Re: [Haskell-cafe] Monads, do and strictness

(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

On Sat, Jan 21, 2012 at 10:08 AM, Roman Cheplyaka wrote:

...

* David Barbour [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

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe