4 ms·

Yeah, the algorithm for parallel decoding you outlined scales linearly in area and power with respect to the speculative look-ahead depth. This is true even if

Yeah, the algorithm for parallel decoding you outlined scales linearly in area and power with respect to the speculative look-ahead depth. This is true even if you speculate on more than the per-byte "boundary or not boundary" condition. A parallel-prefix circuit for processing a DFA with m states where you speculate on all m possible initial states for each of n bytes "only" consumes O(m n) power. [1] In absolute terms this is obviously still a problem as you crank up m or n, but the scaling is certainly not exponential. You do see local exponential scaling if the state space is large enough (think of minimax search in chess) but for these decoding problems the state space is tiny and you don't even need to speculate over all possible states (e.g. you're not going to decode all possible combinations of 4 instructions per cycle, only certain prefixes, etc). [1] The Hillis-Steele paper on data-parallel algorithms from 1986 describes this algorithm for parallel lexing.

FullyFunctional 6y ago
You are right it's x^2 not 2^x. However it's bit worse still because the area grows too which either hurts your timing (longer distances) or forces more stages (power, yet more area, and mispredict penalty). It simply isn't practically scalable much beyond where we are; if it were, you can be sure Intel would have scaled it instead of using µOP caches.
- psykotic 6y ago
  The power scales linearly, not quadratically with the amount of look-ahead. The "m" is the number of states you're speculating on which doesn't grow with look-ahead length. In the case where you're just speculating on whether an instruction starts at a given byte offset, you would have m = 2. I don't think anyone is saying they could scale up the decoder "for free". If they had a fixed-length ISA, I'm sure they would have increased the decoder width sooner (and using different techniques) since with high-end out-of-order cores you're always looking for cheap ways to over-provision your pipeline even if it only helps on some workloads some of the time. Their current use of the uop cache tells us that they consider it the most economical trade-off at that point in the design space (where the decoder can output up to 4 instructions and the uop cache can output up to 6 instructions); you can't infer that they've hit an impassable brick-wall with instruction decoding.

FullyFunctional 6y ago

You are right it's x^2 not 2^x. However it's bit worse still because the area grows too which either hurts your timing (longer distances) or forces more stages (power, yet more area, and mispredict penalty). It simply isn't practically scalable much beyond where we are; if it were, you can be sure Intel would have scaled it instead of using µOP caches.

psykotic 6y ago
The power scales linearly, not quadratically with the amount of look-ahead. The "m" is the number of states you're speculating on which doesn't grow with look-ahead length. In the case where you're just speculating on whether an instruction starts at a given byte offset, you would have m = 2. I don't think anyone is saying they could scale up the decoder "for free". If they had a fixed-length ISA, I'm sure they would have increased the decoder width sooner (and using different techniques) since with high-end out-of-order cores you're always looking for cheap ways to over-provision your pipeline even if it only helps on some workloads some of the time. Their current use of the uop cache tells us that they consider it the most economical trade-off at that point in the design space (where the decoder can output up to 4 instructions and the uop cache can output up to 6 instructions); you can't infer that they've hit an impassable brick-wall with instruction decoding.