[Haskell-cafe] Full strict functor by abusing Haskell exceptions

[Maciej Piechotka]

On Tue, 2010-09-14 at 11:27 +0100, Neil Brown wrote:
> On 13/09/10 17:25, Maciej Piechotka wrote:
> >> import Control.Exception
> >> import Foreign
> >> import Prelude hiding (catch)
> >>
> >> data StrictMonad a = StrictMonad a deriving Show
> >>
> >> instance Monad StrictMonad where
> >>      return x = unsafePerformIO $ do
> >>          (return $! x) `catch` \(SomeException _) ->  return x
> >>          return $! StrictMonad x
> >>      StrictMonad v>>= f = f v
> >>      
> > It seems to be valid IMHO Functor and Monad (I haven't prove it) as long
> > as functions terminates.
> >    
> 
> I'm not sure if I'm allowed to use unsafePerformIO in my 
> counter-example, but you used it so why not ;-)
> The first monad law says: "return a >>= k = k a"
> 
> let k = const (StrictMonad ())
>      a = unsafePerformIO launchMissiles
> 
> In "k a" no missiles will be launched, in "return a >>= k", they will be 
> launched.

I guess we enter a grey area - I did use unsafePerformIO but without
side-effects. 

> You can construct a similar example against "m >>= return = 
> m".

Assuming StrictMonad (constructor) is hidden - I don't think so.

> Although, if you changed your definition of bind to:
> 
> StrictMonad v >>= f = return v >>= f >>= return
> 
> Then as long as "return x >>= return = return x" (which it does for you) 
> then you automatically satisfy the first two monad laws!  Which is an 
> interesting way of solving the problem -- haven't checked the third law 
> though.
> 

My error.

> Thanks,
> 
> Neil.


Regards
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