[libvirt PATCH] qemu: Simplify size check for ppc64 NVDIMMs
Daniel Henrique Barboza
danielhb413 at gmail.com
Mon Dec 7 19:18:19 UTC 2020
On 12/7/20 2:07 PM, Andrea Bolognani wrote:
> We already calculated the guest area, which is what is subject
> to minimum size requirements, a few lines earlier.
>
> Signed-off-by: Andrea Bolognani <abologna at redhat.com>
> ---
> src/qemu/qemu_domain.c | 2 +-
> 1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/src/qemu/qemu_domain.c b/src/qemu/qemu_domain.c
> index f14a15d3b4..4d007bc4a0 100644
> --- a/src/qemu/qemu_domain.c
> +++ b/src/qemu/qemu_domain.c
> @@ -5363,7 +5363,7 @@ qemuDomainNVDimmAlignSizePseries(virDomainMemoryDefPtr mem)
> /* Align down guest_area. 256MiB is the minimum size. Error
> * out if target_size is smaller than 256MiB + label_size,
> * since aligning it up will cause QEMU errors. */
> - if (mem->size < (ppc64AlignSize + mem->labelsize)) {
> + if (guestArea < ppc64AlignSize) {
Makes sense. I suggest to simplify the comment right above it as well:
/* Align down guest_area. We can't align down if guest_area is
* smaller than the 256MiB alignment. */
if (guestArea < ppc64AlignSize) {
(...)
Reviewed-by: Daniel Henrique Barboza <danielhb413 at gmail.com>
> virReportError(VIR_ERR_XML_ERROR, "%s",
> _("minimum target size for the NVDIMM "
> "must be 256MB plus the label size"));
>
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