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

[minh thu]

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
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>