Wolfgang Jeltsch wrote:
> Am Montag, 3. Juli 2006 18:46 schrieb Simon Peyton-Jones:
>> Thanks. Can you confirm that it's fixed in 6.4.2? And if so, can you
>> record that too? No point in us haring after fixed bugs!
>>
>> S
>
> As Chris Kuklewicz pointed out, this bug doesn't seem to be completely fixed
> in 6.4.2. Currently, I neither can confirm this nor can I disprove it since
> at the moment there seem to be no Debian packages for 6.4.2 which I could
> install and I don't have the time to install GHC from source.
>
> Best wishes,
> Wolfgang
I am using ghc-6.4.2 from darwinports on OS X 10.4.7
Can I help confirm / test this with something test cases?
The older messages were:
> Wolfgang Jeltsch wrote:
>> Hello everybody,
>>
>> I wanted to do something like that:
>>
>> data Pair :: (* -> *) -> * where
>> Pair :: a b -> b -> Pair a
>>
>> data Sel :: * -> * where
>> A :: Sel Bool
>> B :: Sel Integer
>>
>> showSnd :: Pair Sel -> String
>> showSnd (Pair A bool)
>> = show bool
>> showSnd (Pair B integer)
>> = show integer
>>
>> However, GHC 6.4.1 with -fglasgow-exts complains in the second last and last
line that there is no instance for (Show b). I don't really understand this
since in my opinion it is clear that bool :: Bool and integer :: Integer and
that therefore bool and integer are showable. What's the problem with my code?
>>
>> Best wishes,
>> Wolfgang
>
> I cut and pasted your code into ghci-6.4.2 with -fglasgow-exts and there was
no error. And I can (print (showSnd (Pair A True)) and (print (showSnd (Pair B
8))) but when I try to compile it with ghc-6.4.2 it complains about the Show
instance. I find this deeply wierd, and it seems like a bug.
>
> Changing the definition of Pair made it compile and run with ghc:
>
>> data Pair :: (* -> *) -> * where
>> Pair :: (Show b) => a b -> b -> Pair a
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