[Haskell-cafe] RFC: Top level mutable state in terms of pure-function initialization?

EatsKittens temporalabstraction at gmail.com
Mon Dec 16 04:30:31 UTC 2013

I don't think you understood me, maybe I wasn't clear enough but what I 
meant was that the ref was identical if and only if the enum seed is 
identical. (enumToIORef 5, enumToIORef 5) consists of two identical 
references. The enumToIORef function is completely reverentially 
transparent and its order of execution should not matter. (enumToIORef 5) 
:: Defaultable a => IORef a should always return the same ref. It will 
return a different ref however for each type a. In this sense, the 
hypothetical program:

x = enumToIORef 0;

y = enumToIORef 0;

main = do {
  writeIORef x "Hello, World"; -- writes to ref x
  message <- readIORef y; -- reads from ref y, which is identical
  putStrLn message;

Should output "Hello, World\n"

However, as noted by Roman, a subtle bug occurs in this case:

x = enumToIORef 0;

y = enumToIORef 0;

main = do {
  writeIORef x "Hello, World"; -- writes to ref x
  message <- readIORef y; -- reads from ref y, which is NOT identical
  print (message + 1);

In this case the compiler deduces (with monomorphic restriction) that (y :: 
IORef Int) while (x :: IORef String), as such since enumToIORef is asked to 
return two different types each time it returns two different references. 
message is in this case read from the default value of 0 and "1\n" is 

On Sunday, 15 December 2013 14:00:17 UTC+1, Joe Quinn wrote:
>  This has the same issue as just using a top-level (IORef b). Assume two 
> use cases:
> let x = enumToIORef 5 in (x, x)
> (enumToIORef 5, enumToIORef 5)
> It's "obvious" in the first case that you have one ref that is used twice, 
> while it is "obvious" in the second case that you have two refs containing 
> the same value. But this breaks the rule that (let x = y in f x) = (f y), 
> because the left is a single ref while the right is two refs.
> There's likely other problems with it too, but this is what stands out to 
> me.
> On 12/15/2013 6:39 AM, EatsKittens wrote:
>  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 listHaskel... at haskell.org <javascript:>http://www.haskell.org/mailman/listinfo/haskell-cafe
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