[Haskell-cafe] Re: Proving stuff about IORefs
Ben Franksen
ben.franksen at online.de
Sun Oct 17 08:15:44 EDT 2010
Derek Elkins wrote:
> On Sun, Oct 17, 2010 at 6:49 AM, Miguel Mitrofanov
> <miguelimo38 at yandex.ru> wrote:
>>
>> On 17 Oct 2010, at 05:21, Ben Franksen wrote:
>>
>>> I want to prove that
>>>
>>> f r == do
>>> s1 <- readIORef r
>>> r' <- newIORef s1
>>> x <- f r'
>>> s3 <- readIORef r'
>>> writeIORef r s3
>>> return x
>>
>> That is not true. Consider the following function:
>>
>> g r1 r2 = writeIORef r1 0 >> writeIORef r2 1 >> readIORef r1
>>
>> This function behaves differently depending on whether r1 and r2 are the
>> same IORef or not. Therefore, the property you want to prove doesn't hold
>> for the partially-applied function
>>
>> f = g r1
>
> I originally was thinking along these lines, and this is an important
> case, but there is an even more trivial example.
>
> Let f be return.
Oh, my god. Of course.
Thanks for pointing me to the obvious ;-)
This means I either have to give up on the second law
callback . embed = id
or find another implementation. Maybe IORef is too powerful. Hmm.
The second law was inspired by the instance for ReaderT:
instance Embed m (ReaderT r m) where
type Content m (ReaderT r m) = r
embed = ReaderT
callback = runReaderT
which is much more general (it does not depend on the inner monad being IO),
and where indeed the second law holds.
Cheers
Ben
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