Re: [Haskell-cafe] how to break foldl' ?

gary ng wrote:

Still a bit confused.

My function is simply

f x y = if x < 100 then x + y else x

Try this: sum_to_100 xs = foldr (\x k a -> if a < 100 then k $! (x+a) else a) id xs 0 As expected, (sum_to_100 [1..]) gives 105, without trying to find the end of an infinite list. You get your early-out semantics combined with strict evaluation. Udo.