[Haskell-cafe] RFC: Top level mutable state in terms of pure-function initialization?
Roman Cheplyaka
roma at ro-che.info
Sun Dec 15 14:47:58 UTC 2013
Ok, it could work, but it can lead to very subtle bugs. You change the
type of IORef during refactoring (and forget to change it in a different
place), and your program now works wrong (but silently so), because now
it's creating two IORefs instead of one.
A better way would be to have phantomly-typed identifiers, like
class IsRefId c where
initialContents :: c a -> a
toInt :: c a -> Int
data MyRefId a where
Ref1 :: RefId Int
Ref2 :: RefId Bool
...
instance IsRefId MyRefId where ...
idToIORef :: IsRefId c => c a -> IORef a
Still can be abused (by not constraining phantom types properly), but
can also provide safety if used right.
(And no, Eq is not strictly necessary, because you convert to integers
anyway.)
Roman
* EatsKittens <temporalabstraction at gmail.com> [2013-12-15 15:05:40+0100]
> 1. Well, since x and y are not of the same type they would in this case
> contain a different IORef. The read in this case would read False since y
> is initialized to false as the default of Bool.
>
> 2. In this example it would again be two different IO refs since the type
> is different. It would again on demand create a different IORef for Int and
> Bool.
>
> 3. No, they refer to a different one as I said, the default-less version
> creates the IORef to be unique if and only if both the seed and
> initialization are the same.
>
> 4. Agreed, I overlooked that not all enum instances are typable. (I believe
> they also need to be an instance of EQ?)
>
>
> On 15 December 2013 14:38, Roman Cheplyaka <roma at ro-che.info> wrote:
>
> > 1. Well, that creates a loophole in the type system:
> >
> > let
> > x :: IORef Int
> > x = enumToIORef 0
> >
> > y :: IORef Bool
> > y = enumToIORef 0
> > in writeIORef x (5 :: Int) >> (readIORef y :: IO Bool)
> >
> > 2. There's also a less obvious loophole with your Defaultable version:
> >
> > let
> > x :: Defaultable a => IORef a
> > x = enumToIORef 0
> > in writeIORef x (5 :: Int) >> (readIORef x :: IO Bool)
> >
> > 3. In your default-less version there's a problem of doing this:
> >
> > x = enumToIORef 0 True
> > y = enumToIORef 0 False
> >
> > Now, x and y presumably refer to the same IORef, but the initial
> > contents of that IORef will depend on whether you access it through x or
> > y.
> >
> > 4. Finally, the implementation of enumToIORef can only inspect it by
> > converting to Int, and cannot distinguish different enums that
> > correspond to the same number.
> >
> > This last problem can be solved by adding a Typeable constraint,
> > because TypeReps carry the information about where (module, package) the
> > type was defined.
> >
> > Roman
> >
> > * EatsKittens <temporalabstraction at gmail.com> [2013-12-15 12:39:20+0100]
> > > A pure function (enumToIORef :: (Enum a, Defaultable b) => a -> IORef b).
> > > This function returns referentially transparently an IORef as a function
> > of
> > > its "seed" with the guarantee that the IORef returned is identical if and
> > > only if the seed is. This function can be used to implement top level
> > > mutable state in a module. The module can specifically create an
> > enumerated
> > > type for this and not export it, thereby removing any possibility of
> > > another module passing the seed type and conflicting.
> > >
> > > The Defaultable class is added in this case to implement only a single
> > > method (defaultValue :: Defaultable a => a -> a). This is conceptionally
> > > the 'simplest' value of the type, such as the empty list, the number 0,
> > > False, the null character &c. The IORef returned by enumToIORef would be
> > > initialized before being written to to this specific default value of its
> > > type. This approach is chosen because it is impossible to initialize it
> > to
> > > user specified value because enumToIORef can be called twice with the
> > same
> > > seed but a different initial value.
> > >
> > > In the alternative it is also possible to do without the default value
> > and
> > > say the IORef returned is the same if and only if the seed and the
> > initial
> > > value given are the same. Allowing the function to remain referentially
> > > transparent as well. This would probably require for good semantics the
> > > underlying type of the IORef to be a member of EQ...?
> > >
> > > All this would of course require that newIORef and enumToIORef never
> > > produce the same IORef.
> > >
> > > Aside its limitations of the type IORef's initialized with this method
> > can
> > > carry, I do believe they cover the vast majority of use cases of top
> > level
> > > mutable state?
> > >
> > > Caveats?
> >
> > > _______________________________________________
> > > Haskell-Cafe mailing list
> > > Haskell-Cafe at haskell.org
> > > http://www.haskell.org/mailman/listinfo/haskell-cafe
> >
> >
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