Re: Use of yield() in the kernel

[Mark Mielke]

Would it be sensible to add a "yield_short()" function to the kernel?

mark

On Sun, Oct 20, 2002 at 11:10:33AM +0200, Duncan Sands wrote:
> > > Hi Pavel, I agree.  I have some questions about the code though:
> > > when you come across a thread with (p->flags & PF_FROZEN), why
> > > break out of the loop?  Why not just skip this thread and go on to
> > > the
> >
> > There's "continue;" in there, and it should "just skip this thread".
> > 									Pavel
> 
> I meant, why not just do as in the following code (I've changed
> INTERESTING also, for same reason, as explained below):
> 
> 	do {
> 		todo = 0;
> 		read_lock(&tasklist_lock);
> 		do_each_thread(g, p) {
> 			unsigned long flags;
> 
>                         if (
> 				!(p->flags & PF_IOTHREAD) &&
> 				(p != current) &&
> 				(p->state != TASK_ZOMBIE) &&
> 				!(p->flags & PF_FROZEN)
> 			) {
> 
> 				/* FIXME: smp problem here: we may not access other process' flags
> 				   without locking */
> 				p->flags |= PF_FREEZE;
> 				spin_lock_irqsave(&p->sig->siglock, flags);
> 				signal_wake_up(p);
> 				spin_unlock_irqrestore(&p->sig->siglock, flags);
> 				todo++;
> 			}
> 		} while_each_thread(g, p);
> 		read_unlock(&tasklist_lock);
> 		yield();
> 		if (time_after(jiffies, start_time + TIMEOUT)) {
> 			printk( "\n" );
> 			printk(KERN_ERR " stopping tasks failed (%d tasks remaining)\n", todo );
> 			return todo;
> 		}
> 	} while(todo);
> 
> The reason is that yield(), which sends the current task to the expired list,
> can take a long time before it runs again.  With the current code, every time
> you meet, for example, a kernel thread you break out of the loop, perform
> a yield (= wait a long time), before going on to the next thread.  This could
> take forever.  With code like that above, you mark as many tasks frozen as
> possible, with as few yields as possible.  Isn't that better?
> 
> Ciao,
> 
> Duncan.
> -
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