Thanks. Could you add to your explanation this one :
*Graph> typeRepArgs (typeOf (+))
[Integer,Integer -> Integer]
In fact, I tried to write a function that would give the types used by
a function,
for instance [Integer, Integer, Integer] for (+) (the last one would
be the 'return' type).
So I applied recursively typeRepArgs to the second element of the list
(if any) (here, Integer -> Integer).
It worked well until I tried it on a function like :: Char -> Int ->
[Char] where
the last recursive call gives [Char] instead of [].
Is it possible to write such a function ?
Thank you,
Thu
2009/2/2 Ross Mellgren
The type of "hello" is String, which is [Char], which is really [] Char (that is, the list type of kind * -> *, applied to Char).
1, 'a', and True are all simple types (I'm sure there's a more particular term, maybe "monomorphic"?) with no type arguments.
[] has a type argument, Char.
Consider:
Prelude Data.Typeable> typeRepArgs (typeOf (Just 1)) [Integer]
and
Prelude Data.Typeable> typeRepArgs (typeOf (Left 'a' :: Either Char Int)) [Char,Int]
-- typeRepArgs is giving you the arguments of the root type application, [] (list) in your case, Maybe and Either for the two examples I gave.
Does this make sense?
-Ross
On Feb 2, 2009, at 3:09 PM, minh thu wrote:
Hello,
With Data.Typeable :
*Graph> typeRepArgs (typeOf 1) [] *Graph> typeRepArgs (typeOf 'a') [] *Graph> typeRepArgs (typeOf True) [] *Graph> typeRepArgs (typeOf "hello") [Char]
I don't understand why the latter is not []. Could someone explain it ?
Thank you, Thu _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe