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.
Stefan
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