llrem.asm

        title   llrem - signed long remainder routine
;***
;llrem.asm - signed long remainder routine
;
;       Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
;       defines the signed long remainder routine
;           __allrem
;
;*******************************************************************************


.xlist
include vcruntime.inc
include mm.inc
.list

;***
;llrem - signed long remainder
;
;Purpose:
;       Does a signed long remainder 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 remainder (dividend%divisor)
;       NOTE: this routine removes the parameters from the stack.
;
;Uses:
;       ECX
;
;Exceptions:
;
;*******************************************************************************

        CODESEG

_allrem PROC NEAR
.FPO (2, 4, 0, 0, 0, 0)

        push    ebx
        push    edi

; 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 lrem(a, b)):
;
;               -----------------
;               |               |
;               |---------------|
;               |               |
;               |--divisor (b)--|
;               |               |
;               |---------------|
;               |               |
;               |--dividend (a)-|
;               |               |
;               |---------------|
;               | return addr** |
;               |---------------|
;               |       EBX     |
;               |---------------|
;       ESP---->|       EDI     |
;               -----------------
;

DVND    equ     [esp + 12]      ; stack address of dividend (a)
DVSR    equ     [esp + 20]      ; 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 bit
        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
        mov     edx,LOWORD(DVSR) ; lo word of b
        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             ; edx <- remainder
        mov     eax,LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
        div     ecx             ; edx <- final remainder
        mov     eax,edx         ; edx:eax <- remainder
        xor     edx,edx
        dec     edi             ; check result sign flag
        jns     short L4        ; negate result, restore stack and return
        jmp     short L8        ; result sign ok, 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

;
; 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.
;

        mov     ecx,eax         ; save a copy of quotient in ECX
        mul     dword ptr HIWORD(DVSR)
        xchg    ecx,eax         ; save product, get quotient in EAX
        mul     dword ptr 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 the original divisor from the result.
;

        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:
        sub     eax,LOWORD(DVSR) ; subtract divisor from result
        sbb     edx,HIWORD(DVSR)
L7:

;
; Calculate remainder by subtracting the result from the original dividend.
; Since the result is already in a register, we will do the subtract in the
; opposite direction and negate the result if necessary.
;

        sub     eax,LOWORD(DVND) ; subtract dividend from result
        sbb     edx,HIWORD(DVND)

;
; Now check the result sign flag to see if the result is supposed to be positive
; or negative.  It is currently negated (because we subtracted in the 'wrong'
; direction), so if the sign flag is set we are done, otherwise we must negate
; the result to make it positive again.
;

        dec     edi             ; check result sign flag
        jns     short L8        ; result is ok, restore stack and return
L4:
        neg     edx             ; otherwise, negate the result
        neg     eax
        sbb     edx,0

;
; Just the cleanup left to do.  edx:eax contains the quotient.
; Restore the saved registers and return.
;

L8:
        pop     edi
        pop     ebx

        ret     16

_allrem ENDP

        end