[Haskell-cafe] I don't understand how ST works

[Silvio Frischknecht]

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> Now comes my question: in the impure values there is always that
> "s". I was thinking that the whole structure should have s as a
> parameter:

Yes

> 
>> data MList s = MList { mlVec :: MVector s Move, mlNextPh :: MList
>> ->
> ST s (Maybe (MList s)) }

you probably meant:

    data MList s = MList { ... , mlNextPh :: Mlist s -> ... }

Now I'm not sure about your exact problem since the following compiles
for me.

    import Data.Vector
    import Data.Vector.Mutable
    import Control.Monad.ST

    type Move = ()
    data MList s = MList {
        mvVec :: MVector s Move,
        mlNextPh :: MList s -> ST s (Maybe (MList s)) }

    splitMove :: MList s -> ST s (Maybe (Move, MList s))
    splitMove ml = do
        m <- unsafeRead (mvVec ml) 0
        undefined

Something you always have to watch out for when dealing with ST is not
to return something that depends on s in the last statement (the one
you use runST on). In other words, if you want to return a vector you
have to freeze it, so it's not mutable anymore.

If you still can't figure it out paste some complete example that
doesn't work.

silvio
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