On Wed, Aug 12, 2009 at 7:00 AM, John Harrop<jharrop...@gmail.com> wrote:
> A very straightforward version, and 875.36796ms/100000000 = 8.7536796ns. > This is on a 2.5GHz machine, so that's only about 22 native instructions per > iteration. The theoretical minimum is over 1/2 of that: > Four fmuls > Three fadds and an fsub > A floating point comparison > An integer add and an integer comparison > A branch back to the start of the loop > So we're within a factor of 2 of *native* code speeds (nevermind Java) here. It's not this straightforward. Superscalar CPUs handle multiple instructions concurrently. I can't give you concrete numbers on the paralellism of current processors, but had a lot of fun trying to saturate both pipelines on my first Pentium, writing i386 assembly ages ago. Isak --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en -~----------~----~----~----~------~----~------~--~---
