Chad,

I was thinking about the dynamic behavior of par, and there's something that's a little confusing to me. Am I right in understanding that (x `par` y) evaluates to y, and may or may not evaluate x along the way?

The reason that par doesn't necessarily evaluate it's 1st argument is that typical parallel Haskell programs contain vast amounts of potential parallelism. So rather than create a relatively heavyweight thread and be forced to administer it, e.g. schedule it for every possible expression that could be evaluated in parallel, the expression is 'sparked', i.e. a lightweight action that simply notes that the expression *could* be evaluated in parallel.

I think it would be easier to reason about if we knew that x would be evaluated, either by a newly-spawned thread or by the parent thread.

Yes, although we usually reason by saying trhat the 'potential parallelism' in two expressions is the same, i.e. making the assumption that all sparked expressions become threads.

If I'm understanding this correctly, this could be done by defining

x `pSeq` y = x `par` y `seq` x `seq` y

This seems plausible. All the best, Phil

From: haskell-bounces@haskell.org [mailto:haskell-bounces@haskell.org] On Behalf Of Scherrer, Chad Sent: 25 April 2006 21:35 To: haskell@haskell.org Subject: [Haskell] parallel seq

I was thinking about the dynamic behavior of par, and there's something that's a little confusing to me. Am I right in understanding that (x `par` y) evaluates to y, and may or may not evaluate x along the way?

I think it would be easier to reason about if we knew that x would be evaluated, either by a newly-spawned thread or by the parent thread. If I'm understanding this correctly, this could be done by defining

x `pSeq` y = x `par` y `seq` x `seq` y (seq has higher precedence than par)

When (x `pSeq` y) is evaluated, x may or may not be evaluated by a new thread. The parent thread continues along evaluating y. Then the parent makes sure x was evaluated before finally returning y.

I've not seen this approach used before -- Is there something I'm missing that makes using pSeq as above just silly?

Thanks!

Chad Scherrer Computational Mathematics Group Pacific Northwest National Laboratory

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-------------------------------------------------- Phil Trinder School of Mathematical and Computer Sciences Heriot-Watt University Edinburgh, EH14 4AS E-mail: trinder@macs.hw.ac.uk Teleph: +44 (0)131 451 3435 Depart: +44 (0)131 451 3328 Fasmly: +44 (0)131 451 3327 Intrnt: http://www.macs.hw.ac.uk/~trinder