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by monocasa 1264 days ago
> The guest is not in control

The guest has it's own page tables above the nested guest phys->host phys tables.

> What context switch time? It takes 5 micros to enter and leave the guest. The rest is just "workload".

And then the kernel doesn't know what to do with nearly every guest exit on KVM, so then you trap out to host user space, which then probably can't do much without the host kernel so you transition back to kernel space to actually perform whatever IO is needed, then back to host user, then back to host kernel to restart the guest, then back from host kernel to guest. So six total context swaps on a good day guest->host_kern->host_user->host_kern->host_user->host_kern->guest.

1 comments

Right, that's very true! It's clear that you know what you're talking about when it comes to KVM and maybe even the internal structure in Linux. However, I/O can be avoided. Imagine a guest that needs no I/O, doesn't have any interrupts enabled, and simply runs a workload straight on the CPU (given that it has all the bits it needs). That is what I have made for $COMPANY, which is in production, and serves a ... purpose. I can't really elaborate more than I already have. But you get the gist of it. It works great. It does the job, and it sandboxes a piece of code at native speed. Lots of ifs and buts and memory sharing and tricks to get it to be fast and low latency. No need for JIT, which is a security and complexity nightmare.

The topic of this thread is about Blink, which happens to be a userspace emulator. Hence my comment.