CODEWITHSUNDEEP
haskellcharquotes
mhl
mhl

Reputation: 705

Haskell char quotes

I've started to learn haskell for real recently, and I'm doing some exercises from wikibooks. I'm doing exercise with RLE encoding, and I've come with solution like this:

import Data.List
rle :: String -> [(Int,Char)]
rle [] = []
rle xs = zip lengths chars
        where
        groups = group xs
        lengths = map length groups
        chars = map head groups

rle_toString :: [(Int, Char)] -> String
rle_toString [] = []
rle_toString (x:xs) = show (fst x ) ++ show (snd x) ++ rle_toString xs`

Not a very elegant solution, but it almost works. The problem is, that I get output like this: "7'a'8'b'7'j'6'q'3'i'7'q'1'p'1'a'16'z'2'n'". The single quotes with chars are not vetry elegant. How can I achieve output like: "7a8b7j6q3i7q1p1a16z2n"?

Upvotes: 1

Views: 917

Answers (2)

Chris
Chris

Reputation: 4231

show is used to print values as they appear in Haskell source code, and thus puts single quotes around characters (and double quotes around strings, and so on). Use [snd x] instead to show just the character.

In Haskell, String is just shorthand for List of Char [Char]. For example, the String "Foo" can also be written like this: ['F','o','o']. So, to convert a single character to a string, just put in in brackets: [char].

Upvotes: 6

Tarmil
Tarmil

Reputation: 11362

The problem is your use of show on a character. show 'a' == "'a'".

The solution is to realize that strings are just lists of characters, so if c is a character, then the one-character string that contains c is just [c].

Upvotes: 1

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