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

LDUR (Q)

Test 1: uops

Code:

  ldur q0, [x6, #1]
  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)
10056551031110301000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000
10045541001110001000821810001000100011000

Test 2: Latency 1->2 roundtrip

Chain cycles: 3

Code:

  ldur q0, [x6, #1]
  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): 7.0047

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)
502051001526010840101100061000130130100141000426691741045485107067150110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602921001710016400051000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004603001001710017400071000040100
502041000496010340101100021000030103100031000426693741045685107085950110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100
502041000496010340101100021000030103100031000426692661045640107081450110302121000410004602241000410004400011000040100

1000 unrolls and 10 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)
500251001486001840011100061000130040100141000026692911046859107190250010300201000010000600201000010000400011000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000601201001710017400051000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010
500241000426001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010
500241000476001240011100011000030010100001001626698581047175107221650074300711001610017600201000010000400011000040010
500241000456001240011100011000030010100001000026693901046957107199450010300201000010000600201000010000400011000040010
500241000436001240011100011000030010100001000026691471046858107189550010300201000010000600201000010000400011000040010

Test 3: throughput

Count: 8

Code:

  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  ldur q0, [x6, #1]
  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 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)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)
802054017280135101080034100080008300248188801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080072180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044005380101101080000100080008300640194801082008001220080012180000100
802044007080101101080000100080008300640410801082008001220080012180000100

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)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)? int output thing (e9)? ldst retires (ed)? int retires (ef)
800254025280041118003010800003064016208001020800000208000018000010
800244004380011118000010800003064009008001020800000208000018000010
800244004780011118000010800003064003608001020800000208000018000010
800244004380011118000010800003063998208001020800000208000018000010
800244004380011118000010800003063998208001020800000208000018000010
800244004380011118000010800003063998208001020800000208000018000010
800244004380011118000010800003063998208001020800000208000018000010
800244004380011118000010800003063998208001020800000208000018000010
800254009680046118003510800593064514008006920800710208000018000010
800244004380011118000010800003063998208001020800000208000018000010