All,
Following definition of notifier_chain_register is from kernel/sys.c
and for kernel 2.6.20.1 .
static int notifier_chain_register(struct notifier_block **nl,
struct notifier_block *n)
{
while ((*nl) != NULL) {
if (n->priority > (*nl)->priority)
break;
nl = &((*nl)->next);
}
n->next = *nl;
rcu_assign_pointer(*nl, n);
return 0;
}
Since the first parameter is passed as a double ptr, the pointer
itself is changed by the traversal and by rcu_assign_pointer(*nl, n) ,
right? And the new value will be address of notifier_block n. This
means the new head of the notifier list is n with the highest
priority( in terms of magnitude atleast ).
Nopes. If you observe closely, *nl (head) is never touched. It is the nl that is changed.
Say,
head = 1000
nl = 5000 (&head)
So, first nl = 5000. Then nl = &(head->next) that will be 1004 (address of 'next' element of the structure)
and so on.
So, 'head' is never touched.
Regards,
- Ratnadeep
My query is ,why are we loosing the notifiers during traversal in the
first while loop, which have a priority > n->priority ?
I am terribly confused here.
Can somebody help me out here?
Thank you
~psr
--
play the game
