Hello, Peter. On Thu, Sep 04, 2025 at 10:28:58PM +0200, Peter Zijlstra wrote: ... > RUNNABLE: > 1) hold both source and target rq->lock. ... > Now, assuming you have a locking order like: > > p->pi_lock > rq->lock > dsq->lock > > When you do something like: > > __schedule() > raw_spin_lock(rq->lock); > next = pick_next_task() -> pick_task_scx() > raw_spin_lock(dsq->lock); > > Then you are, in effect, in the RUNNABLE 1) case above. You hold both > locks. Nothing is going to move your task around while you hold that > dsq->lock. That task is on the dsq, anybody else wanting to also do > anything with that task, will have to first take dsq->lock. > > Therefore, at this point, it is perfectly fine to do: > > set_task_cpu(cpu_of(rq)); // move task here > > There is no actual concurrency. The only thing there is is > set_task_cpu() complaining you're not following the rules -- but you > are, it just doesn't know -- and we can fix that. I can't convince myself this is safe. For example, when task_rq_lock() returns, it should guarantee that the rq that the task is currently associated with is locked and the task can't go anywhere. However, as task_rq_lock() isn't interlocked with dsq lock, this won't hold true. I think this will break multiple things subtly - e.g. the assumptions that task_call_func() makes in the comment wouldn't hold anymore, task_sched_runtime()'s test of task_on_rq_queued() would be racy, and so on. ie. Operations protected by deq/enq pair would be fine but anything which is protected only by task_rq_lock/unlock() would become racy, right? Thanks. -- tejun
