[Haskell-beginners] Re: Advice wanted on parallel processing

[Daniel Fischer]

Am Mittwoch, 18. März 2009 16:49 schrieb Colin Paul Adams:
> >>>>> "Daniel" == Daniel Fischer <daniel.is.fischer at web.de> writes:
>
>     Daniel> If e.g.
>
>     Daniel> data Move = Move {from :: Position, to :: Position}
>
>     Daniel> , the instance would be
>
>     Daniel> instance NFData Move where rnf (Move f t) = rnf f `seq`
>     Daniel> rnf t `seq` ()
>
>     Daniel> That might require NFData instances for Position and its
>     Daniel> components, but specifying these should be automatic.
>
> <switched to beginners list>
>
> Move is somewhat more complicated than that, but it comes down to
> specifying instances for
>
> data Piece_colour
>     = Black
>
>     | White
>
> and
>
> data Piece_type
>     = Lance
>
>     | Reverse_chariot
>     | Side_mover
>     | Vertical_mover
>     | White_horse
>     | Rook
>
>     ... many more
>
> which I'm not sure how to do. I tried added a deriving clause, but the
> compiler complains:
>
>   Can't make a derived instance of `NFData Piece_colour'
>       (`NFData' is not a derivable class)
>     In the data type declaration for `Piece_colour'

For a newtype-wrapper of a type which is an instance of NFData, you can do

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

newtype Thing = T nfdata
	deriving NFData

, but for 

data Thingummy = ...

you have to write the instance yourself.

>
> here I don't have and fields to force.

For an enumeration like Piece_colour or Piece_type,

instance NFData Piece_colour where
	rnf = rwhnf

is all you need, but you don't even necessarily need that, for

data Move = Move {colour :: Piece_colour, typ :: Piece_type, start, end :: 
Position}

instance NFData Move where
	rnf (Move c t s e) = c `seq` t `seq` rnf s `seq` rnf e

is good.