Re: [Haskell-cafe] Newbie question on iterating over IO list

Sebastian, Thanks for your help. I learned two things: o show is not an action, but a function. Using it in the interactive mode of GHCI was making me confused. o To output anything, I need to do 'do'. -Taro Sebastian Sylvan wrote:

On 6/20/06, Taro Ikai wrote:

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Here's a code fragment I have from working with Hal Daume's "Yet Another Haskell Tutorial". I cannot figure out how to iterate over the result from askForNumbers that returns "IO [Integer]", and print them.

The "<-" operator seems to take IO off of the results from actions like getContents, which returns type "IO String", but I cannot seem to use it to take IO off of lists with it. What am I supposed to do?

Also, as a side question, what should I be returning from the do function?

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code>>>>

module Main where

import IO

main = do hSetBuffering stdin LineBuffering s <- askForNumbers {-- This compiles --} map show s {-- But this doesn't. Why? --}

s is a list of numbers, "map show s" is a list of Strings. It is not an IO [Strings] so you can't use it like an action. Think about what you're trying to do here, you map "show" on a list of ints, but what do you do with the result? Nothing. Why would you *want* the above to compile? It would just convert a bunch of numbers into strings and then forget about them! So what you want is probably something like:

let xs = map show s

The rule of thumb is, "use 'let' for regular values, and (<-) for actions".

If you want to print all of them you could do something like:

printList [] = return () printList (x:xs) = do putStrLn x printList xs

On the other hand, this operation seems generally useful, so why not generalise it a bit?

mapAction [] = return () mapAction a (x:xs) = do a x mapAction a xs

And in fact, this function already exists! It's called mapM_ (the M is for Monad). You could use this by passing in "putStrLn" as the action to this function.

/S