From the previous discussion it has been brought to my attention that there is no much difference between
a -> b
and
data F a b = Blank a b -- "-> probably has a blank value constructor"
Therefore it should be trivial to define a label initialized to a function
abstraction without using the -> type constructor. I mean
fun :: F Int Int
fun ? = ?
now what do I write in place of 1st and 2nd question marks? Even if I can do
this, how am I suppose to invoke fun so that it would yield an "Int" not a
"F Int Int"?
Or maybe I should ask the following question: what would the "value
constructor" of -> look like, theoretically?
Thanks
----- Original Message -----
From: "Jon Fairbairn"
For example all functions with Int -> Int are type equivalent However,
data D a b = MkD a b
All objects D Int Int are type equivalent. I'm not sure what your question means, otherwise. If you define data Function a b = F (a -> b) apply:: Function a b -> a -> b apply (F f) a = f a you add an extra level of constructor, but you can still use anything of type Function Int Int where Function Int Int is required: square:: Function Int Int square = F (\ a -> a*a) sqare_root:: Function Int Int square_root = F (\ a -> round (sqrt (fromIntegral a))) E&OE -- it's about my bedtime