Thanks Roman .. I'll try and implement laziness to retain the whole list. Regards, Kashyap On Thu, Jul 25, 2013 at 3:41 AM, Roman Cheplyaka wrote:

To construct such an example, you have to ask yourself: when can we get a list of more than one element?

Consider this:

a = char 'a'

((a >> a) <|> a) >> a

Suppose that our input is "aa". The result of ((a >> a) <|> a) would be the list

[('a', ""), ('a', "a")]

If you proceed with the first element of the list, the overall parse will fail. It can only succeed if you then try the second element.

By the way, you shouldn't confuse Parsec (the library) with the general concept of parser combinators or the implementation from the paper you reference. The above parse would fail in Parsec as well, despite the fact that Parsec allows backtracking.

Roman

* Kashyap CK [2013-07-24 08:38:53-0700]

...

There is reference in the paper that empty list indicates failure...so could we just use it like Maybe? I'd like it very much if I could get an example of a missed match by not using the complete match.

regards, Kashyap

Sent from my Windows Phone From: Roman Cheplyaka Sent: 24/07/2013 8:19 PM To: C K Kashyap Cc: Haskell Cafe Subject: Re: [Haskell-cafe] Parsec question Think about this: if you always take only the first element, why do you need lists at all?

Roman

* C K Kashyap [2013-07-24 19:56:29+0530]

...

Dear Cafe,

I am trying to implement[1] parsec in go using the "Monadic Parser Combinators" paper [2] . I've been able to implement "plus" "bind" and "many" While doing the implementation - I looked at bind closely

bind :: Parser a -> (a -> Parser b) -> Parser b p `bind` f = \inp -> concat [f v inp' | (v,inp') <- p inp]

I wondered if the result needs the complete list - wouldn't just the first successful value suffice? Perhaps - p `bind` f = \inp -> take 1 $ concat [f v inp' | (v,inp') <- p inp]

Will this miss out matches?

Regards, Kashyap

[1] https://github.com/ckkashyap/parsec/blob/master/parsec.go [2] Monadic Parser Combinators: Graham Hutton, Erik Meijer

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