I see there's another answer about this but it misses one key point which is likely the cause of your (and WinHugs') confusion. On 2009 Jan 7, at 19:31, David Schonberger wrote:
listFactorial l = if length l == 0 then return 1 else do fact (head l) listFactorial (tail l)
fact n = if n == 0 then return 1 else return foldr (*) 1 [1..n]
"return" doesn't mean what it does in other languages. It means "wrap this value in a monad". Since neither of the above functions involves a monad at all, the "return"s confuse Hugs into thinking you're doing something more complex than you actually are (hence the scary-looking type inference). Ignoring (most) other optimizations to the code, since the other message hit them:
listFactorial [] = [1] listFactorial (x:xs) = fact x : listFactorial xs
fact 0 = 1 fact n = foldr (*) 1 [1..n]
f = do nums <= askForNums putStr ("Sum is " ++ (show (foldr (+) 0 nums)) ++ "\n") putStr ("Product is " ++ (show (foldr (*) 1 nums)) ++ "\n") print $ listFactorial nums
If you'd rather keep your original definition of "f" and put listFactorial in IO, then:
listFactorial [] = print 1 listFactorial (x:xs) = do print $ fact x listFactorial xs
-- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH