Re: Possible bug from kernel 2.6.22 and above, 2.6.24-rc4

[Ingo Molnar <mingo@elte.hu>]

* Jie Chen <chen@jlab.org> wrote:

>> and then you use this in the measurement loop:
>>
>>    for (k=0; k<=OUTERREPS; k++){
>>      start  = getclock();
>>      for (j=0; j<innerreps; j++){
>>  #ifdef _QMT_PUBLIC
>>        delay((void *)0, 0);
>>  #else
>>        delay(0, 0, 0, (void *)0);
>>  #endif
>>      }
>>      times[k] = (getclock() - start) * 1.0e6 / (double) innerreps;
>>    }
>>
>> the problem is, this does not take the overhead of gettimeofday into 
>> account - which overhead can easily reach 10 usecs (the observed 
>> regression). Could you try to eliminate the gettimeofday overhead from 
>> your measurement?
>>
>> gettimeofday overhead is something that might have changed from .21 to .22 
>> on your box.
>>
>> 	Ingo
>
> Hi, Ingo:
>
> In my pthread_sync code, I first call refer () subroutine which 
> actually establishes the elapsed time (reference time) for 
> non-synchronized delay() using the gettimeofday. Then each 
> synchronization overhead value is obtained by subtracting the 
> reference time from the elapsed time with introduced synchronization. 
> The effect of gettimeofday() should be minimal if the time difference 
> (overhead value) is the interest here. Unless the gettimeofday behaves 
> differently in the case of running 8 threads .vs. running 2 threads.
>
> I will try to replace gettimeofday with a lightweight timer call in my 
> test code. Thank you very much.

gettimeofday overhead is around 10 usecs here:

 2740  1197359374.873214 gettimeofday({1197359374, 873225}, NULL) = 0 <0.000010>
 2740  1197359374.970592 gettimeofday({1197359374, 970608}, NULL) = 0 <0.000010>

and that's the only thing that is going on when computing the reference 
time - and i see a similar syscall pattern in the PARALLEL and BARRIER 
calculations as well (with no real scheduling going on).

	Ingo