[Haskell-cafe] why typeRepArgs (typeOf "hello") is [Char] ?
minh thu
noteed at gmail.com
Mon Feb 2 15:49:57 EST 2009
Thanks a lot !
2009/2/2 Ross Mellgren <rmm-haskell at z.odi.ac>:
> 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 <rmm-haskell at z.odi.ac>:
>>>
>>> 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 at haskell.org
>>>> http://www.haskell.org/mailman/listinfo/haskell-cafe
>>>
>>>
>
>
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