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 fu

Answer 1

You may find

Text-Megaparsec-Lexer.integer :: MonadParsec s m Char => m Integer

does what you want.

The vanilla parsec library seems to be missing a number of obvious parsers, which has led to the rise of "batteries included" parsec derivative packages. I suppose the parsec maintainers will get around to betteries eventually.

https://hackage.haskell.org/package/megaparsec-4.2.0/docs/Text-Megaparsec-Lexer.html

UPDATE

or with vanilla parsec:

Prelude Text.Parsec Text.Parsec.Language Text.Parsec.Token> parse ( integer . makeTokenParser $ haskellStyle ) "integer" "-1234"
Right (-1234)