Am Freitag, 25. August 2006 23:55 schrieb Andrea Rossato:
Il Fri, Aug 25, 2006 at 08:35:01PM +0100, Brian Hulley ebbe a scrivere:
It is maybe easier to just think of a newtype decl as being the same as a data decl except for the fact that you can only have one constructor on the rhs whereas a data decl allows multiple constructors, and a type decl by contrast as just introducing a simple alias for convenience.
First my apologies for being a bit confusing in my question: I'm tired and wrote the code too quickly...;-)
So I'll rephrase:
type T a = Int -> (a, Int) mkT :: a -> T a mkT a = \x -> (a, x)
newtype T1 a = T1 (Int -> (a,Int)) mkT1 :: a -> T1 a mkT1 a = T1 (\x -> (a, x))
data T2 a = T2 (Int -> (a,Int)) mkT2 :: a -> T2 a mkT2 a = T2 (\x -> (a, x))
makeT a b = mkT a b --makeT1 a b = mkT1 a b --makeT2 a b = mkT2 a b
why makeT 1 2 works while makeT1 a makeT2 will not compile? That is to say: why mkT1 and mkT2 cannot be used (even though the compiler does not complain)? That is to say, can someone explain this behaviour? I do not grasp it.
Because T a is a function type, namely Int -> (a,Int), so if a has type ta, mkT a has type Int -> (ta,Int) and so can be applied to a further argument: makeT a b = (mkT a) b is fine. However, neither T1 a nor T2 a is a function type, a value of type T1 a is a function _wrapped by the data (or value) constructor T1_ (the same applies to T2, of course), so before you can apply mkT1 a to an Int, you have to unwrap it: unT1 :: T1 a -> T a unT1 (T1 f) = f makeT1 a b = unT1 (mkT1 a) b will work fine. HTH, Daniel
Thanks for your kind attention. Andrea _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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