[Haskell-cafe] Another monad question...

[Stefan O'Rear]

On Sat, Jul 14, 2007 at 11:26:56PM -0400, David LaPalomento wrote:
> I've been playing with the parsers decribed in "Monadic Parser Combinators"
> (http://www.cs.nott.ac.uk/~gmh/bib.html#monparsing) and I've gotten
> stumped.  I'm trying to get comfortable working monadically, so please
> excuse my ignorance.  Here's the relevant portions of my code:

You don't need to excuse yourself.  If you *ever* feel as though you
might regret a question, we have failed.  The point of this list is for
newbies to ask questions and learn!

>
[reordered]
> -- word parses everything as []
> word :: Parser String
> word = mzero `mplus` do
>  x <- letter;
>  xs <- word;
>  return (x:xs)
>
> As I noted in the code, no matter what inputs I give it,
>> parse word "blah blah"
> always returns [].  Any ideas where where my misunderstanding is?

Your base case is subtly wrong - it should be return [], not mzero.
Mzero always fails - mzero `mplus` x = x, by one of the MonadPlus laws.

> sat :: (Char -> Bool) -> Parser Char
> sat p = Parser (\inp -> [ (v, out) | (v, out) <- parse item inp, p v])
>
> lower :: Parser Char
> lower = Parser (\inp -> parse (sat (\x -> 'a' <= x && x <= 'z')) inp)
>
> upper :: Parser Char
> upper = Parser (\inp -> parse (sat (\x -> 'A' <= x && x <= 'Z')) inp)

These definitions aren't in the best style; while they are correct, I
would prefer to write them as:

sat :: (Char -> Bool) -> Parser Char
sat p = do { ch <- item ; guard (p ch) ; return ch }

lower :: Parser Char
lower = sat isLower

upper :: Parser Char
upper = sat isUpper

Stefan
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