[Haskell-cafe] Parsec: Parenthesized expressions and more!

Vimal j.vimal at gmail.com
Tue Aug 14 04:42:33 EDT 2007


I was trying out some parsing with parsec. I tried:
Accepting proper parenthesized expressions, this was the

parens :: Parser ()
parens = do
  char '('
  char ')'
  <|> return ()

Implementing basically: S -> (S)S | e.

I doubt the fact that 'e' was actually considered, because
the program seems to be accepting all strings of the form
())*. Did I go wrong somewhere? I guess this could be because,
the input was "partially" accepted. Can I force it to derive the
entire input string?

And, btw, is there a method to implement an epsilon production?
I tried
do { string "" ; rules ... }
And it didn't seem to work.

I am also trying out this problem just for fun, but I seem to be getting
wrong answers! http://spoj.pl/problems/FOOL
Maybe I implemented something wrong. So, i will wait for some comments
on the above parsing, and try one more time before I ask questions on that :D

-- Vimal
Department of Computer Science and Engineering
Indian Institute of Technology Madras

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