equals in terms of equalHashValues?

Hi!

I found this comment in the new Hashable trait in the trunk:

* If you define equals in terms of equalHashValues then your hashCode and equals
* methods will never be out of sync.  Something like:
*
* override def equals(other: Any) = other match {
*    case x: YourClass => this equalHashValues x
*    case _            => false
* }

While I think it's a good idea to have convenient hash code support, I think this advice is misleadiung and dangerous: Of course your equals and your hashCode is "in synch" if you follow this advice, but this is certainly not what you want (in 99% of the cases). I mean if you have a large number of objects, there is a realistic chance for a hash-clash: There are only 2^32 Ints. And this could become a really ugly bug to hunt.

This is not only a theoretical question, as Range already *uses* an equals method as described. So I wrote a little test to count the clashes for Range (please correct me if I didn't understood the algorithm correctly):

def hash_code(values:List[Int]): Int = values.foldLeft(0)((x, y) => x * 41 + y)

def main(args:Array[String]) {
    var codes = Map[Int, (Int,Int,Int)]()
    var count = 0
    for(upper <- 2 to 50; lower <- 1 to upper-1; step <- 1 to 5) {
            val code = hash_code(List(lower,upper,step))
            codes.get(code) match {
               case None => codes += code -> ((lower,upper,step))
               case Some(x) => println("Equal hashCode " + code + " for " + x +
                                       " and " + ((lower,upper,step)))
                   count += 1
           }
    }
    println("" + count + " clashes")
}

For the given tiny range 1 <= lower < upper <= 50, 1 <= step <= 5 I got 140 clashes:
Equal hashCode 3486 for (2,3,1) and (1,44,1)
Equal hashCode 3487 for (2,3,2) and (1,44,2)
Equal hashCode 3488 for (2,3,3) and (1,44,3)
...

I wouldn't call Range "hashable" (as in: "Suited as key for a HashMap") anymore.

Cheers,
Daniel