> On 14 May 2019, at 10:29, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote: > > > (please, wrap our emails at 78 chars) > > On Tue, May 14, 2019 at 12:08:23AM +0300, Liran Alon wrote: > >> 3) From (2), we should have theoretically deduced that for every >> #VMExit, there is a need to kick the sibling hyperthread also outside >> of guest until the #VMExit is completed. > > That's not in fact quite true; all you have to do is send the IPI. > Having one sibling IPI the other sibling carries enough guarantees that > the receiving sibling will not execute any further guest instructions. > > That is, you don't have to wait on the VMExit to complete; you can just > IPI and get on with things. Now, this is still expensive, But it is > heaps better than doing a full sync up between siblings. > I agree. I didn’t say you need to do full sync. You just need to IPI the sibling hyperthreads before switching to the full kernel address space. But you need to make sure these sibling hyperthreads don’t get back into the guest until all hyperthreads are running with KVM isolated address space. It is still very expensive if done for every #VMExit. Which as I explained, can be avoided in case we use the KVM isolated address space technique. -Liran
