At 00:42 15/10/03 -0400, Derek Elkins wrote:
If I comment out the indefinitely long list case (test7) in the definition of 'test', the foldl version will pass all these test cases and foldr throws an error on test6. But if I include test7, using Hugs the foldr test case returns a C stack overflow, and the foldl case silently disappears.
foldl never works on infinite lists. foldr isn't tail recursive, so you don't use it with strict functions if you are expecting large input. With lazy (in the second argument) functions it's appropriate.
Ah, thanks :-O That's a usefully concise statement of the problem I was bumping up against. (I also note that foldr in this case messes up the order that operands are applied if the function is non-commutative or non-associative, and foldr1 places other constraints on the type.)
You may also want to look at foldM. Instantiated for an exception-like monad (Maybe,Either,etc.), you can effectively "throw" a result (or lack thereof for Maybe).
Perfect. foldM is just what I was looking for. The corresponding code is: [[ allSame3 :: (Eq a) => [a] -> Bool allSame3 [] = True allSame3 (a:as) = isJust $ foldM nextSame3 a as nextSame3 :: (Eq a) => a -> a -> Maybe a nextSame3 a a1 | a1 == a = Just a | otherwise = Nothing ]] And it does work in the infinite list case. #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact