On Fri, Jul 13, 2007 at 04:29:12PM -0700, Michael Vanier wrote: > Albert, > > Thanks for the very detailed reply! That's the great thing about this > mailing list. > > I find your description of seq somewhat disturbing. Is this behavior > documented in the API? I can't find it there. It suggests that perhaps > there should be a really-truly-absolutely-I-mean-right-now-seq function > that evaluates the first argument strictly no matter what (not that this > should be something that gets used very frequently). Or are there reasons > why this is not feasible? > > Sorry to belabor this. Learning to think lazily is IMO one of the hardest > aspects of learning Haskell.
Can you clarify what you mean by really-truly-absolutely-I-mean-right-now-seq? The entire specification of seq is in <http://haskell.org/onlinereport/basic.html#sect6.2>: The function seq is defined by the equations: seq ⊥ b = ⊥ seq a b = b, if a ≠ ⊥ In particular, seq is not defined in terms of evaluation. Even in a lazy model, you cannot assume any *order* of evaluation, only that both arguments will be demanded if the whole expression is. *if* the whole expression is. Seq is uniform - for ALL data types on the lhs, it evaluates it to WHNF (seeing the top constructor or lambda). A recursive seq would not be uniform, and would require a type class or just specialization. If you can see why (x `seq` x) is redundant, you probably understand laziness. Perhaps it would help to see a definition of seq in Haskell? class Eval a where seq :: a -> b -> b instance Eval (a,b) where seq (x,y) b = b instance Eval [a] where seq [] b = b seq (x:xs) b = b instance Eval (Maybe a) where seq Nothing b = b seq (Just x) b = b instance Eval (Either a b) where seq (Left x) b = b seq (Right x) b = b instance Eval Bool where seq True b = b seq False b = b instance Eval Int where ... seq (-1) b = b seq 0 b = b seq 1 b = b ... ? Stefan
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