Apple Microarchitecture Research by Dougall Johnson

M1/A14 P-core (Firestorm): Overview | Base Instructions | SIMD and FP Instructions
M1/A14 E-core (Icestorm):  Overview | Base Instructions | SIMD and FP Instructions

ST3 (single, post-index, B)

Test 1: uops

Code:

  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop ; nop
  mov x0, 0
  mov x8, 0

(no loop instructions)

1000 unrolls and 1 iteration

Retires (minus 60 nops): 2.000

Issues: 3.000

Integer unit issues: 1.001

Load/store unit issues: 1.000

SIMD/FP unit issues: 1.000

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule simd uop (54)schedule ldst uop (55)dispatch int uop (56)dispatch simd uop (57)dispatch ldst uop (58)int uops in schedulers (59)simd uops in schedulers (5a)ldst uops in schedulers (5b)dispatch uop (78)map ldst uop (7d)map simd uop (7e)map ldst uop inputs (80)map simd uop inputs (81)? int output thing (e9)? ldst retires (ed)? simd retires (ee)
6200629593300710031002100210021002100030008751400030001000100030003000100110001000
6200429290300110011000100010001000100030008751400030001000100030003000100110001000
6200529401300110011000100010001000100030008751400030001000100030003000100110001000
6200429317300110011000100010001000100030008751400030001000100030003000100110001000
6200429306300110011000100010001000100030008751400030001000100030003000100110001000
6200429318300110011000100010001000100030008751400030001000100030003000100110001000
6200429277300110011000100010001000100030008751400030001000100030003000100110001000
6200429321300110011000100010001000100030008751400030001000100030003000100110001000
6200429293300110011000100010001000100030008751400030001000100030003000100110001000
6200429293300110011000100010001000100030008751400030001000100030003000100110001000

Test 2: throughput

Count: 8

Code:

  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  st3 { v0.b, v1.b, v2.b }[1], [x6], x8
  mov x7, x6
  mov x8, x6
  mov x9, x6
  mov x10, x6
  mov x11, x6
  mov x12, x6
  mov x13, x6
  mov x8, 0

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code divided by count): 1.0005

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule simd uop (54)schedule ldst uop (55)dispatch int uop (56)dispatch simd uop (57)dispatch ldst uop (58)int uops in schedulers (59)simd uops in schedulers (5a)ldst uops in schedulers (5b)dispatch uop (78)map int uop (7c)map ldst uop (7d)map simd uop (7e)map int uop inputs (7f)map ldst uop inputs (80)map simd uop inputs (81)? int output thing (e9)? ldst retires (ed)? simd retires (ee)? int retires (ef)
16020680149240185801328003580018801378003780004123760713598423200382401222008000980009200240024240024800018000080000100
16020480039240108801018000780000801098000980003116027513599653200342401192008000880008200240027240027800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240024240024800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240024240024800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240024240024800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240114240114800318000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240024240024800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240018240018800058000080000100
16020480040240105801018000480000801068000680003116027513599653200342401192008000880008200240024240024800018000080000100
16020480039240107801018000680000801088000880003116027513599653200342401192008000880008200240024240024800018000080000100

1000 unrolls and 10 iterations

Result (median cycles for code divided by count): 1.0005

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule simd uop (54)schedule ldst uop (55)dispatch int uop (56)dispatch simd uop (57)dispatch ldst uop (58)int uops in schedulers (59)simd uops in schedulers (5a)ldst uops in schedulers (5b)dispatch uop (78)map int uop (7c)map ldst uop (7d)map simd uop (7e)map int uop inputs (7f)map ldst uop inputs (80)map simd uop inputs (81)? int output thing (e9)? ldst retires (ed)? simd retires (ee)? int retires (ef)
16002680155240095800418003680018800488003880004102413913600393200382400322080009800092024002124002180001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024010824010880031800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010
16002480039240011800118000080000800108000080000115995913599553200002400102080000800002024000024000080001800008000010