[Haskell-beginners] Parser as an instance of the monad class

[Paul Higham]

And thanx to you also, this is starting to make some sense

::paul

On Jan 12, 2011, at 5:49 AM, Brent Yorgey wrote:

> 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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