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Hello everyone,

I found a procedure which converts dwords to ascii charachters. But I couldnt understand it(especially the codes between the dashed lines).
Could someone explain it to me?



dwtoa proc dwValue:DWORD, lpBuffer:DWORD

    push ebx
    push esi
    push edi

    mov eax, dwValue
    mov edi,

    test eax,eax
    jnz sign

  zero:
    mov word ptr ,30h
    jmp dtaexit

  sign:
    jns pos
    mov byte ptr ,'-'
    neg eax
    add edi, 1

;-------------------------------\
  pos:
    mov ecx, 3435973837            ;  what is this number for?
    mov esi, edi

    .while (eax > 0)
      mov ebx,eax
      mul ecx
      shr edx, 3
      mov eax,edx
      lea edx,
      add edx,edx
      sub ebx,edx
      add bl,'0'
      mov ,bl
      add edi, 1
    .endw
-----------------------------------/
    mov byte ptr ,
   
    .while (esi < edi)
      sub edi, 1
      mov al,
      mov ah,
      mov , al
      mov , ah
      add esi, 1
    .endw

  dtaexit:

    pop edi
    pop esi
    pop ebx

    ret

dwtoa endp


Thanks.

Down to the "pos:" label, it prints a "-" if the number is negative (and makes that number positive).
Then, keeps in ECX the value  8 * 2^32 / 10 . It'll be later used as fast division by 10, by simple multiplication with ECX:  "mul ecx | shr edx,3". Notice that after "mul", the EDX part of the 64-bit result is taken.
The "lea edx,  |  add edx,edx"  part is just multiplying EDX by 10.

So basically this segment:

mov ebx,eax
      mul ecx
      shr edx, 3
      mov eax,edx
      lea edx,
      add edx,edx
      sub ebx,edx

means
EBX = EAX % 10;
EAX = EAX / 10;

Notice though, that this calculates the digits in reverse. So, if dwValue=123456, then right now lpBuffer="654321". The ".while (esi < edi)" loop reverses the string to appear correctly.

Here is a good place to start your search for that optimization manual:

http://www.freetechbooks.com/about330.html