(StrictT op) >>= f = StrictT (op >>= \ x -> x `seq` runStrictT (f x))
Are you sure? Here you evaluate the result, and not the computation itself. Wouldn't it be: (StrictT op) >>= f = op ` seq` StrictT (op >>= \x -> runStrictT (f x)) ?? 2012/1/21 David Barbour <dmbarbour@gmail.com>
On Sat, Jan 21, 2012 at 10:08 AM, Roman Cheplyaka <roma@ro-che.info>wrote:
* David Barbour <dmbarbour@gmail.com> [2012-01-21 10:01:00-0800]
As noted, IO is not strict in the value x, only in the operation that generates x. However, should you desire strictness in a generic way, it would be trivial to model a transformer monad to provide it.
Again, that wouldn't be a monad transformer, strictly speaking, because "monads" it produces violate the left identity law.
It meets the left identity law in the same sense as the Eval monad from Control.Strategies.
http://hackage.haskell.org/packages/archive/parallel/3.1.0.1/doc/html/src/Co...
That is, so long as values at each step can be evaluated to WHNF, it remains true that `return x >>= f` = f x.
I did mess up the def of >>=. I think it should be: (StrictT op) >>= f = StrictT (op >>= \ x -> x `seq` runStrictT (f x))
But I'm not interested enough to actually pull out an interpreter and test...
Regards,
Dave
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