Mon, 14 May 2001 20:26:21 -0700, Juan Carlos Arevalo Baeza
class (MonadPlus (p s v)) => Parser p where item :: p s v v force :: p s v a -> p s v a first :: p s v a -> p s v a papply :: p s v a -> s -> [(a,s)]
This MonadPlus superclass can't be written. The Parser class is overloaded only on p and must work uniformly on s and v, which can be expressed for functions (by using s and v as here: type variables not mentioned elsewhere), but can't for superclasses. What you want here is this: class (forall s v. MonadPlus (p s v)) => Parser p where which is not supported by any Haskell implementation (but I hope it will: it's not the first case when it would be useful). This should work on implementations supporting multiparameter type classes (ghc and Hugs): class (MonadPlus (p s v)) => Parser p s v where item :: p s v v force :: p s v a -> p s v a first :: p s v a -> p s v a papply :: p s v a -> s -> [(a,s)] Well, having (p s v) in an argument of a superclass context is not standard too :-( Haskell98 requires types here to be type variables. It requires that each parser type is parametrized by s and v; a concrete parser type with hardwired String can't be made an instance of this class, unless wrapped in a type which provides these parameters. The best IMHO solution uses yet another extension: functional dependencies. class (MonadPlus p) => Parser p s v | p -> s v where item :: p v force :: p a -> p a first :: p a -> p a papply :: p a -> s -> [(a,s)] Having a fundep allows to have methods which don't have v and s in their types. The fundep states that a single parser parses only one type of input and only one type of tokens, so the type will be implicitly deduced from the type of parser itself, basing on available instances. Well, I think that s will always be [v], so it can be simplified thus: class (MonadPlus p) => Parser p v | p -> v where item :: p v force :: p a -> p a first :: p a -> p a papply :: p a -> [v] -> [(a,[v])] Without fundeps it could be split into classes depending on which methods require v: class (MonadPlus p) => BasicParser p where force :: p a -> p a first :: p a -> p a class (BasicParser p) => Parser p v where item :: p v papply :: p a -> [v] -> [(a,[v])] This differs from the fundep solution that sometimes an explicit type constraint must be used to disambiguate the type of v, because the declaration states that the same parser could parse different types of values. Well, perhaps this is what you want and item :: SomeConcreteParser Char could give one character where item :: SomeConcreteParser (Char,Char) gives two? In any case using item in a way which doesn't tell which item type to use is an error.
Ok, last, I wanted to alias a constructor. So:
There is no such thing. A constructor can't be renamed. You would have to wrap the type inside the constructor in a new constructor. -- __("< Marcin Kowalczyk * qrczak@knm.org.pl http://qrczak.ids.net.pl/ \__/ ^^ SYGNATURA ZASTÊPCZA QRCZAK