Re: [Haskell-cafe] why typeRepArgs (typeOf "hello") is [Char] ?

Thanks a lot ! 2009/2/2 Ross Mellgren :

Sure:

(+) :: Integer -> Integer -> Integer (really Num a => a -> a -> a, but we'll use the defaulted one)

Which is really

(+) :: -> Integer (-> Integer Integer) (that is, the function type constructor is * -> * -> * and right associative)

So when you say typeRepArgs (typeOf (+)) you get Integer and (-> Integer Integer), which pretty-prints as (Integer -> Integer)

It is possible, but you have to check if the type constructor is really a function type, e.g.:

import Data.Typeable

funTyCon :: TyCon funTyCon = mkTyCon "->"

argsOf :: TypeRep -> [TypeRep] argsOf ty | typeRepTyCon ty == funTyCon = let (x:y:[]) = typeRepArgs ty in x : argsOf y | otherwise = []

*Main Data.Typeable> let f = (undefined :: Int -> Char -> String -> ()) *Main Data.Typeable> argsOf (typeOf f) [Int,Char,[Char]]

-Ross

On Feb 2, 2009, at 3:27 PM, minh thu wrote:

...

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