;******************************************************************************* ;lldiv.asm - signed long divide routine ; ; Copyright (c) Microsoft Corporation. All rights reserved. ; ;Purpose: ; defines the signed long divide routine ; __alldiv ; ;******************************************************************************* .686p .mmx .xmm .model flat,C .code assume ds:flat assume es:flat assume ss:flat assume fs:nothing assume gs:nothing include hal.inc ;*** ;lldiv - signed long divide ; ;Purpose: ; Does a signed long divide of the arguments. Arguments are ; not changed. ; ;Entry: ; Arguments are passed on the stack: ; 1st pushed: divisor (QWORD) ; 2nd pushed: dividend (QWORD) ; ;Exit: ; EDX:EAX contains the quotient (dividend/divisor) ; NOTE: this routine removes the parameters from the stack. ; ;Uses: ; ECX ; ;Exceptions: ; ;******************************************************************************* _alldiv PROC NEAR push edi push esi push ebx ; Set up the local stack and save the index registers. When this is done ; the stack frame will look as follows (assuming that the expression a/b will ; generate a call to lldiv(a, b)): ; ; ----------------- ; | | ; |---------------| ; | | ; |--divisor (b)--| ; | | ; |---------------| ; | | ; |--dividend (a)-| ; | | ; |---------------| ; | return addr** | ; |---------------| ; | EDI | ; |---------------| ; | ESI | ; |---------------| ; ESP---->| EBX | ; ----------------- ; DVND equ [esp + 16] ; stack address of dividend (a) DVSR equ [esp + 24] ; stack address of divisor (b) ; Determine sign of the result (edi = 0 if result is positive, non-zero ; otherwise) and make operands positive. xor edi,edi ; result sign assumed positive mov eax,HIWORD(DVND) ; hi word of a or eax,eax ; test to see if signed jge short L1 ; skip rest if a is already positive inc edi ; complement result sign flag mov edx,LOWORD(DVND) ; lo word of a neg eax ; make a positive neg edx sbb eax,0 mov HIWORD(DVND),eax ; save positive value mov LOWORD(DVND),edx L1: mov eax,HIWORD(DVSR) ; hi word of b or eax,eax ; test to see if signed jge short L2 ; skip rest if b is already positive inc edi ; complement the result sign flag mov edx,LOWORD(DVSR) ; lo word of a neg eax ; make b positive neg edx sbb eax,0 mov HIWORD(DVSR),eax ; save positive value mov LOWORD(DVSR),edx L2: ; ; Now do the divide. First look to see if the divisor is less than 4194304K. ; If so, then we can use a simple algorithm with word divides, otherwise ; things get a little more complex. ; ; NOTE - eax currently contains the high order word of DVSR ; or eax,eax ; check to see if divisor < 4194304K jnz short L3 ; nope, gotta do this the hard way mov ecx,LOWORD(DVSR) ; load divisor mov eax,HIWORD(DVND) ; load high word of dividend xor edx,edx div ecx ; eax <- high order bits of quotient mov ebx,eax ; save high bits of quotient mov eax,LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend div ecx ; eax <- low order bits of quotient mov edx,ebx ; edx:eax <- quotient jmp short L4 ; set sign, restore stack and return ; ; Here we do it the hard way. Remember, eax contains the high word of DVSR ; L3: mov ebx,eax ; ebx:ecx <- divisor mov ecx,LOWORD(DVSR) mov edx,HIWORD(DVND) ; edx:eax <- dividend mov eax,LOWORD(DVND) L5: shr ebx,1 ; shift divisor right one bit rcr ecx,1 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 neg eax sbb edx,0 ; ; Restore the saved registers and return. ; L8: pop ebx pop esi pop edi ret 16 _alldiv ENDP end