In Haskell, how do you trim whitespace from the beginning and end of a string?

Question

How do you trim whitespace from the start and end of a string?

trim "  abc "   =>  "abc" 

Edit:

Ok, let me be a little clearer. I did not understand that string literals were treated so differently from Strings.

I would like to do this:

import qualified Data.Text as T let s :: String = "  abc  " in T.strip s 

Is this possible in Haskell? I am using -XOverloadedStrings but that appears only to work for literals.

Answer 1

If you have serious text processing needs then use the text package from hackage:

> :set -XOverloadedStrings > import Data.Text > strip "  abc   " "abc" 

If you're too stubborn to use text and don't like the inefficiency of the reverse method then perhaps (and I mean MAYBE) something like the below will be more efficient:

import Data.Char  trim xs = dropSpaceTail "" $ dropWhile isSpace xs  dropSpaceTail maybeStuff "" = "" dropSpaceTail maybeStuff (x:xs)         | isSpace x = dropSpaceTail (x:maybeStuff) xs         | null maybeStuff = x : dropSpaceTail "" xs         | otherwise       = reverse maybeStuff ++ x : dropSpaceTail "" xs   > trim "  hello this \t should trim ok.. .I  think  ..  \t " "hello this \t should trim ok.. .I  think  .." 

I wrote this on the assumption that the length of spaces would be minimal, so your O(n) of ++ and reverse is of little concern. But once again I feel the need to say that if you actually are concerned about the performance then you shouldn't be using String at all - move to Text.

EDIT making my point, a quick Criterion benchmark tells me that (for a particularly long string of words with spaces and ~200 pre and post spaces) my trim takes 1.6 ms, the trim using reverse takes 3.5ms, and Data.Text.strip takes 0.0016 ms...

Answer 2

From: http://en.wikipedia.org/wiki/Trim_(programming)#Haskell

import Data.Char (isSpace)  trim :: String -> String trim = f . f    where f = reverse . dropWhile isSpace