On 2/14/22 19:50, Tony Krowiak wrote:
/**
- * vfio_ap_mdev_verify_no_sharing - verifies that the AP matrix is not configured
+ * vfio_ap_mdev_verify_no_sharing - verify APQNs are not shared by matrix mdevs
*
- * @matrix_mdev: the mediated matrix device
+ * @mdev_apm: mask indicating the APIDs of the APQNs to be verified
+ * @mdev_aqm: mask indicating the APQIs of the APQNs to be verified
*
- * Verifies that the APQNs derived from the cross product of the AP adapter IDs
- * and AP queue indexes comprising the AP matrix are not configured for another
+ * Verifies that each APQN derived from the Cartesian product of a bitmap of
+ * AP adapter IDs and AP queue indexes is not configured for any matrix
* mediated device. AP queue sharing is not allowed.
*
- * Return: 0 if the APQNs are not shared; otherwise returns -EADDRINUSE.
+ * Return: 0 if the APQNs are not shared; otherwise return -EADDRINUSE.
*/
-static int vfio_ap_mdev_verify_no_sharing(struct ap_matrix_mdev *matrix_mdev)
+static int vfio_ap_mdev_verify_no_sharing(unsigned long *mdev_apm,
+ unsigned long *mdev_aqm)
{
- struct ap_matrix_mdev *lstdev;
+ struct ap_matrix_mdev *matrix_mdev;
DECLARE_BITMAP(apm, AP_DEVICES);
DECLARE_BITMAP(aqm, AP_DOMAINS);
- list_for_each_entry(lstdev, &matrix_dev->mdev_list, node) {
- if (matrix_mdev == lstdev)
+ list_for_each_entry(matrix_mdev, &matrix_dev->mdev_list, node) {
+ /*
+ * If the input apm and aqm belong to the matrix_mdev's matrix,
+ * then move on to the next.
+ */
+ if (mdev_apm == matrix_mdev->matrix.apm &&
+ mdev_aqm == matrix_mdev->matrix.aqm)
continue;
We may have a problem here. This check seems like it exists to stop you from
comparing an mdev's apm/aqm with itself. Obviously comparing an mdev's newly
updated apm/aqm with itself would cause a false positive sharing check, right?
If this is the case, I think the comment should be changed to reflect that.
Aside from the comment, what stops this particular series of if statements from
allowing us to configure a second mdev with the exact same apm/aqm values as an
existing mdev? If we do, then this check's continue will short circuit the rest
of the function thereby allowing that 2nd mdev even though it should be a
sharing violation.
--
-- Jason J. Herne (jjherne@xxxxxxxxxxxxx)