- .title vax_bn_mul_add_word unsigned multiply & add, 32*32+32+32=>64
+ .title vax_bn_mul_add_words unsigned multiply & add, 32*32+32+32=>64
;
; w.j.m. 15-jan-1999
;
movl r6,r0 ; return c
ret
\f
- .title vax_bn_mul_word unsigned multiply & add, 32*32+32=>64
+ .title vax_bn_mul_words unsigned multiply & add, 32*32+32=>64
;
; w.j.m. 15-jan-1999
;
; }
;
; Using EDIV would be very easy, if it didn't do signed calculations.
-; Therefore, som extra things have to happen around it. The way to
-; handle that is to shift all operands right one step (basically dividing
-; them by 2) and handle the different cases depending on what the lowest
-; bit of each operand was.
+; Any time any of the input numbers are signed, there are problems,
+; usually with integer overflow, at which point it returns useless
+; data (the quotient gets the value of l, and the remainder becomes 0).
;
-; To start with, let's define the following:
+; If it was just for the dividend, it would be very easy, just divide
+; it by 2 (unsigned), do the division, multiply the resulting quotient
+; and remainder by 2, add the bit that was dropped when dividing by 2
+; to the remainder, and do some adjustment so the remainder doesn't
+; end up larger than the divisor. For some cases when the divisor is
+; negative (from EDIV's point of view, i.e. when the highest bit is set),
+; dividing the dividend by 2 isn't enough, and since some operations
+; might generate integer overflows even when the dividend is divided by
+; 4 (when the high part of the shifted down dividend ends up being exactly
+; half of the divisor, the result is the quotient 0x80000000, which is
+; negative...) it needs to be divided by 8. Furthermore, the divisor needs
+; to be divided by 2 (unsigned) as well, to avoid more problems with the sign.
+; In this case, a little extra fiddling with the remainder is required.
;
-; a' = l & 1
-; a2 = <h,l> >> 1 # UNSIGNED shift!
-; b' = d & 1
-; b2 = d >> 1 # UNSIGNED shift!
+; So, the simplest way to handle this is always to divide the dividend
+; by 8, and to divide the divisor by 2 if it's highest bit is set.
+; After EDIV has been used, the quotient gets multiplied by 8 if the
+; original divisor was positive, otherwise 4. The remainder, oddly
+; enough, is *always* multiplied by 8.
+; NOTE: in the case mentioned above, where the high part of the shifted
+; down dividend ends up being exactly half the shifted down divisor, we
+; end up with a 33 bit quotient. That's no problem however, it usually
+; means we have ended up with a too large remainder as well, and the
+; problem is fixed by the last part of the algorithm (next paragraph).
;
-; Now, use EDIV to calculate a quotient and a remainder:
+; The routine ends with comparing the resulting remainder with the
+; original divisor and if the remainder is larger, subtract the
+; original divisor from it, and increase the quotient by 1. This is
+; done until the remainder is smaller than the divisor.
;
-; q'' = a2/b2
-; r'' = a2 - q''*b2
+; The complete algorithm looks like this:
;
-; If b' is 0, the quotient is already correct, we just need to adjust the
-; remainder:
+; d' = d
+; l' = l & 7
+; [h,l] = [h,l] >> 3
+; [q,r] = floor([h,l] / d) # This is the EDIV operation
+; if (q < 0) q = -q # I doubt this is necessary any more
;
-; if (b' == 0)
-; {
-; r = 2*r'' + a'
-; q = q''
-; }
-;
-; If b' is 1, we need to do other adjustements. The first thought is the
-; following (note that r' will not always have the right value, but an
-; adjustement follows further down):
-;
-; if (b' == 1)
-; {
-; q' = q''
-; r' = a - q'*b
-;
-; However, one can note the folowing relationship:
-;
-; r'' = a2 - q''*b2
-; => 2*r'' = 2*a2 - 2*q''*b2
-; = { a = 2*a2 + a', b = 2*b2 + b' = 2*b2 + 1,
-; q' = q'' }
-; = a - a' - q'*(b - 1)
-; = a - q'*b - a' + q'
-; = r' - a' + q'
