Combining string modification and concatenation in haskell

Question

So my problem is to take a string in haskell and to modify it so that if there are certain characters, they are changed to other characters, and I have created a helper function to do this, however there is one case where if the character is '!' then it become '!!!111oneone', so i figure to do this you would need to concatenate the current string with '!!111oneone', the trouble is that my function was working with chars however to do this we would need to work with the string, how would you combine this, ie a helper to modify the chars if necessary and implementing the conversion if there is a '!'.

Here is what i have so far

convert :: String -> String
convert [] = []
convert (x:xs) =
| x == '!'  = !helper
| otherwise = converthelper x

You can use ++ to concatenate strings. If c is a single char, you can use [c] to turn it into a string (if you need to). Instead c : s prepends char c to the string s. — chi
– chi, Commented Mar 25, 2019 at 19:15

chepner · Accepted Answer · 2019-03-25 19:18:42Z

3

Assuming your helper is something like

helper :: Char -> String
helper '!' = "!!!111oneone"
helper c = [c]

then you can use concatMap to map helper over each character in your string, and then concatenate the results into a single string.

convert :: String -> String
convert = concatMap helper
-- convert msg = concatMap helper msg

The trick is that your helper promotes every character to a list of characters; most characters just become the corresponding one-character string, but ! becomes something more.

(Note that concatMap forms the basis of the Monad instance for lists. You could also write convert msg = msg >>= helper.)

edited Mar 25, 2019 at 19:18

answered Mar 25, 2019 at 19:12

chepner

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6 Comments

Rein Henrichs Over a year ago

Congrats, you've just defined an L-System. (Ok, technically you also need a starting string.)

chepner Over a year ago

Heh, I don't think I ever knew there was a formal name for that process.

chepner Over a year ago

(Of course, in this case we're stopping after one round of replacements.)

chi Over a year ago

This is a very nice approach, but I fear it is too high-level for the OP who is a beginner and did not know about ++, which is more basic than concatMap.

Rein Henrichs Over a year ago

Yep. iterate (>>= helper) will give you the whole sequence.

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