With Target in entity defined as:

public class Entity {     

   public ISomeInterface Target {get; private set;}    

}

My editor then gives me a warning:

Possible unintended reference comparison; to get a value comparison cast the LHS to type `Object`

Although it compiles, this is the first time I've had this warning. Should I not be comparing interface references like this?

Answer 1

As you already stated it is a warning not an error. And the warning tells you what the compiler really does and maybe your intentionnally guess what should happen is wrong.

So being said that, lets further dig what the problem could be. Here is some sample code (that what's being meant with please give an complete example ):

using System;

public interface ISomeInterface
{
}

public class ClassA : ISomeInterface
{
    public ClassA(int id)
    {
        Id = id;
    }

    public int Id { get; }

    public ISomeInterface Entity { get; set; }
    
    public static bool operator== (ClassA x, ClassA y)
    {
        return true;
    }
    
    public static bool operator!= (ClassA x, ClassA y)
    {
        return false;
    }

    public override bool Equals(object obj)
    {
        return true;
    }
    
    public override int GetHashCode()
    {
        return 72;
    }

    public void Test()
    {
        if (Entity == this)
        {
            Console.WriteLine("same");
        }
        else
        {
            Console.WriteLine("different");
        }
    }
}

public class Program
{
    public static void Main()
    {
        var first = new ClassA(1);
        var second = new ClassA(1);
        first.Entity = second;
        first.Test(); // writes "different"
    }
}

As you can see the code implements the equality comparers and Equals() . All these methods are saying that all instance are the same, but the call in if(Entity == this) still tells that they are different.

This is the meaning of the warning and the root cause is, that an equality check of an interface always falls back to the implementation within the type object . And that implementation is ReferenceEquals(x, y) .

So to avoid this warning and make your intention clear you should either explicitly write if(ReferenceEquals(Entity, this)) or implement a stand-alone class that implements IEqualityComparer<ISomeInterface> and use an instance of it if(new MyComparer().Equals(Enity, this)) (do not instantiate the comparer within the if statement in real code) to make your intentation of comparison obvious.