提交 0d905bca 编写于 作者: I Ingo Molnar

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>
上级 ba77813a
......@@ -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__ */
......
......@@ -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)
{
......
......@@ -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;
}
......
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