2 ms·
Because bounded loops are functionally a compression scheme on the graph. It's a tradeoff for sure though. Personally (and going way off into the weeds, these
by monocasa 5y ago
Because bounded loops are functionally a compression scheme on the graph. It's a tradeoff for sure though.
Personally (and going way off into the weeds, these are my thoughts and not at all representative of people working on BPF) I'd also like to see other schemes eventually that also don't necessarily require verifiable halting semantics for correctness. One example: something like io_uring, but instead there's a program running just barely in kernel space, accounted to the host user thread's scheduling, issuing new syscalls already on the kernel side of the syscall boundary.