Through some experimentation and thinking, I think I finally got the

[openssl.git] / crypto / bn / asm / vms.mar

diff --git a/crypto/bn/asm/vms.mar b/crypto/bn/asm/vms.mar
index a2ac41a556ee6a267c505c5e27fd890266f343be..402ff87aef706c90cd17b8ad129a86f6f3ea60b4 100644 (file)
--- a/crypto/bn/asm/vms.mar
+++ b/crypto/bn/asm/vms.mar
@@ -172,39 +172,54 @@ n=12 ;(AP)        n       by value (input)
 ; }
 ;
 ; Using EDIV would be very easy, if it didn't do signed calculations.
 ; }
 ;
 ; 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.  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.

 ;
 ;

-; 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 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, 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 & 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 (r')
-;   {
-;     r = r - d
-;     q = q + 1
-;   }
-; while (r >= d)
+; r'    = r >> 30
+; if (d' > 0) q = q << 1
+; q     = q << 1
+; r     = (r << 2) + l'
+;
+; while ([r',r] >= d)

 ;   {
 ;   {

-;     r = r - d
+;     [r',r] = [r',r] - d

 ;     q = q + 1
 ;   }
 ;
 ;     q = q + 1
 ;   }
 ;


@@ -216,63 +231,69 @@ d=12 ;(AP)        d       by value (input)
 
 ;lprim=r5
 ;rprim=r6
 
 ;lprim=r5
 ;rprim=r6

+;dprim=r7

 
 
        .psect  code,nowrt
 
 
 
        .psect  code,nowrt
 

-.entry bn_div_words,^m<r2,r3,r4,r5,r6>
+.entry bn_div_words,^m<r2,r3,r4,r5,r6,r7>

        movl    l(ap),r2
        movl    h(ap),r3
        movl    d(ap),r4
 
        movl    l(ap),r2
        movl    h(ap),r3
        movl    d(ap),r4
 

-       movl    #0,r5
-       movl    #0,r6
+       bicl3   #^XFFFFFFFC,r2,r5 ; l' = l & 3
+       bicl3   #^X00000003,r2,r2

 
 

-       rotl    #-1,r2,r2       ; l = l >> 1 (almost)
-       rotl    #-1,r3,r3       ; h = h >> 1 (almost)
+       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    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...
        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
        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 r'
-4$:
+3$:     
+       tstl    r7
+       blss    4$

        ashl    #1,r2,r2        ; q = q << 1
        ashl    #1,r2,r2        ; q = q << 1

-       ashl    #1,r3,r3        ; r = r << 1
-       addl    r5,r3           ; r = r + a'
+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'

 
 

-       tstl    r6
-       beql    5$              ; if r'
-       subl    r4,r3           ;   r = r - d
-       incl    r2              ;   q = q + 1

 5$:
 5$:

-       cmpl    r3,r4
-       blssu   42$             ; while r >= d
-       subl    r4,r3           ;   r = r - d
+       tstl    r6
+       bneq    6$
+       cmpl    r3,r7
+       blssu   42$             ; while [r',r] >= d'
+6$:
+       subl    r7,r3           ;   r = r - d
+       sbwc    #0,r6

        incl    r2              ;   q = q + 1
        brb     5$      
 42$:
 ;      movl    r3,r1
        movl    r2,r0
        incl    r2              ;   q = q + 1
        brb     5$      
 42$:
 ;      movl    r3,r1
        movl    r2,r0

+       ret

 666$:
 666$:

+       movl    #^XFFFFFFFF,r0

        ret
 \f
        .title  vax_bn_add_words  unsigned add of two arrays
        ret
 \f
        .title  vax_bn_add_words  unsigned add of two arrays