1 files changed, 141 insertions, 113 deletions

diff --git a/src/lib/libcrypto/bn/asm/vms.mar b/src/lib/libcrypto/bn/asm/vms.mar
index 465f2774b6..aefab15cdb 100644
--- a/src/lib/libcrypto/bn/asm/vms.mar
+++ b/src/lib/libcrypto/bn/asm/vms.mar
@@ -1,4 +1,4 @@
-        .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
 ;
@@ -59,7 +59,7 @@ w=16 ;(AP)	w	by value (input)
         movl    r6,r0                   ; return c
         ret
 
-        .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
 ;
@@ -172,147 +172,175 @@ n=12 ;(AP)	n	by value (input)
 ; }
 ;
 ; 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$:
+        subl    r7,r3           ;   [r',r] = [r',r] - d'
+        sbwc    #0,r6
+        incl    r2              ;   [q',q] = [q',q] + 1
+        adwc    #0,r8
+        brb     5$      
 
         .title  vax_bn_add_words  unsigned add of two arrays
 ;