On Fri, 27 Oct 2000, Matthias Höchsmann wrote:

Hello,

I have the following problem:

basic datatypes

...

type Sequence a = [a] data Tree a = N a (Forest a) deriving (Ord,Eq,Show) type Forest a = Sequence (Tree a)

i want to construct a class Xy

...

class Xy s a where test :: s a -> a

and make an instance for list of characters

...

instance Xy [] Char where test [a] = a

this works, and an instance for a forest and tried something like this

...

instance ([] Tree) Char where test x@(N a xs):txs = a

Don't you mean test (N a xs:txs) = a ? /Lars L

Matthias Höchsmann wrote:

...

type Sequence a = [a] data Tree a = N a (Forest a) deriving (Ord,Eq,Show) type Forest a = Sequence (Tree a)

i want to construct a class Xy

...

class Xy s a where test :: s a -> a

[...]

...

instance ([] Tree) Char where test x@(N a xs):txs = a

To make it syntactically correct this should at least be something like

instance Xy ([] Tree) Char where test (N a xs:txs) = a

But the real problem is in the expression ([] Tree), which is the same as writing [Tree]. This is not a legal type expression, since Tree is a type constructor, not a ground type, so you cannot apply it to the list constructor. What you are trying to say is probably something like this:

instance Xy (\a . [Tree a]) Char -- not Haskell

But unfortunately there are no lambdas on the type level - they would render the type system undecidable. For the same reason it is not allowed to use a type synonym in an instance declaration:

instance Xy Forest Char -- illegal

The only thing you can do is turning Forest into a data type:

data Tree a = N a (Forest a) deriving (Ord,Eq,Show) data Forest a = Forest [Tree a]

instance Xy Forest Char where test (Forest (N a xs:txs)) = a

HTH, - Andreas -- Andreas Rossberg, rossberg@ps.uni-sb.de :: be declarative. be functional. just be. ::