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Brent Jenkins wrote:

http://geektyper.com/

Chris Maunder wrote:

because all hackers are hot, buff, emotionally and physically mature geniuses

Member 4194593 wrote:

Don't tell me MS

Member 4194593 wrote:

don't ask why not

Member 4194593 wrote:

Don't tell me MS

Member 4194593 wrote:

and don't ask why not

mov eax, [ebx]

This is the source code:

;*******************************************************************************
;
;   SUB a LONG_NUMB value from another. This is a routine called to subtract a
;   long number from another after all scaling and validations have been done.
;
;       esi has the source OFFSET of the data.
;       edi has the already scaled OFFSET of the destination data.
;       ecx has the DWORD count to subtract.
;
;   Returns nothing with edi pointing to the last word modified. Note that this
;   may not be the highest DWORD in the destination if a short source is
;   subtracted from a long destination and there are no borrows past the last
;   DWORD in the destination or source. This is primarily a check for a size
;   extension due to a final borrow.
;
;*******************************************************************************

ALIGN   OWORD

SUBIt   PROC PRIVATE USES eax ebx esi

    pushfd
    clc
;
;   subtract all the value DWORDS.
;
ALIGN      OWORD

DoAll:
    mov    ebx,[esi]
    lea    esi,[esi + (SIZEOF DWORD)]
    sbb    [edi],ebx
    lea    edi,[edi + (SIZEOF DWORD)]
    dec    ecx
    jnz    DoAll
    jnc    Exit
    mov    eax,0
;
;   Propogate the borrow.
;
ALIGN      OWORD

Again:
    sbb    [edi],eax
    lea    edi,[edi + (SIZEOF DWORD)]
    jc     Again
;
;   Exit SUBIt, adjust the destination pointer to the last modified DWORD.
;
Exit:
    sub    edi,(SIZEOF DWORD)
    popfd
    test   dTestZero,0
    ret
SUBIt   ENDP

This is the generated code from MASM 9.0 (Visual Studio "Disassembly" tab):

ALIGN   OWORD

SUBIt   PROC PRIVATE USES eax ebx esi
00404FC0  push        eax  
00404FC1  push        ebx  
00404FC2  push        esi  

    pushfd
00404FC3  pushfd           
    clc
00404FC4  clc              
00404FC5  lea         esp,[esp] 
00404FCC  lea         esp,[esp] 
;
;   subtract all the value DWORDS.
;
ALIGN      OWORD

DoAll:
    mov    ebx,[esi]
00404FD0  mov         ebx,dword ptr [esi] 
    lea    esi,[esi + (SIZEOF DWORD)]
00404FD2  lea         esi,[esi+4] 
    sbb    [edi],ebx
00404FD5  sbb         dword ptr [edi],ebx 
    lea    edi,[edi + (SIZEOF DWORD)]
00404FD7  lea         edi,[edi+4] 
    dec    ecx
00404FDA  dec         ecx  
    jnz    DoAll
00404FDB  jne         DoAll (404FD0h) 
    jnc    Exit
00404FDD  jae         Exit (404FF7h) 
    mov    eax,0
00404FDF  mov         eax,0 
00404FE4  lea         esp,[esp] ************************** alignment inserted instructions 
00404FEB  add         eax,0 ****************************** think this might reset the carry Flag?
;
;   Propogate the borrow.
;
ALIGN      OWORD ***************************************** just added this to help speed up the code

Again:
    sbb    [edi],eax
00404FF0  sbb         dword ptr [edi],eax **************** and cause the SBB here to fail?
    lea    edi,[edi + (SIZEOF DWORD)]
00404FF2  lea         edi,[edi+4] 
    jc     Again
00404FF5  jb          Again (404FF0h) 
;
;   Exit SUBIt, adjust the destination pointer to the last modified DWORD.
;
Exit:
    sub    edi,(SIZEOF DWORD)
00404FF7  sub         edi,4 
    popfd
00404FFA  popfd            
    test   dTestZero,0
00404FFB  test        dword ptr [dTestZero (40C6BCh)],0 
    ret
00405005  pop         esi  
00405006  pop         ebx  
00405007  pop         eax  
00405008  ret              
00405009  lea         esp,[esp] 
SUBIt   ENDP