On 22/06/07, Michael T. Richter
So, I'm now going over the code in the 'Report with a fine-toothed comb because a) I'm actually able to read it now pretty fluently and b) I want to know what's there in detail for a project I'm starting. I stumbled across this code:
any :: (a -> Bool) -> [a] -> Bool any p = or . map p
or :: [Bool] -> Bool or = foldr (||) False
Now I see how this works and it's all elegant and clear and all that. But I have two nagging problems with it (that are likely related):
Using foldr means I'll be traversing the whole list no matter what. This implies (perhaps for a good reason) that it can only work on a finite list. I don't see any early bale-out semantics. The way I read this it's going to expand a whole list of n and perform n comparisons (including the one with the provided False).
Well, try it: Prelude> any (>10) [1..] True By way of contrast, this (doesn't) work as you expected: Prelude> let any' p = foldl (||) False . map p Prelude> any' (>10) [1..] ^C Interrupted. A left fold will keep on going with an infinite list in this case. D.