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Home/ Questions/Q 7730933
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T06:21:13+00:00 2026-06-01T06:21:13+00:00

I am trying to parse user entered string like A12, into a Haskell tuple,

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I am trying to parse user entered string like “A12”, into a Haskell tuple, like (‘A’, 12).

Here’s what I have tried:

import Data.Maybe

type Pos = (Char, Int)

parse :: String -> Maybe Pos
parse u = do
  (c, rest) <- (listToMaybe.reads) u
  (r, _) <- (listToMaybe.reads) rest
  return $ (c, r)

But this always returns Nothing. Why does this happen, and what is the correct way to parse this string? Since this is fairly simple, I’d like to avoid using Parsec or a similar advanced parsing library.

EDIT (to clarify):
Sample Input and Output:

"A12" gives Just ('A', 12)

"J5" gives Just ('J', 5)

"A" gives Nothing

"2324" gives Nothing

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  1. Editorial Team
    Editorial Team
    2026-06-01T06:21:14+00:00Added an answer on June 1, 2026 at 6:21 am

    read is usually the opposite of show and they both generally use Haskell syntax to represent the given values. This means that since the Haskell syntax for characters uses single quotes, show on a character will add single quotes around it, and read will expect the single quotes to be there.

    In other words, your function expects syntax like 'A' 42, and indeed it works if you try that:

    > parse "'A' 42"
    Just ('A',42)
    

    For your format, I would instead use pattern matching for the first character and then reads for the rest, e.g. something like this:

    parse :: String -> Maybe Pos
    parse [] = Nothing
    parse (c:rest) = do
      (r, _) <- listToMaybe $ reads rest
      return (c, r)
    
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