3 ms·

Based on the information from [1] we have something like this for both loops: .LBB0_2: eor x13, x9, x9, lsr #30 # 2 \* p1-6

by pbsd 6y ago

Based on the information from [1] we have something like this for both loops: .LBB0_2: eor x13, x9, x9, lsr #30 # 2 \* p1-6 mul x13, x13, x11 # 1 \* p5-6 eor x13, x13, x13, lsr #27 # 2 \* p1-6 mul x13, x13, x12 # 1 \* p5-6 eor x13, x13, x13, lsr #31 # 2 \* p1-6 str x13, [x0, x10, lsl #3] # 1 \* p7-8 add x13, x10, #2 # 1 \* p1-6 add x9, x9, x8 # 1 \* p1-6 mov x10, x13 # none cmp x13, x1 # b.lo .LBB0_2 # Fused into 1 \* p1-3 # Total: 11 uops .LBB1_2: mul x13, x9, x11 # 1 \* p5-6 umulh x14, x9, x11 # 1 \* p5-6 eor x13, x14, x13 # 1 \* p1-6 mul x14, x13, x12 # 1 \* p5-6 umulh x13, x13, x12 # 1 \* p5-6 eor x13, x13, x14 # 1 \* p1-6 str x13, [x0, x10, lsl #3] # 1 \* p7-8 add x13, x10, #2 # 1 \* p1-6 add x9, x9, x8 # 1 \* p1-6 mov x10, x13 # none cmp x13, x1 # b.lo .LBB1_2 # Fused into 1 \* p1-3 # Total: 10 uops Purely based on number of uops, there's a slight win for wyhash, all other things being equal. However, I doubt that you're really getting one iteration per second here; there are 6 integer units, and even if you perfectly exploited instruction parallelism you're limited to 6 ALU instructions per cycle, which are less than the extent of either loop. It would be possible if the mul-umulh pairs are getting fused, which would bring it down to 8 uops per iteration. Taking into account the port distribution, each iteration of wyhash involves 4 uops being dispatched to ports 5 and 6, which means you should be getting at least 2 cycles/iteration purely for the multiplications. If it's much lower than that, the whole multiplication being fused into a single port 5-6 uop might be right. However I can neither confirm nor deny that the loops behave like that on the M1, as I don't have one. [1] https://dougallj.github.io/applecpu/firestorm.html https://dougallj.github.io/applecpu/firestorm.html

gpderetta 6y ago
I think you are right that mull/h are fused. I think that M1 has 128 ALUs for the vector unit, so it would be a good way to make use of them. M1 is far from the first iteration of the architecture and Apple has likely picked most if not all low hanging fruits. It also helps x86 emulation I guess. edit: but see the comment else thread about the loop iteration time being off by a factor of 2.
- pbsd 6y ago
  Oh yeah, I thought the add r,r,2 was odd but didn't investigate. This brings things back to ~2+ cycles per iteration, which strictly speaking does not require fusion. It would be easier to test this explicitly instead of inside some unrelated RNG.

by pbsd 6y ago

gpderetta 6y ago

I think you are right that mull/h are fused. I think that M1 has 128 ALUs for the vector unit, so it would be a good way to make use of them. M1 is far from the first iteration of the architecture and Apple has likely picked most if not all low hanging fruits. It also helps x86 emulation I guess. edit: but see the comment else thread about the loop iteration time being off by a factor of 2.

pbsd 6y ago
Oh yeah, I thought the add r,r,2 was odd but didn't investigate. This brings things back to ~2+ cycles per iteration, which strictly speaking does not require fusion. It would be easier to test this explicitly instead of inside some unrelated RNG.