> (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 <dmbarb...@gmail.com> > On Sat, Jan 21, 2012 at 10:08 AM, Roman Cheplyaka <r...@ro-che.info>wrote: > >> * David Barbour <dmbarb...@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/Control-Parallel-Strategies.html#Eval > > 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 > >
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