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

8 Jun 2012

      Hi,

I created a gist with a minimal (still 111 lines) module:

https://gist.github.com/2898128

I still get the errors:

WhatsWrong.hs:53:5:
     Couldn't match type `s' with `PrimState (ST s)'
       `s' is a rigid type variable bound by
           a type expected by the context: ST s [Move] at 
WhatsWrong.hs:48:21
     In a stmt of a 'do' block: listMoves ml
     In the second argument of `($)', namely
       `do { v <- U.new maxMovesPerPos;
             let ml = ...;
             listMoves ml }'
     In the expression:
       runST
       $ do { v <- U.new maxMovesPerPos;
              let ml = ...;
              listMoves ml }

WhatsWrong.hs:65:44:
     Couldn't match type `s' with `PrimState (ST s)'
       `s' is a rigid type variable bound by
           the type signature for nextPhaseOnlyCapts :: GenPhase s
           at WhatsWrong.hs:64:1
     Expected type: U.MVector (PrimState (ST s)) Move
       Actual type: U.MVector s Move
     In the return type of a call of `mlVec'
     In the third argument of `genCapts', namely `(mlVec ml)'

Thanks,
Nicu

Am 08.06.2012 02:47, schrieb 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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Re: [Haskell-cafe] I don't understand how ST works

Nicu Ionita