How do I get Parsec to let me call `read` :: Int?

Question

I've got the following, which type-checks:

p_int = liftA read (many (char ' ') *> many1 digit <* many (char ' '))

Now, as the function name implies, I want it to give me an Int. But if I do this:

p_int = liftA read (many (char ' ') *> many1 digit <* many (char ' ')) :: Int

I get this type error:

Couldn't match expected type `Int' with actual type `f0 b0'
In the return type of a call of `liftA'
In the expression:
    liftA read (many (char ' ') *> many1 digit <* many (char ' ')) ::
      Int
In an equation for `p_int':
    p_int
      = liftA read (many (char ' ') *> many1 digit <* many (char ' ')) ::
          Int

Is there a simpler, cleaner way to parse integers that may have whitespace? Or a way to fix this?

Ultimately, I want this to be part of the following:

betaLine = string "BETA " *> p_int <*> p_int  <*> p_int <*>
           p_int <*> p_parallel <*> p_exposure <* eol

which is to parse lines that look like this:

BETA  6 11 5 24 -1 oiiio

So I can eventually call a BetaPair constructor which will need those values (some as Int, some as other types like [Exposure] and Parallel)

(if you're curious, this is a parser for a file format that represents, among other things, hydrogen-bonded beta-strand pairs in proteins. I have no control over the file format!)

stephen tetley stephen tetley · Accepted Answer · 2012-05-23T19:45:25

How do I get Parsec to let me call read :: Int?

A second answer is "Don't use read".

Using read is equivalent to re-parsing data you have already parsed - so using it within a Parsec parser is a code smell. Parsing natural numbers is harmless enough, but read has different failure semantics to Parsec and it is tailored to Haskell's lexical syntax so using it for more complicated number formats is problematic.

If you don't want to go to the trouble of defining a LanguageDef and using Parsec's Token module here is a natural number parser that doesn't use read:

-- | Needs @foldl'@ from Data.List and 
-- @digitToInt@ from Data.Char.
--
positiveNatural :: Stream s m Char => ParsecT s u m Int
positiveNatural = 
    foldl' (\a i -> a * 10 + digitToInt i) 0 <$> many1 digit

How do I get Parsec to let me call `read` :: Int?

3 Answers