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

LDR (Q)

Test 1: uops

Code:

  ldr q0, [x6]
  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)
10056471031110301000816410001000100011000
10045511001110001000816410001000100011000
10045531001110001000823610001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045571001110001000816410001000100011000
10045511001110001000834410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000

Test 2: Latency 1->2 roundtrip

Chain cycles: 3

Code:

  ldr q0, [x6]
  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): 7.0049

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)? simd retires (ee)? int retires (ef)
5020510015260108401011000610001301301001410004266921210456181070792501103021210004100046029410017100164000510000040100
5020410006160103401011000210000301031000310016267202810467621071972501643025010017100176022410004100044000110000040100
5020410004760103401011000210000301031000310004266921210456181070792501103021210004100046022410004100044000110000040100
5020410015260116401091000510002301361001610004266921210456181070792501103021210004100046029410016100174000710000040100
5020410006060103401011000210000301031000310016267103710463701071534501643025110016100176022410004100044000110000040100
5020410014860116401091000510002301361001610004266921210456181070792501103021210004100046022410004100044000110000040100
5020410004760103401011000210000301031000310004266937410456851070859501103021210004100046029410015100164000910000040100
5020410004760103401011000210000301031000310004266921210456181070792501103021210004100046022410004100044000110000040100
5020410005660103401011000210000301031000310016267152110465921071782501693025010017100176022410004100044000110000040100
5020410004760103401011000210000301031000310004266921210456181070792501103021210004100046022410004100044000110000040100

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)
500251001526001840011100061000130040100141000426693831047000107204050020300321000410004600201000010000400011000040010
500241000496001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241000496001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241000566001340011100021000030010100001001426697501047129107216150072300671001610017600201000010000400011000040010
500241000496001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241000496001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241000496001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241000536001340011100021000030010100001000326692991046855107190250019300301000410004600441000410004400011000040010
500241000746001440011100031000030010100001000026713881047771107280850010300201000010000600201000010000400011000040010
500241000506001340011100021000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010

Test 3: throughput

Count: 8

Code:

  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  ldr q0, [x6]
  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)
8020540187801351018003410080008300455196801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100
8020440047801011018000010080008300640028801082008001220080012180000100

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)
8002540354800451180034108000830640248800182080012208001218000010
8002440058800111180000108000030640218800102080000208000018000010
8002440060800111180000108000030640218800102080000208000018000010
8002440054800111180000108000030640326800102080000208000018000010
8002440063800111180000108000030640434800102080000208000018000010
8002440054800111180000108000030640218800102080000208007118000010
8002440066800111180000108000030640362800102080000208000018000010
8002440072800111180000108000030640272800102080000208000018000010
8002440063800111180000108000030640380800102080000208000018000010
8002440063800111180000108000030640380800102080000208000018000010