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 (D)

Test 1: uops

Code:

  ldr d0, [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)
10056671031110301000816410001000100011000
10045431001110001000816410001000100011000
10045431001110001000798210001000100011000
10045541001110001000798210001000100011000
10045431001110001000798210001000100011000
10045431001110001000798210001000100011000
10045431001110001000798210001000100011000
10045431001110001000798210001000100011000
10045431001110001000799010001000100011000
10045431001110001000798210001000100011000

Test 2: Latency 1->2 roundtrip

Chain cycles: 3

Code:

  ldr d0, [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): 7.0042

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)
502041000476010340101100021000030103100031000426689781045465107064550110302121000410004602241000410004400011000040100
502041000406010240101100011000030103100031000426690771045565107073750110302121000410004602241000410004400011000040100
502041000416010240101100011000030103100031001626712461046438107160050164302461001710016603761002810030400131000040100
502051000926011240107100041000130133100151000426689851045408107059450110302121000410004603001001710017400071000040100
502041000496010340101100021000030103100031000426690771045563107073750110302121000410004602241000410004400011000040100
502041000446010240101100011000030103100031000426721011046848107198350110302121000410004602241000410004400011000040100
502041000426010240101100011000030103100031001526694441046429107091450164302501001710017602241000410004400011000040100
502041000426010240101100011000030103100031000426690771045563107073750110302121000410004602241000410004400011000040100
502041000426010240101100011000030103100031000426690771045563107073750110302121000410004602241000410004400011000040100
502041000426010240101100011000030103100031000426690771045563107073750110302121000410004602241000410004400011000040100

1000 unrolls and 10 iterations

Result (median cycles for code, minus 3 chain cycles): 7.0047

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)
500251001466001840011100061000130040100141000426692571046891107193750020300321000410004600201000010000400011000040010
500251000776002140015100041000230043100151000026695521047027107206450010300201000010000600201000010000400011000040010
500241000496001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000476001340011100021000030010100001001526696661047095107213950074300701001710017600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010
500241000576001340011100021000030010100001001426696661047093107212550072300671001610017600201000010000400011000040010
500241000476001340011100021000030010100001000026692821046913107195050010300201000010000600201000010000400011000040010

Test 3: throughput

Count: 8

Code:

  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [x6]
  ldr d0, [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): 0.5007

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)
8020540164801351018003410080008300483840801082008001220080012180000100
8020440053801011018000010080008300640194801082008001220080012180000100
8020440069801011018000010080059300436287801592008007120080012180000100
8020440064801051018000410080008300640248801082008001220080012180000100
8020440062801011018000010080008300640302801082008001220080012180000100
8020440056801011018000010080008300640248801082008001220080012180000100
8020440056801011018000010080008300640248801082008001220080012180000100
8020440056801011018000010080008300640248801082008001220080012180000100
8020440056801011018000010080008300640248801082008001220080012180000100
8020440056801011018000010080008300640248801082008001220080012180000100

1000 unrolls and 10 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)
8002540264800451180034108000830400188800182080012208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208000018000010
8002440051800111180000108000030640164800102080000208006918000010