Re: [RFC][-mm] Memory controller add mm->owner

[Paul Menage]

On Wed, Mar 26, 2008 at 3:29 AM, Balbir Singh <balbir@xxxxxxxxxxxxxxxxxx> wrote:
>  >>
>  >> - in the worst case, it's not going to be worse than doing a
>  >> for_each_thread() loop
>  >>
>
>  This will have to be the common case, since you never know what combination of
>  clone calls did CLONE_VM and what did CLONE_THREAD. At exit time, we need to pay
>  a for_each_process() overhead.

I'm not convinced of this. All we have to do is find some other
process p where p->mm == current->mm and make it the new owner.
Exactly what sequence of clone() calls was used to cause the sharing
isn't really relevant. I really think that a suitable candidate will
be found amongst your children or your first sibling in 99.9% of those
cases where more than one process is using an mm.

The actual sequence would have to go something like:

static inline bool need_new_owner(struct mm_struct *mm) {
  return (mm && mm->owner == current && atomic_read(&mm->users) > 1);
}
static inline void try_give_mm_ownership(
    struct task_struct *task,
    struct mm_struct *mm) {
  if (task->mm != mm) return;
  task_lock(task);
  if (task->mm == mm) {
    mm->owner = task;
  }
  task_unlock(task);
}

struct mm_struct *mm = current->mm;
task_lock(current);
current->mm = NULL;
task_unlock(current);

/* First try my children */
if (need_new_owner(mm)) {
  for_each_child(current, c) {
    try_give_mm_ownership(c);
    if (!need_new_owner(mm)) break;
  }
}

/* Then try my siblings */
if (need_new_owner(mm)) {
  for_each_child(current->real_parent, c) {
    try_give_mm_ownership(c);
    if (!need_new_owner(mm)) break;
  }
}

if (need_new_owner(mm)) {
  /* We'll almost never get here */
  for_each_process(p) {
    try_give_mm_ownership(p);
    if (!need_new_owner(mm)) break;
  }
}

Paul
--
To unsubscribe from this list: send the line "unsubscribe linux-kernel" in
the body of a message to majordomo@xxxxxxxxxxxxxxx
More majordomo info at  http://vger.kernel.org/majordomo-info.html
Please read the FAQ at  http://www.tux.org/lkml/