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

AND (register, lsl, 64-bit)

Test 1: uops

Code:

  and x0, x0, x1, lsl #17
  mov x0, 1
  mov x1, 2

(no loop instructions)

1000 unrolls and 1 iteration

Retires: 1.000

Issues: 2.000

Integer unit issues: 2.001

Load/store unit issues: 0.000

SIMD/FP unit issues: 0.000

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)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)
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000
1004203020012001100052265100010002000002001001000

Test 2: Latency 1->2

Code:

  and x0, x0, x1, lsl #17
  mov x0, 1
  mov x1, 2

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code): 2.0030

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
10204200302010120101101045291431010410210202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100

1000 unrolls and 10 iterations

Result (median cycles for code): 2.0030

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
10024200302002120021100255292201002510034201262002610010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010

Test 3: Latency 1->3

Code:

  and x0, x1, x0, lsl #17
  mov x0, 1
  mov x1, 2

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code): 2.0030

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
10204200302010120101101045291001010410210202202000110100
10204200302010120101101045291861010410212202202000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100
10205200602011620116101385291861010410212202242000110100
10204200302010120101101045291861010410212202242000110100

1000 unrolls and 10 iterations

Result (median cycles for code): 2.0030

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
10024200302002120021100255292761002510032200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292531002010020200202001110010
10024200302002120021100205292761002510032200202001110010

Test 4: throughput

Count: 8

Code:

  and x0, x8, x9, lsl #17
  and x1, x8, x9, lsl #17
  and x2, x8, x9, lsl #17
  and x3, x8, x9, lsl #17
  and x4, x8, x9, lsl #17
  and x5, x8, x9, lsl #17
  and x6, x8, x9, lsl #17
  and x7, x8, x9, lsl #17
  mov x8, 9
  mov x9, 10

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code divided by count): 0.6675

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
8020453429160118160118801311360297801298023416038416006580100
8020453413160115160115801291360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100
8020453404160117160117801301360838801308023616027216001780100

1000 unrolls and 10 iterations

Result (median cycles for code divided by count): 0.6671

retire uop (01)cycle (02)schedule uop (52)schedule int uop (53)dispatch int uop (56)int uops in schedulers (59)dispatch uop (78)map int uop (7c)map int uop inputs (7f)? int output thing (e9)? int retires (ef)
8002453390160039160039800511359975800518005816009216002980010
8002453390160039160039800511359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002553410160085160085801001359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010
8002453371160021160021800201359903800208002016002016001180010