; }
;
; 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. 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, the divisor is so large,
+; from an unsigned point of view, that the dropped lowest bit is
+; insignificant for the operation, and therefore doesn't need
+; bothering with. The remainder might end up incorrect, bit that's
+; adjusted at the end of the routine anyway.
;
-; 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 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.
;
-; 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 & 3
+; [h,l] = [h,l] >> 2
+; [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'
-;
-; This enables us to use r'' instead of discarding and calculating another
-; modulo:
+; r' = r >> 30
+; if (d' > 0) q = q << 1
+; q = q << 1
+; r = (r << 2) + l'
;
-; if (b' == 1)
+; while ([r',r] >= d)
; {
-; 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' + b
-; q' = q' - 1
-; }
+; [r',r] = [r',r] - d
+; q = q + 1
; }
;
-; return q'
+; 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''
+;lprim=r5
+;rprim=r6
+;dprim=r7
.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>
movl l(ap),r2
movl h(ap),r3
movl d(ap),r4
- movl #0,r5
- movl #0,r8
- movl #0,r0
-; movl #0,r1
-
- rotl #-1,r2,r6 ; a20 = l >> 1 (almost)
- rotl #-1,r3,r7 ; a21 = h >> 1 (almost)
- rotl #-1,r4,r9 ; b2 = d >> 1 (almost)
-
- tstl r6
- bgeq 1$
- xorl2 #^X80000000,r6 ; fixup a20 so highest bit is 0
- incl r5 ; a' = 1
-1$:
+ bicl3 #^XFFFFFFFC,r2,r5 ; l' = l & 3
+ bicl3 #^X00000003,r2,r2
+
+ bicl3 #^XFFFFFFFC,r3,r6
+ bicl3 #^X00000003,r3,r3
+
+ addl r6,r2
+ rotl #-2,r2,r2 ; l = l >> 2
+ rotl #-2,r3,r3 ; h = h >> 2
+
+ movl #0,r6
+ movl r4,r7 ; d' = d
+
+ tstl r4
+ beql 666$ ; 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$:
tstl r7
- bgeq 2$
- 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$:
- tstl r9
- beq 666$ ; 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$
+ blss 4$
+ ashl #1,r2,r2 ; q = q << 1
+4$:
+ ashl #1,r2,r2 ; q = q << 1
+ rotl #2,r3,r3 ; r = r << 2
+ bicl3 #^XFFFFFFFC,r3,r6 ; r' gets the high bits from r
+ bicl3 #^X00000003,r3,r3
+ addl r5,r3 ; r = r + l'
+
5$:
- ashl #1,r11,r11
- subl2 r10,r11
- addl2 r5,r11
- bgeq 7$
+ tstl r6
+ bneq 6$
+ cmpl r3,r7
+ blssu 42$ ; while [r',r] >= d'
6$:
- decl r10
- addl2 r4,r11
- blss 6$
-7$:
-; movl r11,r1
+ subl r7,r3 ; r = r - d
+ sbwc #0,r6
+ incl r2 ; q = q + 1
+ brb 5$
42$:
- movl r10,r0
+; movl r3,r1
+ movl r2,r0
+ ret
666$:
+ movl #^XFFFFFFFF,r0
ret
\f
.title vax_bn_add_words unsigned add of two arrays
projects / openssl.git / blobdiff