Dan Gibson wrote:
> >From man psrset:
> The default processor set (0) always exists and may not be destroyed.
> All processes and processors at system init time start out in the
> system default processor set. For this reason processor 0 may never be
> removed from the default group. (Hence this feature is of no value on
> a single processor system.)
>
> Hence, in order to have two bound processes, you need at least three
> processors. What you have probably done is inadvertently bound both
> processes to CPU 1.
What I do is launch the benchmarks and then run a C program which
basically does:
processor_bind(P_PID, pid_of_bench1, 0, NULL);
processor_bind(P_PID, pid_of_bench2, 1, NULL);
I think this way you avoid the processor sets. I did this a while ago
and can't remember the details, but I think this works and effectively
binds the pids to processors 0 and 1. Is it wrong?
Regards,
Javi