Pattern matching and infinite streams

Question

So, I\'m working to teach myself Scala, and one of the things I\'ve been playing with is the Stream class. I tried to use a naïve translation of the classic Haskell

Answer 1

a match {case x#::xs =>... is about the same as val (x, xs) = (a.head, a.tail). So the difference between the bad version and the good one, is that in that in the bad version, you're calling a.tail and b.tail right at the start, instead of just use them to build the tail of the resulting stream. Furthermore when you use them at the right of #:: (not pattern matching, but building the result, as in #:: merge(a.b.tail) you are not actually calling merge, that will be done only later, when accessing the tail of the returned Stream. So in the good version, a call to merge does not call tail at all. In the bad version, it calls it right at start.

Now if you consider numbers, or even a simplified version, say 1 #:: merge(numbers, anotherStream), when you call you call tail on that (as take(10) will), merge has to be evaluated. You call tail on numbers, which call merge with numbers as parameters, which calls tails on numbers, which calls merge, which calls tail...

By contrast, in super lazy Haskell, when you pattern match, it does barely any work. When you do case l of x:xs, it will evaluate l just enough to know whether it is an empty list or a cons. If it is indeed a cons, x and xs will be available as two thunks, functions that will eventually give access, later, to content. The closest equivalent in Scala would be to just test empty.

Note also that in Scala Stream, while the tail is lazy, the head is not. When you have a (non empty) Stream, the head has to be known. Which means that when you get the tail of the stream, itself a stream, its head, that is the second element of the original stream, has to be computed. This is sometimes problematic, but in your example, you fail before even getting there.