Hi!
> >> >> --- a/kernel/exit.c
> >> >> +++ b/kernel/exit.c
> >> >> @@ -835,7 +835,7 @@ void do_exit(long code)
> >> >> /*
> >> >> * Make sure we are holding no locks:
> >> >> */
> >> >> - debug_check_no_locks_held(tsk);
> >> >> + debug_check_no_locks_held();
> >> >
> >> > Is task guaranteed == current?
> >>
> >> Yes, the first line of do_exit is:
> >> struct task_struct *tsk = current;
> >
> > Aha, I understand it now.
> >
> > Accessing current is slower than local variable. So your "new" code
> > will work but will be slower. Please revert this part.
>
> Using current instead of passing in tsk was done at Andrew Morton's
> suggestion, and makes no difference from the freezer's perspective
> since it would have to use current to get the task to pass in, so I'm
> going to leave it as is.
Well, current is:
static inline struct thread_info *current_thread_info(void)
{
register unsigned long sp asm ("sp");
return (struct thread_info *)(sp & ~(THREAD_SIZE - 1));
}
#define get_current() (current_thread_info()->task)
#define current get_current()
Instead of passing computed value to debug_check_no_locks_held(), you
force it to be computed again. do_exit() performance matters for
configure scripts, etc.
I'd say it makes sense to keep the optimalization. akpm can correct
me.
Pavel
--
(english) http://www.livejournal.com/~pavelmachek
(cesky, pictures) http://atrey.karlin.mff.cuni.cz/~pavel/picture/horses/blog.html
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