[Haskell-cafe] folds with escapes

Michael Vanier mvanier at cs.caltech.edu
Wed Jul 4 19:20:20 EDT 2007

I'm sure this has been done a hundred times before, but a simple generalization of foldl just 
occurred to me and I wonder if there's anything like it in the standard libraries (I couldn't find 
anything).  Basically, I was trying to define the "any" function in terms of a fold, and my first 
try was this:

 > any :: (a -> Bool) -> [a] -> Bool
 > any p = foldl (\b x -> b || p x) False

This is inefficient, because if (p x) is ever True the rest of the list is scanned unnecessarily. 
So I wrote a more general foldl with an "escape" predicate which terminates the evaluation, along 
with a function which tells what to return in that case (given an argument of the running total 'z'):

 > foldle :: (b -> Bool) -> (a -> a) -> (a -> b -> a) -> a -> [b] -> a
 > foldle _ _ _ z [] = z
 > foldle p h f z (x:xs) = if p x then h z else foldle p h f (f z x) xs

Using this function, "foldl" is:

 > foldl' = foldle (const False) id

and "any" is just:

 > any p = foldle p (const True) const False

I also thought of an even more general fold:

 > foldle' :: (b -> Bool) -> (a -> b -> [b] -> a) -> (a -> b -> a) -> a -> [b] -> a
 > foldle' _ _ _ z [] = z
 > foldle' p h f z (x:xs) = if p x then h z x xs else foldle' p h f (f z x) xs

Using this definition, you can write "dropWhile" as:

 > dropWhile :: (a -> Bool) -> [a] -> [a]
 > dropWhile p = foldle' (not . p) (\_ x xs -> x:xs) const []

Again, I'm sure this has been done before (and no doubt better); I'd appreciate any pointers to 
previous work along these lines.


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