sched: Add debug check to task_of()
A frequent mistake appears to be to call task_of() on a scheduler entity that is not actually a task, which can result in a wild pointer. Add a check to catch these mistakes. Suggested-by: Ingo Molnar <mingo@elte.hu> Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl> LKML-Reference: <new-submission> Signed-off-by: Ingo Molnar <mingo@elte.hu>
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Ingo Molnar
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00aec93d10
commit
8f48894fcc
@@ -3,15 +3,18 @@
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* policies)
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*/
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static inline struct task_struct *rt_task_of(struct sched_rt_entity *rt_se)
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{
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return container_of(rt_se, struct task_struct, rt);
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}
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#ifdef CONFIG_RT_GROUP_SCHED
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#define rt_entity_is_task(rt_se) (!(rt_se)->my_q)
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static inline struct task_struct *rt_task_of(struct sched_rt_entity *rt_se)
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{
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#ifdef CONFIG_SCHED_DEBUG
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WARN_ON_ONCE(!rt_entity_is_task(rt_se));
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#endif
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return container_of(rt_se, struct task_struct, rt);
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}
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static inline struct rq *rq_of_rt_rq(struct rt_rq *rt_rq)
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{
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return rt_rq->rq;
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@@ -26,6 +29,11 @@ static inline struct rt_rq *rt_rq_of_se(struct sched_rt_entity *rt_se)
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#define rt_entity_is_task(rt_se) (1)
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static inline struct task_struct *rt_task_of(struct sched_rt_entity *rt_se)
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{
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return container_of(rt_se, struct task_struct, rt);
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}
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static inline struct rq *rq_of_rt_rq(struct rt_rq *rt_rq)
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{
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return container_of(rt_rq, struct rq, rt);
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