Haskell filter function laziness

Question

Given:

take 5 (filter p xs)

say if filter p xs would return 1K match, would Haskell only filter out 5 matches and without producing a large intermediate result?

Answer 1

It will scan xs only as much as needed to produce 5 matches, evaluating p only on this prefix of xs.

To be more precise, it can actually perform less computation, depending on how the result is used. For instance,

main = do
   let p x = (x==3) || (x>=1000000)
       list1 = [0..1000000000]
       list2 = take 5 (filter p list1)
   print (head list2)

will only scan list1 until 3 is found, and no more, despite take asking for five elements. This is because head is demanding only the first of these five, so laziness causes to evaluate just that.