On Tue, Oct 23, 2012 at 8:25 PM, Stefan Hajnoczi <stefanha@xxxxxxxxx> wrote:
> On Fri, Sep 21, 2012 at 01:17:21PM +0200, Vasilis Liaskovitis wrote:
>> +static void dimm_populate(DimmDevice *s)
>> +{
>> + DeviceState *dev= (DeviceState*)s;
>> + MemoryRegion *new = NULL;
>> +
>> + new = g_malloc(sizeof(MemoryRegion));
>> + memory_region_init_ram(new, dev->id, s->size);
>> + vmstate_register_ram_global(new);
>> + memory_region_add_subregion(get_system_memory(), s->start, new);
>> + s->mr = new;
>> +}
>> +
>> +static void dimm_depopulate(DimmDevice *s)
>> +{
>> + assert(s);
>> + vmstate_unregister_ram(s->mr, NULL);
>> + memory_region_del_subregion(get_system_memory(), s->mr);
>> + memory_region_destroy(s->mr);
>> + s->mr = NULL;
>> +}
>
> How is dimm hot unplug protected against callers who currently have RAM
> mapped (from cpu_physical_memory_map())?
>
> Emulated devices call cpu_physical_memory_map() directly or indirectly
> through DMA emulation code. The RAM pointer may be held for arbitrary
> lengths of time, across main loop iterations, etc.
>
> It's not clear to me that it is safe to unplug a DIMM that has network
> or disk I/O buffers, for example. We also need to be robust against
> malicious guests who abuse the hotplug lifecycle. QEMU should never be
> left with dangling pointers.
>
Not sure about the block layer. But I think those thread are already
out of big lock, so there should be a MemoryListener to catch the
RAM-unplug event, and if needed, bdrv_flush.
Regards,
pingfan
> Stefan
>
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