Put string in to list in Haskell

Question

I need to create my own words functions. It takes string and puts into list where ever there is space. For example string "i need help" would result in ["i","need","help"]. The definitions must be exactly

anything :: String -> [String]

I currently came up with stupid solution which look like this ( also it doesn't work)

test :: String -> [String]
test d = beforep d : (test (afterp d)) : []

beforep :: String -> String
beforep d = takeWhile (/=' ') d
afterp :: String -> String
afterp d = if (dropWhile (/=' ') d)==[] then []
      else tail(dropWhile (/=' ') d)

test -> uses tail recursion

beforep -> get everything till first space

afterp -> gets everything after space

Any ideas ? If you have any other solutions to this problem it would help. Thank you

yes but i cant use the words function. I need to write it without the words function. — Andrius
– Andrius, Commented Oct 29, 2013 at 14:10

Neil Forrester · Accepted Answer · 2013-10-29 14:11:50Z

5

You've very nearly got it. If I attempt to run your code as is, I get:

test.hs:2:23:
    Couldn't match expected type `Char' with actual type `String'
    Expected type: String
      Actual type: [String]
    In the return type of a call of `test'
    In the first argument of `(:)', namely `(test (afterp d))'

So examine line 2:

test d = beforep d : (test (afterp d)) : []
--                                      ^
-- This is the problem -----------------|

The type of the cons operator is:

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

Your test function returns a [String] already, you don't want to try to cons it onto an empty list. That would imply that the return type would be [[String]].

Try this instead:

test d = beforep d : (test (afterp d))

After that change, it compiles, but when you run test "i need help" you get the infinite list:

["i","need","help","","","","","","","",""...

The problem is that you need to include a base case in test that stops when you pass it an empty list. Here's the working code:

test :: String -> [String]
test [] = []
test d = beforep d : (test (afterp d))

beforep :: String -> String
beforep d = takeWhile (/=' ') d

afterp :: String -> String
afterp d = if (dropWhile (/=' ') d)==[]     -- Slightly reformatted
             then []                        -- to improve readability,
             else tail(dropWhile (/=' ') d) -- no real change.

answered Oct 29, 2013 at 14:11

Neil Forrester

5,2911 gold badge32 silver badges33 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Andrius Over a year ago

Thank you very much, great explanation. Exactly what i needed u just made my day ^^

Collectives™ on Stack Overflow

Put string in to list in Haskell

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related