Overwriting GetHashCode (Enum plus Integer comparison)

Question

I want to overwrite a GetHashCode method with following code:

public override int GetHashCode()
{
    var magicNumber1 = 5;
    var magicNumber2 = 3;

    return intType * magicNumber1 + 
          (int)enumType * magicNumber2;
}

Is this a proper way of returning GetHashCode when one value is Integer and second an Enum?

What are the **ranges** of `intType` and `enumType`, please? — Dmitry Bychenko, Apr 12 '18 at 12:20
Why these magic numbers? Use prime numbers like here: https://stackoverflow.com/a/263416/284240 — Tim Schmelter, Apr 12 '18 at 12:21
I am not experienced and I am not sure if using magic number is best solution. Also there is a risk of overflowing, isnt it? — helpME1986, Apr 12 '18 at 12:21
As for the enum part of your question, yes, that's fine. The rest of the implementation can be found in the duplicate question. — Tim Schmelter, Apr 12 '18 at 12:23

Dmitry Bychenko · Accepted Answer · 2018-04-12T12:36:01.867

Well, you current implementation is quite OK, but you can be better off if intType and enumType are of small range. Suppose, intType (allmost always) is in the [0..1000000] and enumType is in the [0..10] range then

   public override int GetHashCode() {
      return unchecked(intType * 10 + (int)enumType);
   }

will be a better choice: many hash collisions that a present in your current code are not such in the code above. For instance

   intType | enumType | old hash | suggested
   -----------------------------------------
         0          5         15           5
         3          0         15          30

Edit: In your case (please, see comments)

it should not be more than 100

assuming that you don't have negative values, you can try

   public override int GetHashCode() {
     // * 128 == << 7 may be faster than * 100 at some systems 
     return unchecked(intType * 128 + (int)enumType);
   }

and expect to have no collisions at all

Overwriting GetHashCode (Enum plus Integer comparison)

1 Answers1