;*******************************************************************************
;ullrem.asm - unsigned long remainder routine
;
;       Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
;       defines the unsigned long remainder routine
;           __aullrem
;
;*******************************************************************************

include hal.inc
        
;***
;ullrem - unsigned long remainder
;
;Purpose:
;       Does a unsigned 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:
;
;*******************************************************************************

__aullrem        PROC NEAR

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

DVND    equ     [esp + 8]       ; stack address of dividend (a)
DVSR    equ     [esp + 16]      ; stack address of divisor (b)

; 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.
;

        mov     eax,HIWORD(DVSR) ; check to see if divisor < 4194304K
        or      eax,eax
        jnz     short L1        ; 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, eax <- quotient
        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
        jmp     short L2        ; restore stack and return

;
; Here we do it the hard way.  Remember, eax contains DVSRHI
;

L1:
        mov     ecx,eax         ; ecx:ebx <- divisor
        mov     ebx,LOWORD(DVSR)
        mov     edx,HIWORD(DVND) ; edx:eax <- dividend
        mov     eax,LOWORD(DVND)
L3:
        shr     ecx,1           ; shift divisor right one bit; hi bit <- 0
        rcr     ebx,1
        shr     edx,1           ; shift dividend right one bit; hi bit <- 0
        rcr     eax,1
        or      ecx,ecx
        jnz     short L3        ; loop until divisor < 4194304K
        div     ebx             ; 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         ; put partial product in ECX, get quotient in EAX
        mul     dword ptr LOWORD(DVSR)
        add     edx,ecx         ; EDX:EAX = QUOT * DVSR
        jc      short L4        ; 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're ok, otherwise
; subtract the original divisor from the result.
;

        cmp     edx,HIWORD(DVND) ; compare hi words of result and original
        ja      short L4        ; if result > original, do subtract
        jb      short L5        ; if result < original, we're ok
        cmp     eax,LOWORD(DVND) ; hi words are equal, compare lo words
        jbe     short L5        ; if less or equal we're ok, else subtract
L4:
        sub     eax,LOWORD(DVSR) ; subtract divisor from result
        sbb     edx,HIWORD(DVSR)
L5:

;
; Calculate remainder by subtracting the result from the original dividend.
; Since the result is already in a register, we will perform the subtract in
; the opposite direction and negate the result to make it positive.
;

        sub     eax,LOWORD(DVND) ; subtract original dividend from result
        sbb     edx,HIWORD(DVND)
        neg     edx             ; and negate it
        neg     eax
        sbb     edx,0

;
; Just the cleanup left to do.  dx:ax contains the remainder.
; Restore the saved registers and return.
;

L2:

        pop     ebx

        ret     16

__aullrem        ENDP

        end
