- .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.
-; It doesn't accept a signed dividend, but accepts a signed divisor.
-; So, shifting down the dividend right one bit makes it positive, and
-; just makes us lose the lowest bit, which can be used afterwards as
-; an addition to the remainder. All that needs to be done at the end
-; is a little bit of fiddling; shifting both quotient and remainder
-; one step to the left, and deal with the situation when the remainder
-; ends up being larger than the divisor.
+; 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).
;
-; We end up doing something like this:
+; 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.
;
-; l' = l & 1
-; [h,l] = [h,l] >> 1
-; [q,r] = floor([h,l] / d)
-; if (q < 0) q = -q # Because EDIV thought d was negative
+; 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, we need to adjust back by multiplying quotient and remainder with 2,
-; and add the bit that dropped out when dividing by 2:
+; 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.
;
-; r' = r & 0x80000000
-; q = q << 1
-; r = (r << 1) + a'
+; The complete algorithm looks like this:
;
-; And now, the final adjustment if the remainder happens to get larger than
-; the divisor:
+; 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 (r')
+; r' = r >> 29
+; if (d' >= 0)
+; q' = q >> 29
+; q = q << 3
+; else
+; q' = q >> 30
+; q = q << 2
+; r = (r << 3) + l'
+;
+; if (d' < 0)
; {
-; r = r - d
-; q = q + 1
+; [r',r] = [r',r] - q
+; while ([r',r] < 0)
+; {
+; [r',r] = [r',r] + d
+; [q',q] = [q',q] - 1
+; }
; }
-; while (r > d)
+;
+; while ([r',r] >= d')
; {
-; r = r - d
-; q = q + 1
+; [r',r] = [r',r] - d'
+; [q',q] = [q',q] + 1
; }
;
; return q
l=8 ;(AP) l by value (input)
d=12 ;(AP) d by value (input)
-;lprim=r5
-;rprim=r6
-
+;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>
+.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,r6
+ bicl3 #^XFFFFFFF8,r2,r5 ; l' = l & 7
+ bicl3 #^X00000007,r2,r2
- rotl #-1,r2,r2 ; l = l >> 1 (almost)
- rotl #-1,r3,r3 ; h = h >> 1 (almost)
+ bicl3 #^XFFFFFFF8,r3,r6
+ bicl3 #^X00000007,r3,r3
+
+ addl r6,r2
+
+ 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 r2
- bgeq 1$
- xorl2 #^X80000000,r2 ; fixup l so highest bit is 0
- incl r5 ; l' = 1
-1$:
- tstl r3
- bgeq 2$
- xorl2 #^X80000000,r2 ; fixup l so highest bit is 1,
- ; since that's what was lowest in h
- xorl2 #^X80000000,r3 ; fixup h so highest bit is 0
-2$:
tstl r4
beql 666$ ; Uh-oh, the divisor is 0...
-
+ 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$:
- tstl r3
- bgeq 4$
- incl r6 ; since the high bit in r is set, set rprim
-4$:
- ashl #1,r2,r2
- ashl #1,r3,r3
- addl r5,r3
-
- tstl r6
- beql 5$
- subl r4,r3
- incl r2
-5$:
- cmpl r3,r4
- blequ 42$
- subl r4,r3
- incl r2
- brb 5$
+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$:
; movl r3,r1
movl r2,r0
+ ret
666$:
+ 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
;