On Wed, 2007-01-03 at 21:33 +0200, Michael S. Tsirkin wrote:

quoted

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.

quoted

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.

quoted

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?

I think having a 2nd function to set the udata seems onerous.