Reverse all bits of a DWORD

[HSE]

@HSE: Why 11 "active lines"? It's 8 instructions.

Hi JJ!
 
See last Nidud post, you forget some lines. You can see your results in previous post

Active lines are lines with instructions, lines with only labels or comments (or nothing) don't count. Some assemblers require only labels lines, others no. (I prefer to use only labes lines always)

[jj2007]

you forget some lines

New version. The input in eax remains unchanged, while the result is returned in edx. Actually, 7 lines are enough, and no labels needed - see attachment

Code: [Select]

input before    b:eax           11111111111111111111111111111111
reversed once   b:edx           11111111111111111111111111111111
reversed twice  b:edx           11111111111111111111111111111111

input before    b:eax           11110000111100001111000011110000
reversed once   b:edx           00001111000011110000111100001111
reversed twice  b:edx           11110000111100001111000011110000

input before    b:eax           11101110111011101110111011101110
reversed once   b:edx           01110111011101110111011101110111
reversed twice  b:edx           11101110111011101110111011101110

input before    b:eax           11001100110011001100110011001100
reversed once   b:edx           00110011001100110011001100110011
reversed twice  b:edx           11001100110011001100110011001100

input before    b:eax           10101010101010101010101010101010
reversed once   b:edx           01010101010101010101010101010101
reversed twice  b:edx           10101010101010101010101010101010

input before    b:eax           00000000000000000000000000000000
reversed once   b:edx           00000000000000000000000000000000
reversed twice  b:edx           00000000000000000000000000000000

[TimoVJL]

without m2m MACRO it could be 7.

[jj2007]

without m2m MACRO it could be 7.

Right, mov ecx, 32 is ok, too. If it's not speed-critical, I always use m2m for the range -128...127.

[aw27]

This is the smallest without using the loop instruction 

Code: [Select]

mov eax, 0
mov esi, 00100110011110001001100110010101b
reverse:
lzcnt edx, esi
bts eax, edx
not dl
and dl, 1fh
btc esi, edx
jnz reverse


[aw27]

For people with Haswell because this requires Bit Manipulation Instruction Set 2 (BMI2):

Code: [Select]

mov eax, 0
mov esi, 00100110011110001001100110010101b
reverse:
lzcnt ecx, esi
bsr edx, esi
bts eax, ecx
bzhi esi, esi, edx ; requires BMI2
jnz reverse


[HSE]

This is the smallest without using the loop instruction 

Apparently lzcnt is a lot simpler CISC algorithm than loop (but this machine only have SS3  ) . I think that replacing loop with dec ecx / jnl @B is simpler and universal.

Code: [Select]

edx = 00100110011110001001100110010101y, challenge [awch1.asm, 49]
eax = 10101001100110010001111001100100y, Nidud (4 active lines, value in edx) [awch1.asm, 56]
eax = 10101001100110010001111001100100y, AW (8 active lines, value in esi) [awch1.asm, 70]
edx = 10101001100110010001111001100101y, JJ Oops (if wrong no matter active lines, value in eax) [awch1.asm, 83]
edx = 10101001100110010001111001100100y, JJ-Nidud (11 active lines, value in eax) [awch1.asm, 101]
edx = 10101001100110010001111001100100y, JJ Another (7 active lines, value in eax) [awch1.asm, 114]


note: asmc /Zne /c /coff awch1.asm

[aw27]

lzcnt is similar to bsr but bsr brought me a headache with the flags.

[TimoVJL]

This was modified from C code

Code: [Select]

  mov esi, 26789995h
  xor edi, edi

  xor eax, eax
  mov edx, 1Fh
L1:
  bt esi, eax
  ;setb cl
  ;test cl, cl
  jnb @F ; jz @F
  bts edi, edx
@@:
  ;setb cl
  ;add eax, 1h
  inc eax
  ;sub edx, 1h
  dec edx
  jnz L1


[daydreamer]

why shouldnt a single 1/x with fdiv with double work on reversing bits,as 2 reversed=0.5,4 reversed =0.25,8 reversed=0.125 ...?

[jj2007]

why shouldnt a single 1/x with fdiv with double work on reversing bits,as 2 reversed=0.5,4 reversed =0.25,8 reversed=0.125 ...?

Why don't you test it and post some code?