Re: [Haskell-beginners] Functional Parses

Thank you very much. I'll check out the code on the website tomorrow and hopefully this will solve the problem. Whenever monads apper things seem to get tricky - sadly there won't be a lecture on functional programming in the next semester at the university I study at :( Am Tue, 3 Jul 2012 18:01:32 +0100 schrieb Stephen Tetley :

On 3 July 2012 15:13, Christian Maeder wrote:

...

This type synonym is unsuitable for a Monad instance. Better would be: newtype Parser a = Parser (String -> [(a, String)]) but that would require to change your code below.

This is alluded to in the closing chapter remarks (section 8.9) of Graham Hutton's book and there is code available on the website that accompanies the book that "solves" the problem. Unfortunately, this chapter does seem to trip people up who use the book for self study.

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