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) { }
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__ */
.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)
{
#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>
* 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).
alloc_bootmem_cpumask_var(&cpu_isolated_map);
#endif /* SMP */
+ perf_counter_init();
+
scheduler_running = 1;
}