On 10/07/2015 17:32, Frederic Konrad wrote:
>>>
>
> I think something like that can work because we don't have two
> flush_queued_work at the same time on the same CPU?
Yes, this works; there is only one consumer.
Holding locks within a callback can be very painful, especially if there
is a chance that the callback will take a very coarse lock such as big
QEMU lock. It can cause AB-BA deadlocks.
Paolo
> static void flush_queued_work(CPUState *cpu)
> {
> struct qemu_work_item *wi;
>
> if (cpu->queued_work_first == NULL) {
> return;
> }
>
> qemu_mutex_lock(&cpu->work_mutex);
> while ((wi = cpu->queued_work_first)) {
> cpu->queued_work_first = wi->next;
> qemu_mutex_unlock(&cpu->work_mutex);
> wi->func(wi->data);
> qemu_mutex_lock(&cpu->work_mutex);
> wi->done = true;
> if (wi->free) {
> g_free(wi);
> }
> }
> cpu->queued_work_last = NULL;
> qemu_mutex_unlock(&cpu->work_mutex);
>
> qemu_cond_broadcast(&qemu_work_cond);
> }
