[Haskell-cafe] A probably-stupid question about a Prelude implementation.

[Dougal Stanton]

On 22/06/07, Michael T. Richter <ttmrichter at gmail.com> wrote:
>
>  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.