Split strings into lists with Haskell

Question

Back Story: I've been struggling with this problem for the last week now, reading through Learn You A Haskell, and tutorials online, but I can't figure it out.

I've made a lot of progress with understanding list comprehension and recursion, but this one problem is the one thing still standing in my way.

Question: I'm trying to convert a String (or what I think is a String) into a divided list. Here's the code that I have so far (thanks again Jan for the help). It imports the contents of a .txt file.

import System.Environment

main :: IO ()
main = do
     args <- getArgs
     if null args
       then putStrLn "usage: ./pattern dataset.txt"
       else do contents <- readFile $ head args
               putStrLn $ "Filer1: " ++ filterLower(contents)
               convert' contents

filterLower :: String -> String
filterLower st = [ c | c <- st, c `elem` ['A'..'Z']]

I tried making my own convert function:

convert' :: String -> [String]
convert' x = (x:[])

It works, but it doesn't work for this problem.

Any help would be appreciated.

Error: This is the error that I regularly get.

Couldn't match expected type `IO ()' with actual type `[String]'
    In the return type of a call of `convert''
    In the expression: convert' contents
    In the expression:
      do { contents <- readFile $ head args;
             putStrLn $ "-: " ++ filterLower contents;
           convert' contents }

What are you expecting to happen after you call convert'? It looks like you're just trying to throw the result away: you don't try to print it or anything. — dave4420
– dave4420, Commented Jan 12, 2012 at 18:22
By "divided string" I mean take something like this "A sentence in quotes" and turn it into something like this ["A", "sentence", "in", "a", "list"]. — subtlearray
– subtlearray, Commented Jan 12, 2012 at 18:52

ehird · Accepted Answer · 2012-01-12 19:07:45Z

6

The problem is that the result of convert' is a list of strings, but you're trying to use it as an IO action. You have to do something with the result. For instance,

print $ convert' contents

would print out the result to the screen. Or you could use, e.g. mapM_ putStrLn to print each element out on its own line. Or you could give it a name, and continue writing statements, like:

let converted = convert' contents
...

Going by your comment, it seems that the way you're trying to divide the string is to turn it into a list of its words, e.g. convert' "Hello world" ≡ ["Hello", "world"]. In that case, you don't need convert' at all — the standard words function does just that! So you can just use words contents and not define convert' at all.

edited Jan 12, 2012 at 19:07

answered Jan 12, 2012 at 18:27

ehird

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2 Comments

subtlearray Over a year ago

I remember Jan telling me about words, but I didn't realize it was a function. But I should have known by now that everything in Haskell is a function. :D I'm back on track now. Thank you for your help.

luqui Over a year ago

@SubtleArray, FWIW, it's not true that everything in Haskell is a function. For example, the string "this is not a function" is truthful. Only functions are functions ;-)

Daniel Fischer · Accepted Answer · 2012-02-16 14:22:05Z

5

convert' :: String -> [String], so convert' contents :: [String] is just a list of strings. But you have placed it in an IO-block,

do contents <- readFile whatever
   putStrLn $ ...
   convert' contents

where only expressions of type IO something can appear. There's a simple way to make an IO-less expression into an IO thingy, just return it:

do contents <- readFile ...
   putStrLn $ ...
   return $ convert' contents

is correctly typed. But this do-block has type IO [String], so it doesn't match the declared type of main. How to fix that depends on what you want your programme to do.

edited Feb 16, 2012 at 14:22

answered Jan 12, 2012 at 18:29

Daniel Fischer

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1 Comment

subtlearray Over a year ago

I see. It's like a C++ function, limited in functionality by its type. Thank you for the clarification. I'm VERY new to Haskell, and I'm still trying to get around the syntax.

Luis Casillas · Accepted Answer · 2012-01-12 18:39:24Z

1

I don't understand what you're trying to do here, but the biggest problem is you're not using the do-block correctly. main has type IO (), which implies two constraints:

Every "bare" statement (statement without <-) in the do-block must have type IO a for some a.
The final statement in the do-block must have type IO (); your final statement has type [String] (and thus the error you're getting). The simplest statement with that type would be return ().

edited Jan 12, 2012 at 18:39

answered Jan 12, 2012 at 18:30

Luis Casillas

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