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

LDUR (S)

Test 1: uops

Code:

  ldur s0, [x6, #1]
  mov x0, 1
  mov x1, 2
  mov x8, 0

(no loop instructions)

1000 unrolls and 1 iteration

Retires: 1.000

Issues: 1.000

Integer unit issues: 0.001

Load/store unit issues: 1.000

SIMD/FP unit issues: 0.000

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule ldst uop (55)dispatch ldst uop (58)simd uops in schedulers (5a)dispatch uop (78)map ldst uop (7d)map ldst uop inputs (80)? int output thing (e9)? ldst retires (ed)
10057181031110301000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000

Test 2: Latency 1->2 roundtrip

Chain cycles: 3

Code:

  ldur s0, [x6, #1]
  fmov x0, d0
  eor x8, x8, x0
  eor x8, x8, x0
  add x6, x6, x8
  mov x0, 1
  mov x1, 2
  mov x8, 0

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code, minus 3 chain cycles): 7.0042

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)? int retires (ef)
502051001536010840101100061000130130100141000426691291045470107065650110302121000410004602241000410004400011000040100
502041000506010340101100021000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031001426693641045667107083050162302481001610017602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690231045541107071550110302121000410004602241000410004400011000040100

1000 unrolls and 10 iterations

Result (median cycles for code, minus 3 chain cycles): 7.0049

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)? int retires (ef)
500251001566001840011100061000130040100141000426693561046989107202950020300321000410004601961002810030400131000040010
500241000586001340011100021000030013100031000026693361046937107197450010300201000010000601141001710016400051000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241001346001340011100021000030010100001000026693091046924107196150010300201000010000600201000010000400011000040010
500251000796002240017100041000130043100161000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010

Test 3: throughput

Count: 8

Code:

  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  ldur s0, [x6, #1]
  mov x7, x6
  mov x8, x6
  mov x9, x6
  mov x10, x6
  mov x11, x6
  mov x12, x6
  mov x13, x6

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code divided by count): 0.5006

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule ldst uop (55)dispatch int uop (56)dispatch ldst uop (58)int uops in schedulers (59)simd uops in schedulers (5a)dispatch uop (78)map int uop (7c)map ldst uop (7d)map int uop inputs (7f)map ldst uop inputs (80)? int output thing (e9)? ldst retires (ed)? int retires (ef)
8020540558801371018003610080008300248188801082008001220080012180000100
8020440065801011018000010080008300640464801082008001220080012180000100
8020440050801011018000010080008300640194801082008001220080072180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640356801082008001220080012180000100
8020440051801011018000010080008300640302801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440037801051018000410080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100

1000 unrolls and 10 iterations

Result (median cycles for code divided by count): 0.5007

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)schedule ldst uop (55)dispatch int uop (56)dispatch ldst uop (58)int uops in schedulers (59)simd uops in schedulers (5a)dispatch uop (78)map int uop (7c)map ldst uop (7d)map int uop inputs (7f)map ldst uop inputs (80)? int output thing (e9)? ldst retires (ed)? int retires (ef)
8002540292800451180034108000830320242800182080012208000018000010
8002440054800111180000108000030640218800102080000208000018000010
8002440054800111180000108000030640362800102080000208000018000010
8002440054800111180000108000030640272800102080000208000018000010
8002440063800111180000108000030640218800102080000208000018000010
8002440054800111180000108000030640218800102080000208000018000010
8002440054800111180000108000030640218800102080000208007118000010
8002440054800111180000108000030640218800102080000208000018000010
8002440060800111180000108000030640218800102080000208000018000010
8002440054800111180000108000030640218800102080000208000018000010