Got it. Many thanks to Thomas Davie's and Andrew Wagner's replies.  ^_^

On Fri, Mar 20, 2009 at 9:00 PM, Thomas Davie <tom.davie@gmail.com> wrote:

On 20 Mar 2009, at 13:45, ZelluX wrote:

Hi, all

I'm new to haskell and currently reading yaht. I find some problems when trying to solve exercise 3.10.

The exersices asks to read a list of numbers terminated by a zero, and figure out the sum and product of the list. My program is as follows:

ex3_10 = do
 hSetBuffering stdin LineBuffering
 numbers <- getNumber
 let sum = foldr (+) 0 numbers
     product = foldr (*) 1 numbers
 putStrLn "The sum is " ++ show(sum)
 putStrLn "The product is " ++ show(product)

getNumber = do
 putStrLn "Give me a number (or 0 to stop):"
 num <- getLine
 if read num == 0
    then return []
    else do
      rest <- getNumber
      return (read num : rest)

But when i load the program, ghci reports error:
   Couldn't match expected type `[a]' against inferred type `IO ()'
   In the first argument of `(++)', namely `putStrLn "The sum is "'
   In a stmt of a 'do' expression:
         putStrLn "The sum is " ++ show (sum)

And i just don't understand the first sentence. Could you tell what does it mean?

It means that you can't use ++ on an IO action, ++'s type reveals why:

(++) :: [a] -> [a] -> [a]

It accepts two lists, not a list and an IO action.  So the question then is, which IO action are you using ++ on?  The answer is (putStrLn "The sum is").  Your last two statements are being parsed as:

(putStrLn "The sum is ") ++ (show sum)

(putStrLn "The product is ") ++ (show product)

As a random aside, it's usually a good plan in Haskell to get out of IO based computations as fast as possible, and use pure functions instead.  You might want to consider something like this:

ex3_10 = do
 hSetLineBuffering stdin LineBuffering
 interact sumAndProduct

sumAndProduct :: String -> String
sumAndProduct = (\ns -> sumText ns ++ productText ns) . map read . lines

sumText :: (Num a, Show a) => [a] -> String
sumText = ("The sum is " ++) . show . sum

productText :: (Num a, Show a) => [a] -> String
productText = ("The product is " ++) . show . product

Why might you want to do this?  Well, firstly, the code becomes more readable, secondly, a computation that was not sequential is no longer described as being sequential, and finally, the results are more composible.  We can now use sumText and productText, safe in the knowledge that they will never have side effects etc.

Bob