The eventfd context is used as our irqbypass token, therefore if an
eventfd is re-used, our token is the same. The irqbypass code will
return an -EBUSY in this case, but we'll still attempt to unregister
the producer, where if that duplicate token still exists, results in
removing the wrong object. Clear the token of failed producers so
that they harmlessly fall out when unregistered.
Fixes: 6d7425f109d2 ("vfio: Register/unregister irq_bypass_producer")
Reported-by: guomin chen <guomin_chen@xxxxxxxx>
Tested-by: guomin chen <guomin_chen@xxxxxxxx>
Signed-off-by: Alex Williamson <alex.williamson@xxxxxxxxxx>
---
drivers/vfio/pci/vfio_pci_intrs.c | 4 +++-
1 file changed, 3 insertions(+), 1 deletion(-)
diff --git a/drivers/vfio/pci/vfio_pci_intrs.c b/drivers/vfio/pci/vfio_pci_intrs.c
index 1d9fb2592945..869dce5f134d 100644
--- a/drivers/vfio/pci/vfio_pci_intrs.c
+++ b/drivers/vfio/pci/vfio_pci_intrs.c
@@ -352,11 +352,13 @@ static int vfio_msi_set_vector_signal(struct vfio_pci_device *vdev,
vdev->ctx[vector].producer.token = trigger;
vdev->ctx[vector].producer.irq = irq;
ret = irq_bypass_register_producer(&vdev->ctx[vector].producer);
- if (unlikely(ret))
+ if (unlikely(ret)) {
dev_info(&pdev->dev,
"irq bypass producer (token %p) registration fails: %d\n",
vdev->ctx[vector].producer.token, ret);
+ vdev->ctx[vector].producer.token = NULL;
+ }
vdev->ctx[vector].trigger = trigger;
return 0;
