; }
;
; Using EDIV would be very easy, if it didn't do signed calculations.
-; Any time, any of the input numbers are signed, there are problems,
+; 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).
;
; 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. This method works as long as the
-; divisor is positive, so we'll keep that (with a small adjustment)
-; as the main method.
-; 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, it needs to be divided by 4. 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.
+; 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.
;
; So, the simplest way to handle this is always to divide the dividend
-; by 4, and to divide the divisor by 2 if it's highest bit is set.
-; After EDIV has been used, the quotient gets multiplied by 4 if the
-; original divisor was positive, otherwise 2. The remainder, oddly
-; enough, is *always* multiplied by 4.
+; 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).
;
; The routine ends with comparing the resulting remainder with the
; original divisor and if the remainder is larger, subtract the
; The complete algorithm looks like this:
;
; d' = d
-; l' = l & 3
-; [h,l] = [h,l] >> 2
+; 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
;
-; r' = r >> 30
-; if (d' >= 0) q = q << 1
-; q = q << 1
-; r = (r << 2) + l'
+; 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)
; {
; while ([r',r] < 0)
; {
; [r',r] = [r',r] + d
-; q = q - 1
+; [q',q] = [q',q] - 1
; }
; }
;
-; while ([r',r] >= d)
+; while ([r',r] >= d')
; {
-; [r',r] = [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
-;dprim=r7
-
+;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>
+.entry bn_div_words,^m<r2,r3,r4,r5,r6,r7,r8>
movl l(ap),r2
movl h(ap),r3
movl d(ap),r4
- bicl3 #^XFFFFFFFC,r2,r5 ; l' = l & 3
- bicl3 #^X00000003,r2,r2
+ bicl3 #^XFFFFFFF8,r2,r5 ; l' = l & 7
+ bicl3 #^X00000007,r2,r2
- bicl3 #^XFFFFFFFC,r3,r6
- bicl3 #^X00000003,r3,r3
+ bicl3 #^XFFFFFFF8,r3,r6
+ bicl3 #^X00000007,r3,r3
addl r6,r2
- rotl #-2,r2,r2 ; l = l >> 2
- rotl #-2,r3,r3 ; h = h >> 2
+
+ rotl #-3,r2,r2 ; l = l >> 3
+ rotl #-3,r3,r3 ; h = h >> 3
- movl #0,r6
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 1$
3$:
tstl r7
blss 4$
- ashl #1,r2,r2 ; if d' >= 0, q = q << 1
-4$:
- ashl #1,r2,r2 ; q = q << 1
- rotl #2,r3,r3 ; r = r << 2
- bicl3 #^XFFFFFFFC,r3,r6 ; r' gets the high bits from r
- bicl3 #^X00000003,r3,r3
+ rotl #3,r2,r2 ; q = q << 3
+ bicl3 #^XFFFFFFF8,r2,r8 ; q' gets the high bits from q
+ bicl3 #^X00000007,r2,r2
+ bsb 41$
+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
- sbwc #0,r6
+ subl r2,r3 ; [r',r] = [r',r] - [q',q]
+ sbwc r8,r6
45$:
bgeq 5$ ; while r < 0
- decl r2 ; q = q - 1
- addl r7,r3 ; [r',r] = [r',r] + d
+ 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$:
- subl r7,r3 ; [r',r] = [r',r] - d
+ subl r7,r3 ; [r',r] = [r',r] - d'
sbwc #0,r6
- incl r2 ; q = q + 1
+ incl r2 ; [q',q] = [q',q] + 1
+ adwc #0,r8
brb 5$
-42$:
-; movl r3,r1
- movl r2,r0
- ret
-666$:
- movl #^XFFFFFFFF,r0
- ret
\f
.title vax_bn_add_words unsigned add of two arrays
;
projects / openssl / commitdiff