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Today's Topics:
1. Re: Parser as an instance of the monad class (Brent Yorgey)
----------------------------------------------------------------------
Message: 1
Date: Wed, 12 Jan 2011 08:49:20 -0500
From: Brent Yorgey <byor...@seas.upenn.edu>
Subject: Re: [Haskell-beginners] Parser as an instance of the monad
class
To: beginners@haskell.org
Message-ID: <20110112134920.ga...@seas.upenn.edu>
Content-Type: text/plain; charset=us-ascii
On Tue, Jan 11, 2011 at 11:39:48PM -0800, Paul Higham wrote:
> I am working my way through Graham Hutton's book and find that his
> approach to introducing monads is rather nice. Trouble is that the
> code there does not work "out of the book". Specifically, if you
> define the Parser type as follows:
>
> type Parser a :: String -> [(a,String)]
>
> you can then define the return and bind functions as
>
> return v = \x -> [(v,x)]
> p >>= f = \x -> case p x of
> [] -> []
> [(v,y)] -> (f v) y
>
> but you get name conflicts with the functions of the same names in
> the Prelude. Fair enough.
>
> There are a number of ugly things that you can do at this point that
> are just plain wrong, so I thought that the right thing to do would
> be to get the Parser type to be manifested as an instance of the
> Monad class. Ok, now what? Since Parser is only a type synonym it
> cannot be used directly in the following way:
>
> instance Monad Parser where
> return v = . . .
> p >>= f = . . .
>
> but since Parser is not an algebraic type as it is defined that won't
> work either. So how do you do it? help .
Make Parser a newtype, like so:
newtype Parser a = Parser (String -> [(a, String)])
Now you will have some extra Parser constructors to deal with, but you
can make this an instance of Monad just fine.
-Brent
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End of Beginners Digest, Vol 31, Issue 11
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