[Haskell-cafe] Re: Long running Haskell program

[Heinrich Apfelmus]

David Menendez wrote:
> I think replacing "put s" with "put $! s" should guarantee that the
> state is evaluated.
> 
> If you're using get and put in many place in the code, you could try
> something along these lines:
> 
> newtype SStateT s m a = S { unS :: StateT s m a } deriving (Monad, etc.)
> 
> instance (Monad m) => MonadState s (SStateT s m) where
>     get = S get
>     put s = S (put $! s)
> 

Interestingly, this is different from  Control.Monad.State.Strict . The
latter never forces the state itself, just the pair constructor of the
(result,state) pair.

Here the different cases:

    evalLazy   m = Control.Monad.State.Lazy.evalState m 0
    evalStrict m = Control.Monad.State.Strict.evalState m 0


    -- Pair constructor non-bottom
    GHCi> evalLazy   $ put undefined
    ()
    GHCi> evalStrict $ put undefined
    ()

    -- Pair constructor bottom
    GHCi> evalLazy   $ put $! undefined
    *** Exception: Prelude.undefined
    GHCi> evalStrict $ put $! undefined
    *** Exception: Prelude.undefined

    -- Last pair constructor non-bottom
    GHCi> evalLazy   $ (put $! undefined) >> put 1
    ()
    -- Everything bottom
    GHCi> evalStrict $ (put $! undefined) >> put 1
    *** Exception: Prelude.undefined



Regards,
apfelmus

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