CODEWITHSUNDEEP
haskell
Morgan Schofield
Morgan Schofield

Reputation: 11

How to remove a specific last character in a String

I want to remove the last character of a String, if it's a "!"

I know I want base cases, ie. if the string is empty:

remove [] = []

How do I index the last character? Using:

last

And, what's the best way to create an 'if, then' loop for this problem?

Should I use guards? |

Upvotes: 1

Views: 549

Answers (2)

Will Ness
Will Ness

Reputation: 71055

You don't have to do this with a loop. Instead of such imperative thinking you can take a higher-level view and compose the program from several higher-order functions, getting this one-liner,

import Data.List (tails)

foo :: String -> String
foo  =  (take 1 =<<) . takeWhile (/= "!") . tails
 --  =  concat . map (take 1) . takeWhile (/= "!") . tails

or with list comprehensions,

foo xs  =  [c | (c:_) <- takeWhile (/= "!") $ tails xs]

tails repeatedly takes tails of the input list, takeWhile gets rid of the final '!', if any, and concatMap (take 1) reconstitutes the string back from its tails:

concatMap (take 1) . tails

is an identity operation on lists, and we're getting in on the action, in the middle.

Upvotes: 1

bereal
bereal

Reputation: 34272

The Haskell way of solving this is with a recursion and pattern matching:

remove "" = ""                   -- obvious
remove "!" = ""                  -- that's what we're here for
remove (x:xs) = (x : remove xs)  -- keep the head, proceed to the tail

Any attempts with last and conditional constructions will be less elegant and efficient.

Upvotes: 7

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