Re: [libvirt PATCH] qemu: Simplify size check for ppc64 NVDIMMs
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(a)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(a)gmail.com>
virReportError(VIR_ERR_XML_ERROR, "%s",
_("minimum target size for the NVDIMM "
"must be 256MB plus the label size"));