* Eric W. Biederman <ebiederm@xmission.com> wrote:
yeah - by all means get rid of it, but first separate the data
structures along uses. Then all the 'why should we iterate two lists in
sequence' questions vanish.
this can be zapped today. The patch below does it - the scheduler use
was purely historic. oldest_child/older_sibling used to have a role but
it has none today.
hm, this use is pretty valid although not user-detectable.
zappable too.
i think you are right.
Ingo
Subject: [patch] sched: get rid of p->children use in show_task()
From: Ingo Molnar <mingo@elte.hu>
the p->parent PID printout gives us all the information about the
task tree that we need - the eldest_child()/older_sibling()/
younger_sibling() printouts are mostly historic and i do not
remember ever having used those fields. (IMO in fact they confuse
the SysRq-T output.) So remove them.
This code has sentimental value though, those fields and
printouts are one of the oldest ones still surviving from
Linux v0.95's kernel/sched.c:
if (p->p_ysptr || p->p_osptr)
printk(" Younger sib=%d, older sib=%d\n\r",
p->p_ysptr ? p->p_ysptr->pid : -1,
p->p_osptr ? p->p_osptr->pid : -1);
else
printk("\n\r");
written 15 years ago, in early 1992.
Signed-off-by: Ingo Molnar <mingo@elte.hu>
---
kernel/sched.c | 35 +----------------------------------
1 file changed, 1 insertion(+), 34 deletions(-)
Index: linux/kernel/sched.c
===================================================================
--- linux.orig/kernel/sched.c
+++ linux/kernel/sched.c
@@ -4687,27 +4687,6 @@ out_unlock:
return retval;
}
-static inline struct task_struct *eldest_child(struct task_struct *p)
-{
- if (list_empty(&p->children))
- return NULL;
- return list_entry(p->children.next,struct task_struct,sibling);
-}
-
-static inline struct task_struct *older_sibling(struct task_struct *p)
-{
- if (p->sibling.prev==&p->parent->children)
- return NULL;
- return list_entry(p->sibling.prev,struct task_struct,sibling);
-}
-
-static inline struct task_struct *younger_sibling(struct task_struct *p)
-{
- if (p->sibling.next==&p->parent->children)
- return NULL;
- return list_entry(p->sibling.next,struct task_struct,sibling);
-}
-
static const char stat_nam[] = "RSDTtZX";
static void show_task(struct task_struct *p)
@@ -4738,19 +4717,7 @@ static void show_task(struct task_struct
free = (unsigned long)n - (unsigned long)end_of_stack(p);
}
#endif
- printk("%5lu %5d %6d ", free, p->pid, p->parent->pid);
- if ((relative = eldest_child(p)))
- printk("%5d ", relative->pid);
- else
- printk(" ");
- if ((relative = younger_sibling(p)))
- printk("%7d", relative->pid);
- else
- printk(" ");
- if ((relative = older_sibling(p)))
- printk(" %5d", relative->pid);
- else
- printk(" ");
+ printk("%5lu %5d %6d", free, p->pid, p->parent->pid);
if (!p->mm)
printk(" (L-TLB)\n");
else
-