> Without extra param (1000 iterations in cycles):
> ave 101.283 min 91 max 247
> With extra param (1000 iterations in cycles):
> ave 103.311 min 91 max 221
A 2% hit then. Not huge, but 0 either.
> Convert cycles to ns (3466.727 MHz CPU):
>
> Without: 101.283 / 3466.727 = .02922us == 29.22ns
> With: 103.311 / 3466.727 = .02980us == 29.80ns
>
> So I measure a .58ns average increase for passing in the additional
> parameter.
That depends on CPU speed though. Percentage is likely to be more universal.
> Here is a snipit of the test:
>
> spin_lock_irq(&lock);
> do_gettimeofday(&start_tv);
> for (i=0; i<1000; i++) {
> cycles_start[i] = get_cycles();
> ib_req_notify_cq(cb->cq, IB_CQ_NEXT_COMP);
> cycles_stop[i] = get_cycles();
> }
> do_gettimeofday(&stop_tv);
> spin_unlock_irq(&lock);
>
> if (stop_tv.tv_usec < start_tv.tv_usec) {
> stop_tv.tv_usec += 1000000;
> stop_tv.tv_sec -= 1;
> }
>
> for (i=0; i < 1000; i++) {
> cycles_t v = cycles_stop[i] - cycles_start[i];
> sum += v;
> if (v > max)
> max = v;
> if (min == 0 || v < min)
> min = v;
> }
>
> printk(KERN_ERR PFX "FOO delta sec %lu usec %lu sum %llu min %llu max
> %llu\n",
> stop_tv.tv_sec - start_tv.tv_sec,
> stop_tv.tv_usec - start_tv.tv_usec,
> (unsigned long long)sum, (unsigned long long)min,
> (unsigned long long)max);
Good job, the test looks good, thanks.
So what does this tell you?
To me it looks like there's a measurable speed difference,
and so we should find a way (e.g. what I proposed) to enable chelsio userspace
without adding overhead to other low level drivers or indeed chelsio kernel
level code.
What do you think? Roland?
--
MST
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