On Thu, Aug 22, 2013 at 09:08:48PM +0200, Oleg Nesterov wrote:
On 08/20, Richard Guy Briggs wrote:
>
> static inline int is_global_init(struct task_struct *tsk)
> {
> - return tsk->pid == 1;
> + return task_pid_nr(tsk) == 1;
> }
Probably it would be better to simply kill it. Almost every usage is
wrong.
Can you be more clear? I don't follow. It should instead return a
boolean. Usage of is_global_init() or task_pid_nr()?
If is_global_init(), is that because they could be unaware of pid
namespaces?
If task_pid_nr(), is that for the same reason?
> static inline bool is_idle_task(const struct task_struct *p)
> {
> - return p->pid == 0;
> + return task_pid(p) == &init_struct_pid;
> }
hmm. there should be a simpler check for this...
Other than the original, this one is pretty simple. What did you have
in mind?
> static inline int has_group_leader_pid(struct task_struct *p)
> {
> - return p->pid == p->tgid;
> + return task_pid(p) == task_tgid(p);
> }
>
> static inline
> int same_thread_group(struct task_struct *p1, struct task_struct *p2)
> {
> - return p1->tgid == p2->tgid;
> + return task_tgid(p1) == task_tgid(p2);
This is suboptinal. See the attached
include-linux-schedh-dont-use-task-pid-tgid-in-same_thread_group-has_group_leader_pid.patch
from -mm below.
I'm fine with that if it is deemed better. The point was to remove the
dependence on task_struct::tgid.
Oleg.
- RGB
--
Richard Guy Briggs <rbriggs(a)redhat.com>
Senior Software Engineer
Kernel Security
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