[Haskell-cafe] Noob error: Type b -> c b Does not match IO a

[Daniel Fischer]

Am Sonntag, 26. Juni 2005 06:06 schrieb Scott Turner:
> On 2005 June 25 Saturday 17:49, kynn at panix.com wrote:
> >    Simplified:
> >    prodList xs = foldl (*) 1 xs
> >
> > But my original at least made some provision for short circuiting the
> > whole operation if the list contained a 0.  As far as I can figure,
> > fold, map, etc., are not suitable for any situation in which
> > short-circuiting would be desirable (e.g. and, or, etc.).  Am I wrong?
>
> Actually, foldr is applicable to short-circuiting.
>    foldr (&&) True
> works fine for detecting whether all elements of a list are True, and does
> not evaluate the list beyond the first occurrence of False.  Similarly, if
> `m` is defined as a short-circuiting multiplication operator, i.e.
>    m x y = if x==0 then 0 else x*y
> then
>    foldr m 1
> does not evaluate the list beyond the first occurrence of 0. Unfortunately,
> `m` does not work as well with foldr as &&.  The function (foldr m 1)
> creates a stack containing one (*) operation for every list element to be
> multiplied, which can cause undesirable memory usage.
>
> It's still possible to use fold and get short circuiting with good memory
> usage.
>   upTo pred = foldr (\a -> \xs -> if pred a then [a] else a:xs) []
>   prodList = foldl' (*) 1 . upTo (== 0)
> It might be considered cheating, but AFAICT the test for ==0 needs to be
> separated from the multiplication proper.
>

Plain

foldr m 1 

does fine, in fact much better than

foldl' (*) 1 . upTo (== 0),

both in hugs and ghc, regarding speed and memory usage.

Cheers,
Daniel