On Tue, Dec 15, 2020 at 03:19:47PM +0800, Keqian Zhu wrote: > When handle dirty log, we face qemu_real_host_page_size and > TARGET_PAGE_SIZE. The first one is the granule of KVM dirty > bitmap, and the second one is the granule of QEMU dirty bitmap. > > Generally speaking, qemu_real_host_page_size >= TARGET_PAGE_SIZE, > so misuse TARGET_PAGE_SIZE to init kvmslot dirty_bmap may waste > memory. For example, when qemu_real_host_page_size is 64K and > TARGET_PAGE_SIZE is 4K, this bugfix can save 93.75% memory. > > Signed-off-by: Keqian Zhu <zhukeqi...@huawei.com> > --- > accel/kvm/kvm-all.c | 6 +++++- > 1 file changed, 5 insertions(+), 1 deletion(-) > > diff --git a/accel/kvm/kvm-all.c b/accel/kvm/kvm-all.c > index baaa54249d..c5e06288eb 100644 > --- a/accel/kvm/kvm-all.c > +++ b/accel/kvm/kvm-all.c > @@ -620,8 +620,12 @@ static void kvm_memslot_init_dirty_bitmap(KVMSlot *mem) > * too, in most cases). > * So for now, let's align to 64 instead of HOST_LONG_BITS here, in > * a hope that sizeof(long) won't become >8 any time soon. > + * > + * Note: the granule of kvm dirty log is qemu_real_host_page_size. > + * And mem->memory_size is aligned to it (otherwise this mem can't > + * be registered to KVM). > */ > - hwaddr bitmap_size = ALIGN(((mem->memory_size) >> TARGET_PAGE_BITS), > + hwaddr bitmap_size = ALIGN(mem->memory_size / qemu_real_host_page_size, > /*HOST_LONG_BITS*/ 64) / 8;
Yes I think this is correct. Thanks. So one thing I failed to notice is cpu_physical_memory_set_dirty_lebitmap() will "amplify" the host dirty pages into guest ones; seems we're all good then. Reviewed-by: Peter Xu <pet...@redhat.com> -- Peter Xu
