3 ms·
Your point is based on a fundamental mis-understanding of what the reorder buffer (ROB) is doing. Any time there is a fault, it is detected by the retirement s
by brandmeyer 6y ago
Your point is based on a fundamental mis-understanding of what the reorder buffer (ROB) is doing.
Any time there is a fault, it is detected by the retirement stage as it inspects the ROB. There are already hundreds of state changes that must be rolled back: all of the succeeding entries in the ROB. So the work performed by the recovery mechanism isn't changed by the amount of clobbered state because it is completely clobbered! The ROB must rewind it all.
The register file doesn't need any additional ports, because the fault recovery mechanism doesn't touch the physical registers. It only updates the register alias table (RAT) to point to the appropriate registers. But that machine must be capable of restoring 100% of the register pointers very quickly no matter what. Whether the faulting instruction touched one, two, or 16 registers doesn't matter. The 100+ updates following it are enough state that the answer must be: "Everything".
- alain94040 6y agoYou are correct for complex CPUs with actual ROB with hundreds of entries. For those micro-architecture, LDM is not particularly stressful. I was focused on the micro-controller case, with much simpler pipelines. LDM is one of the instructions causing the pipeline a lot more headaches than it's worth.
- brandmeyer 6y agoMicrocontrollers don't need additional ports either. Take a look a the ARMv7-M status/control register. It has a few bits of state to allow ldm, stm, and if-then to be resumed in-place. At that scale, those mechanisms aren't micro-coded, they are state-machined. You just have to restore the state machine.