sched/core: Remove pointless printout in sched_show_task()

In sched_show_task() we print out a useless hex number, not even a
symbol, and there's a big question mark whether this even makes sense
anyway, I suspect we should just remove it all.

Signed-off-by: Linus Torvalds <[email protected]>
Acked-by: Andy Lutomirski <[email protected]>
Cc: Peter Zijlstra <[email protected]>
Cc: Tetsuo Handa <[email protected]>
Cc: Thomas Gleixner <[email protected]>
Cc: [email protected]
Cc: [email protected]
Cc: [email protected]
Cc: [email protected]
Cc: [email protected]
Cc: [email protected]
Link: http://lkml.kernel.org/r/CA+55aFzphURPFzAvU4z6Moy7ZmimcwPuUdYU8bj9z0J+S8X1rw@mail.gmail.com
Signed-off-by: Ingo Molnar <[email protected]>

Author: torvalds

kernel/sched/core.c

@@ -5198,17 +5198,8 @@ void sched_show_task(struct task_struct *p)
 		state = __ffs(state) + 1;
 	printk(KERN_INFO "%-15.15s %c", p->comm,
 		state < sizeof(stat_nam) - 1 ? stat_nam[state] : '?');
-#if BITS_PER_LONG == 32
-	if (state == TASK_RUNNING)
-		printk(KERN_CONT " running  ");
-	else
-		printk(KERN_CONT " %08lx ", thread_saved_pc(p));
-#else
 	if (state == TASK_RUNNING)
 		printk(KERN_CONT "  running task    ");
-	else
-		printk(KERN_CONT " %016lx ", thread_saved_pc(p));
-#endif
 #ifdef CONFIG_DEBUG_STACK_USAGE
 	free = stack_not_used(p);
 #endif