-; => r' = 2*r'' - q' + a'
+; r' = r >> 29
+; if (d' >= 0)
+; q' = q >> 29
+; q = q << 3
+; else
+; q' = q >> 30
+; q = q << 2
+; r = (r << 3) + l'
;
-; This enables us to use r'' instead of discarding and calculating another
-; modulo:
-;
-; if (b' == 1)
+; if (d' < 0)
; {
-; q' = q''
-; r' = (r'' << 1) - q' + a'
-;
-; Now, all we have to do is adjust r', because it might be < 0:
-;
-; while (r' < 0)
+; [r',r] = [r',r] - q
+; while ([r',r] < 0)
; {
-; r' = r' + b
-; q' = q' - 1
+; [r',r] = [r',r] + d
+; [q',q] = [q',q] - 1
; }
; }
;
-; return q'
+; while ([r',r] >= d')
+; {
+; [r',r] = [r',r] - d'
+; [q',q] = [q',q] + 1
+; }
+;
+; return q
h=4 ;(AP) h by value (input)
l=8 ;(AP) l by value (input)
d=12 ;(AP) d by value (input)
-;aprim=r5
-;a2=r6
-;a20=r6
-;a21=r7
-;bprim=r8
-;b2=r9
-;qprim=r10 ; initially used as q''
-;rprim=r11 ; initially used as r''
-
+;r2 = l, q
+;r3 = h, r
+;r4 = d
+;r5 = l'
+;r6 = r'
+;r7 = d'
+;r8 = q'
.psect code,nowrt
-.entry bn_div_words,^m<r2,r3,r4,r5,r6,r7,r8,r9,r10,r11>
+.entry bn_div_words,^m<r2,r3,r4,r5,r6,r7,r8>
movl l(ap),r2
movl h(ap),r3
movl d(ap),r4
- movl #0,r5
- movl #0,r8
- movl #0,r0
-; movl #0,r1
+ bicl3 #^XFFFFFFF8,r2,r5 ; l' = l & 7
+ bicl3 #^X00000007,r2,r2
- rotl #-1,r2,r6 ; a20 = l >> 1 (almost)
- rotl #-1,r3,r7 ; a21 = h >> 1 (almost)
- rotl #-1,r4,r9 ; b2 = d >> 1 (almost)
+ bicl3 #^XFFFFFFF8,r3,r6
+ bicl3 #^X00000007,r3,r3
+
+ addl r6,r2
- tstl r6
- bgeq 1$
- xorl2 #^X80000000,r6 ; fixup a20 so highest bit is 0
- incl r5 ; a' = 1
-1$:
- tstl r7
- bgeq 2$
- xorl2 #^X80000000,r6 ; fixup a20 so highest bit is 1,
- ; since that's what was lowest in a21
- xorl2 #^X80000000,r7 ; fixup a21 so highest bit is 1
-2$:
- tstl r9
+ rotl #-3,r2,r2 ; l = l >> 3
+ rotl #-3,r3,r3 ; h = h >> 3
+
+ movl r4,r7 ; d' = d
+
+ movl #0,r6 ; r' = 0
+ movl #0,r8 ; q' = 0
+
+ tstl r4
beql 666$ ; Uh-oh, the divisor is 0...
- bgtr 3$
- xorl2 #^X80000000,r9 ; fixup b2 so highest bit is 0
- incl r8 ; b' = 1
-3$:
- tstl r9
- bneq 4$ ; if b2 is 0, we know that b' is 1
- tstl r3
- bneq 666$ ; if higher half isn't 0, we overflow
- movl r2,r10 ; otherwise, we have our result
- brb 42$ ; This is a success, really.
-4$:
- ediv r9,r6,r10,r11
-
- tstl r8
- bneq 5$ ; If b' != 0, go to the other part
-; addl3 r11,r11,r1
-; addl2 r5,r1
- brb 42$
-5$:
- ashl #1,r11,r11
- subl2 r10,r11
- addl2 r5,r11
- bgeq 7$
-6$:
- decl r10
- addl2 r4,r11
- blss 6$
-7$:
-; movl r11,r1
+ bgtr 1$
+ rotl #-1,r4,r4 ; If d is negative, shift it right.
+ bicl2 #^X80000000,r4 ; Since d is then a large number, the
+ ; lowest bit is insignificant
+ ; (contradict that, and I'll fix the problem!)
+1$:
+ ediv r4,r2,r2,r3 ; Do the actual division
+
+ tstl r2
+ bgeq 3$
+ mnegl r2,r2 ; if q < 0, negate it
+3$:
+ tstl r7
+ blss 4$
+ rotl #3,r2,r2 ; q = q << 3
+ bicl3 #^XFFFFFFF8,r2,r8 ; q' gets the high bits from q
+ bicl3 #^X00000007,r2,r2
+ bsb 41$
+4$: ; else
+ rotl #2,r2,r2 ; q = q << 2
+ bicl3 #^XFFFFFFFC,r2,r8 ; q' gets the high bits from q
+ bicl3 #^X00000003,r2,r2
+41$:
+ rotl #3,r3,r3 ; r = r << 3
+ bicl3 #^XFFFFFFF8,r3,r6 ; r' gets the high bits from r
+ bicl3 #^X00000007,r3,r3
+ addl r5,r3 ; r = r + l'
+
+ tstl r7
+ bgeq 5$
+ bitl #1,r7
+ beql 5$ ; if d' < 0 && d' & 1
+ subl r2,r3 ; [r',r] = [r',r] - [q',q]
+ sbwc r8,r6
+45$:
+ bgeq 5$ ; while r < 0
+ decl r2 ; [q',q] = [q',q] - 1
+ sbwc #0,r8
+ addl r7,r3 ; [r',r] = [r',r] + d'
+ adwc #0,r6
+ brb 45$
+
+; The return points are placed in the middle to keep a short distance from
+; all the branch points
42$:
- movl r10,r0
+; movl r3,r1
+ movl r2,r0
+ ret
666$:
+ movl #^XFFFFFFFF,r0
ret
+
+5$:
+ tstl r6
+ bneq 6$
+ cmpl r3,r7
+ blssu 42$ ; while [r',r] >= d'
+6$:
+ subl r7,r3 ; [r',r] = [r',r] - d'
+ sbwc #0,r6
+ incl r2 ; [q',q] = [q',q] + 1
+ adwc #0,r8
+ brb 5$
\f
.title vax_bn_add_words unsigned add of two arrays
;
projects / openssl.git / blobdiff