Thanks Tomasz and Cale, it worked now ! On 11/4/05, Cale Gibbard wrote:

On 04/11/05, Abhijit Ray wrote:

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Hi I just started on haskell , i am using the yet another haskell tutorial by Hal Daume

I wrote the following program and it dint compile. what's wrong with it

module Main where

import IO

main = do hSetBuffering stdin LineBuffering let testList = makeList let sum = foldr (+) 0 testList putStrLn "Sum is " ++ show(sum)

makeList = do putStrLn "enter a num" num <- getLine let nbr = read num if nbr == 0 then do return [] else do all <- makeList return (nbr : all) ---------------------------------------------- the error i get is

Chasing modules from: Sum.hs Compiling Main ( Sum.hs, Sum.o )

Sum.hs:6: Couldn't match `[a]' against `IO ()' Expected type: [a] Inferred type: IO () In the application `putStrLn "Sum is "' In the first argument of `(++)', namely `putStrLn "Sum is "'

Sum.hs:9: Couldn't match `[b]' against `IO [a]' Expected type: [b] Inferred type: IO [a] In the third argument of `foldr', namely `testList' In the definition of `sum': sum = foldr (+) 0 testList -----------------------------

Thanks, Abhijit Ray

Hello,

The first error is related to the way in which the following line parses:

...

putStrLn "Sum is " ++ show(sum)

which is like: (putStrLn "Sum is") ++ (show sum) It's complaining that you're passing an IO action to (++). It's an easy fix, add some parens, or a well-placed ($): putStrLn $ "Sum is" ++ show sum

The second problem arises due to not running the IO action you wanted to:

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main = do hSetBuffering stdin LineBuffering let testList = makeList let sum = foldr (+) 0 testList putStrLn $ "Sum is " ++ show sum What is the type of testList? Well, it must be the same as makeList -- what is makeList's type? The definition of makeList is in the form of a do-block which carries out an IO computation producing a list of numbers, so it's (Num a) => IO [a]. Note that this is not a list, so it's unsuitable for passing to foldr.

What do we do? We run the IO action using "<-":

...

main = do hSetBuffering stdin LineBuffering testList <- makeList let sum = foldr (+) 0 testList putStrLn $ "Sum is " ++ show sum This will cause the running of main to *run* the IO action makeList, getting the result into the list testList, which now has type (Num a) => [a] (here a is going to be defaulted to Integer, since nothing says otherwise, and one needs to pick something to be able to read).

Hope this helps, and good luck with Haskell :) - Cale