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

SBFIZ (64-bit)

Test 1: uops

Code:

  sbfiz x0, x0, #3, #7
  mov x0, 1
  mov x1, 2

(no loop instructions)

1000 unrolls and 1 iteration

Retires: 1.000

Issues: 1.000

Integer unit issues: 1.001

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)? int output thing (e9)? int retires (ef)
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000
100410301001100110002560310001000100010011000

Test 2: Latency 1->2

Code:

  sbfiz x0, x0, #3, #7
  mov x0, 1
  mov x1, 2

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code): 1.0030

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)
10204100301010110101101082593301010710214102141000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100
10204100301010110101101072595391010710212102121000110100

1000 unrolls and 10 iterations

Result (median cycles for code): 1.0030

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)
10024100301002110021100282594741002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010
10024100301002110021100202595911002010020100201001110010

Test 3: throughput

Count: 8

Code:

  sbfiz x0, x8, #3, #7
  sbfiz x1, x8, #3, #7
  sbfiz x2, x8, #3, #7
  sbfiz x3, x8, #3, #7
  sbfiz x4, x8, #3, #7
  sbfiz x5, x8, #3, #7
  sbfiz x6, x8, #3, #7
  sbfiz x7, x8, #3, #7
  mov x8, 9

(fused SUBS/B.cc loop)

100 unrolls and 100 iterations

Result (median cycles for code divided by count): 0.3342

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)
80204268658011580115801202403608012080224802248001580100
80204267428011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100
80204267378011580115801202403608012080224802248001580100

1000 unrolls and 10 iterations

Result (median cycles for code divided by count): 0.3339

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)
80024280058003780037800422774848004380046800468002880010
80024267898003880038800432774048002080020800208001180010
80024267178002180021800202774048002080020800208001180010
80024267178002180021800202774048002080020800208001180010
80024279638003780037800422827388002080020800208001180010
80024267138002180021800202774048002080020800208001180010
80024267138002180021800202774048002080020800208001180010
80024267138002180021800202774048002080020800208001180010
80024267138002180021800202774048002080020800208001180010
80024267138002180021800202774048002080020800208001180010