CODEWITHSUNDEEP
javaequalsimplementationhashcode
user692168
user692168

Reputation:

Is this a good implementation of hashCode()?

I have a class that will be used in a HashSet. It only contains two members, and both are of the same type interface. This is what it looks like:

class MyClass{
    MyInterface a;
    MyInterface b;

    public int hashCode(){
         return a.hashCode() + b.hashCode();
    }

    public boolean equals(Object obj){
         if(!(obj instanceof MyClass)
              return false;
         MyClass other (MyClass) obj;
         return (this.a == other.a && this.b == other.b) || (this.a == other.b && this.b == other.a);
    }
}

As you can see, two instances of MyClass are "equal" if they contain the same two instances of MyInterface.

Now, I was thinking that for hashCode(), I could just add up the default hashcodes of its members. Is this good enough? If not, what is a proper implementation of hashCode() for this case?

Upvotes: 4

Views: 147

Answers (3)

Dan Forbes
Dan Forbes

Reputation: 2824

I would say no. Wouldn't this mean that these two instances of MyClass would hash to the same value:

MyClass {
  a.hashCode = 2;
  b.hashCode = 3;
}

and

MyClass {
  a.hashCode = 1;
  b.hashCode = 4;
}

Upvotes: 1

Luca Fagioli
Luca Fagioli

Reputation: 13349

Yes it is.

Because even with hashCode collision, the docs state:

It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results.

Upvotes: 0

Louis Wasserman
Louis Wasserman

Reputation: 198014

Yes, this is perfectly fine. It's equivalent to the implementation of Set.hashCode() for two-element sets.

Upvotes: 3

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