Does Scala tuple type uses all of its elements to compute its hash code?

Question

Or does Scala compute tuple hashcode using something independent of its elements, like memory address?

In other words, given two Tuple2s (a, b) and (c, d), does a == c && b == d imply (a, b).hashCode == (c, d).hashCode?

Answer 1

does a == c && b == d imply (a, b).hashCode == (c, d).hashCode?

Yes, it does. This is the contract between == and hashCode. (*)

Does Scala tuple type uses all of its elements to compute its hash code?

Yes, it does. Scala Tuple2 is just a case class:

final case class Tuple2[@specialized(Int, Long, Double, Char, Boolean/*, AnyRef*/) +T1, @specialized(Int, Long, Double, Char, Boolean/*, AnyRef*/) +T2](_1: T1, _2: T2)
  extends Product2[T1, T2]
{
 ...
}

In Scala hashCode for a case class is calculated in the following way:

hashCode in case classes in Scala

What code is generated for an equals/hashCode method of a case class?

---

(*) From the following it can be seen that for case classes (including tuples) the contract is fulfilled.

Answer 2

Scala tuples (and case classes for that matter) implement hashCode and == based on all of their contents and nothing else.

So yes, a == c && b == d implies (a, b).hashCode == (c, d).hashCode, assuming the hashCode methods for a, b, c, and d are also well-behaved.