[Haskell-cafe] problem with IO, strictness, and "let"

Stefan O'Rear stefanor at cox.net
Fri Jul 13 19:59:53 EDT 2007

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

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


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