Reversing : The Hacker's Guide to Reverse Engineering

shr edx,1 ; shift dividend right one bit

rcr eax,1

or ebx,ebx

jnz short L5 ; loop until divisor < 4194304K

div ecx ; now divide, ignore remainder

mov esi,eax ; save quotient

;

; We may be off by one, so to check, we will multiply the quotient

; by the divisor and check the result against the orignal dividend

; Note that we must also check for overflow, which can occur if the

; dividend is close to 2**64 and the quotient is off by 1.

;

mul dword ptr HIWORD(DVSR) ; QUOT * HIWORD(DVSR)

mov ecx,eax

mov eax,LOWORD(DVSR)

mul esi ; QUOT * LOWORD(DVSR)

add edx,ecx ; EDX:EAX = QUOT * DVSR

jc short L6 ; carry means Quotient is off by 1

;

; do long compare here between original dividend and the result of the

; multiply in edx:eax. If original is larger or equal, we are ok,

; otherwise subtract one (1) from the quotient.

;

cmp edx,HIWORD(DVND) ; compare hi words of result and

original

ja short L6 ; if result > original, do subtract

jb short L7 ; if result < original, we are ok

cmp eax,LOWORD(DVND); hi words are equal, compare lo words

jbe short L7 ; if less or equal we are ok, else

;subtract

L6:

dec esi ; subtract 1 from quotient

L7:

xor edx,edx ; edx:eax <- quotient

mov eax,esi

;

; Just the cleanup left to do. edx:eax contains the quotient. Set the

; sign according to the save value, cleanup the stack, and return.

;

L4:

dec edi ; check to see if result is negative

jnz short L8 ; if EDI == 0, result should be negative

neg edx ; otherwise, negate the result

Listing B.2 (continued)

Understanding Compiled Arithmetic 533

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