perf_counter: initialize the per-cpu context earlier

percpu scheduling for perfcounters wants to take the context lock,
but that lock first needs to be initialized. Currently it is an
early_initcall() - but that is too late, the task tick runs much
sooner than that.

Call it explicitly from the scheduler init sequence instead.

[ Impact: fix access-before-init crash ]

LKML-Reference: <new-submission>
Signed-off-by: NIngo Molnar <mingo@elte.hu>

include/linux/perf_counter.h

@@ -573,6 +573,8 @@ extern struct perf_callchain_entry *perf_callchain(struct pt_regs *regs);
 
 extern int sysctl_perf_counter_priv;
 
+extern void perf_counter_init(void);
+
 #else
 static inline void
 perf_counter_task_sched_in(struct task_struct *task, int cpu)		{ }
@@ -600,9 +602,10 @@ perf_counter_mmap(unsigned long addr, unsigned long len,
 
 static inline void
 perf_counter_munmap(unsigned long addr, unsigned long len,
-		    unsigned long pgoff, struct file *file) 		{ }
+		    unsigned long pgoff, struct file *file)		{ }
 
 static inline void perf_counter_comm(struct task_struct *tsk)		{ }
+static inline void perf_counter_init(void)				{ }
 #endif
 
 #endif /* __KERNEL__ */

kernel/perf_counter.c

@@ -3265,15 +3265,12 @@ static struct notifier_block __cpuinitdata perf_cpu_nb = {
 	.notifier_call		= perf_cpu_notify,
 };
 
-static int __init perf_counter_init(void)
+void __init perf_counter_init(void)
 {
 	perf_cpu_notify(&perf_cpu_nb, (unsigned long)CPU_UP_PREPARE,
 			(void *)(long)smp_processor_id());
 	register_cpu_notifier(&perf_cpu_nb);
-
-	return 0;
 }
-early_initcall(perf_counter_init);
 
 static ssize_t perf_show_reserve_percpu(struct sysdev_class *class, char *buf)
 {

kernel/sched.c

@@ -39,6 +39,7 @@
 #include <linux/completion.h>
 #include <linux/kernel_stat.h>
 #include <linux/debug_locks.h>
+#include <linux/perf_counter.h>
 #include <linux/security.h>
 #include <linux/notifier.h>
 #include <linux/profile.h>
@@ -8996,7 +8997,7 @@ void __init sched_init(void)
 		 * 1024) and two child groups A0 and A1 (of weight 1024 each),
 		 * then A0's share of the cpu resource is:
 		 *
-		 * 	A0's bandwidth = 1024 / (10*1024 + 1024 + 1024) = 8.33%
+		 *	A0's bandwidth = 1024 / (10*1024 + 1024 + 1024) = 8.33%
 		 *
 		 * We achieve this by letting init_task_group's tasks sit
 		 * directly in rq->cfs (i.e init_task_group->se[] = NULL).
@@ -9097,6 +9098,8 @@ void __init sched_init(void)
 	alloc_bootmem_cpumask_var(&cpu_isolated_map);
 #endif /* SMP */
 
+	perf_counter_init();
+
 	scheduler_running = 1;
 }