On Saturday, 12 May 2007 01:29, Linus Torvalds wrote:
>
> On Sat, 12 May 2007, Rafael J. Wysocki wrote:
> >
> > We use this function (ie. kernel/power/process.c:is_user_space()) to
> > distinguish kernel threads from user space processes. Therefore we make it
> > always return true for user space processes and always return false for kernel
> > threads. In the latter case we need to use the task_lock() to ensure that the
> > result is as desired (ie. false), because otherwise it might be racing with
> > either fs/aio.c:use_mm() or fs/aio.c:unuse_mm().
>
> But there is no race protection in the *caller*, so if it can ever return
> one or the other, what protects it from changing once the caller returns?
>
> And if the value can change (because some thread uses "use_mm()"), then
> the caller cannot rely on the value that got returned.
The value cannot change because of that. There only is a small window inside
unuse_mm() (or use_mm()) in which the value may be wrong. Namely:
static void unuse_mm(struct mm_struct *mm)
{
struct task_struct *tsk = current;
task_lock(tsk);
tsk->flags &= ~PF_BORROWED_MM;
---
--- If is_user_space() without the task_lock() is called right here, it will
--- return 'true', although it should return 'false'.
---
tsk->mm = NULL;
/* active_mm is still 'mm' */
enter_lazy_tlb(mm, tsk);
task_unlock(tsk);
}
IOW, quoting Andrew, "is_user_space() requires that the state of p->mm and
p->flags be consistent".
> So you migt as well not return any value at all, since the returned value
> is apparently meaningless once the lock has been released.
No, it is not meaningless.
Rafael
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