[Haskell-cafe] why typeRepArgs (typeOf "hello") is [Char] ?
noteed at gmail.com
Mon Feb 2 15:27:47 EST 2009
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
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 ->
the last recursive call gives [Char] instead of .
Is it possible to write such a function ?
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.
> Prelude Data.Typeable> typeRepArgs (typeOf (Just 1))
> Prelude Data.Typeable> typeRepArgs (typeOf (Left 'a' :: Either 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?
> On Feb 2, 2009, at 3:09 PM, minh thu wrote:
>> With Data.Typeable :
>> *Graph> typeRepArgs (typeOf 1)
>> *Graph> typeRepArgs (typeOf 'a')
>> *Graph> typeRepArgs (typeOf True)
>> *Graph> typeRepArgs (typeOf "hello")
>> I don't understand why the latter is not . Could someone explain it ?
>> Thank you,
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
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