[Haskell-beginners] Some confusion still with a YAHT example

[Brandon S. Allbery KF8NH]

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 at kf8nh.com
system administrator [openafs,heimdal,too many hats] allbery at ece.cmu.edu
electrical and computer engineering, carnegie mellon university    KF8NH


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