Re: sys_sched_yield fast path

[Davide Libenzi]

On 10-Mar-2001 Andi Kleen wrote:
> Davide Libenzi <[e-mail]> writes:
> 
> 
>> Probably the rate at which is called sys_sched_yield() is not so high to let
>> the performance improvement to be measurable.
> 
> LinuxThreads mutexes call sched_yield() when a lock is locked, so when you 
> have a  multithreaded program with some lock contention it'll be called a
> lot.

This is the linux thread spinlock acquire :

static void __pthread_acquire(int * spinlock)
{
  int cnt = 0;
  struct timespec tm;

  while (testandset(spinlock)) {
    if (cnt < MAX_SPIN_COUNT) {
      sched_yield();
      cnt++;
    } else {
      tm.tv_sec = 0;
      tm.tv_nsec = SPIN_SLEEP_DURATION;
      nanosleep(&tm, NULL);
      cnt = 0;
    }
  }
}

Yes, it calls sched_yield() but this is not a std wait for mutex but for
spinlocks that are hold a very short time.
Real wait are implemented using signals.
More, with the new implementation of sys_sched_yield() the task release all its
time quantum so, even in a case where a task repeatedly calls sched_yield() the
call rate is not so high if there is at least one process to spin.
And if there isn't one task with goodness() > 0, nobody cares about
sched_yield() performance.

- Davide