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 (S)

Test 1: uops

Code:

  ldr s0, [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)
10056591031110301000816410001000100011000
10045511001110001000821810001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000
10045511001110001000816410001000100011000

Test 2: Latency 1->2 roundtrip

Chain cycles: 3

Code:

  ldr s0, [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): 7.0044

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)
5020510015760108401011000610001301301001410004266949110456811070859501103021210004100046022410004100044000110000040100
5020410005160103401011000210000301031000310004266918510456111070781501103021210004100046022410004100044000110000040100
5020410004560102401011000110000301031000310004266910410455751070749501103021210004100046022410004100044000110000040100
5020410004460102401011000110000301031000310004266918510456081070781501103021210004100046022410004100044000110000040100
5020510016060110401051000310002301331001510004266926610456401070814501103021210004100046022410004100044000110000040100
5020410004260102401011000110000301031000310004266907710455631070737501103021210004100046022410004100044000110000040100
5020410004460102401011000110000301031000310004266907710455631070737501103021210004100046022410004100044000110000040100
5020410004060102401011000110000301031000310004266902310455411070715501103021210004100046022410004100044000110000040100
5020410004060102401011000110000301031000310004266902310455411070715501103021210004100046022410004100044000110000040100
5020410004460102401011000110000301031000310004266907710455631070737501103021210004100041355770884230152086146574662093852742601

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)
500251008926007040043100161001130166100641000426692581046891107193750020300321000410004601201001710017400051000040010
500251000966002340015100061000230043100151000026693181046873107191650010300201000010000600201000010000400011000040010
500241000406001240011100011000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000566001240011100011000030010100001004926758641049598107464550243301721004910052600201000010000400011000040010
500241000406001240011100011000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000406001240011100011000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000566001240011100011000030010100001005026758431049587107474250240301761005010052600201000010000400011000040010
500241000406001240011100011000030010100001000026693361046935107197250010300201000010000600201000010000400011000040010
500241004266006440043100121000930139100511000026698491047154107218850010300201000010000600201000010000400011000040010
500251000826002240017100041000130043100161000026692821046913107195050010300201000010000600201000010000400011000040010

Test 3: throughput

Count: 8

Code:

  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [x6]
  ldr s0, [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)
8020540336801311018003010080008300280188801082008001220080012180000100
8020440053801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440051801011018000010080008300640194801082008001220080012180000100
8020440048801011018000010080008300640194801082008001220080012180000100
8020440045801011018000010080008300640194801082008001220080072180000100
8020440045801011018000010080008300640194801082008001220080012180000100
8020440048801011018000010080008300640356801082008001220080012180000100

1000 unrolls and 10 iterations

Result (median cycles for code divided by count): 0.5005

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)
8002540249800471180036108000030319982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010
8002440043800111180000108000030639982800102080000208000018000010