[luau] getting system time in milisecs

[Lockhart, Charles]

> Do you really need to check it per iteration?  I mean, in your case, 
> could you move the timing outside
> the loop and divide the result by the number of iterations?  
> If you do 
> that, then you will get an average
> iteration which will minimize the time spent timing.

I need to be able to perform each cycle within a very small period of time
on average, and guarantee that the period of a single cycle doesn't exceed a
certain threshold.  

> Not quite sure what you mean by "debug".  You mean the timing code is 
> skewing your results?

The timing code isn't necesary for the operation, just in order to
characterize the operation.  The extra overhead impacts the overall period.

> Here's an idea:  say get_time() is your timing function (be it from 
> gettimeofday() or clock() or whatever),
> before you start your loop do a little profiling of the 
> timing function, 
> something like
> 
> before = get_time();
> for (i = 0, total = 0; i < 10000; i++) {
> get_time();
> }
> timing_cost = (get_time() - before) / 10000
> 
> Then you have a rough idea of what get_time() is going to take during 
> your loop, so you can
> subtract it out from the results.

Jeez, I hadn't thought of that.  Thanks, I'll try that tomorrow.

> On a side note, there is a syscall called times() which fills a 
> structure that contains clock ticks since the
> start of the  program.  It has two fields. One of the fields is the 
> clock ticks spent in kernel-space and the other is
> the clock ticks spent in userspace.  If you might be able to 
> use it to 
> isolate the times from parts of you program,
> too.

Cool, thanks Ray.

-Charles