diff options
| author | Richard Levitte <levitte@openssl.org> | 2002-12-02 21:31:45 +0000 |
|---|---|---|
| committer | Richard Levitte <levitte@openssl.org> | 2002-12-02 21:31:45 +0000 |
| commit | 6ab285bf4c9e0f48f3e8af1ad952cd5224c7d461 (patch) | |
| tree | 25216a96a96ce24fe750dbba893fc9abe69bc2ea /crypto/bn/asm/vms.mar | |
| parent | 1d3159bccaa400d6966005b9bc49cca1f6719962 (diff) | |
I think I got it now. Apparently, the case of having to shift down
the divisor was a bit more complex than I first saw. The lost bit
can't just be discarded, as there are cases where it is important.
For example, look at dividing 320000 with 80000 vs. 80001 (all
decimals), the difference is crucial. The trick here is to check if
that lost bit was 1, and in that case, do the following:
1. subtract the quotient from the remainder
2. as long as the remainder is negative, add the divisor (the whole
divisor, not the shofted down copy) to it, and decrease the
quotient by one.
There's probably a nice mathematical proof for this already, but I
won't bother with that, unless someone requests it from me.
Diffstat (limited to 'crypto/bn/asm/vms.mar')
| -rw-r--r-- | crypto/bn/asm/vms.mar | 34 |
1 files changed, 27 insertions, 7 deletions
diff --git a/crypto/bn/asm/vms.mar b/crypto/bn/asm/vms.mar index 6c1665a9be..2a752489f5 100644 --- a/crypto/bn/asm/vms.mar +++ b/crypto/bn/asm/vms.mar @@ -187,11 +187,8 @@ n=12 ;(AP) n by value (input) ; 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. +; 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. @@ -213,10 +210,20 @@ n=12 ;(AP) n by value (input) ; if (q < 0) q = -q # I doubt this is necessary any more ; ; r' = r >> 30 -; if (d' > 0) q = q << 1 +; if (d' >= 0) q = q << 1 ; q = q << 1 ; r = (r << 2) + l' ; +; if (d' < 0) +; { +; [r',r] = [r',r] - q +; while ([r',r] < 0) +; { +; [r',r] = [r',r] + d +; q = q - 1 +; } +; } +; ; while ([r',r] >= d) ; { ; [r',r] = [r',r] - d @@ -278,13 +285,26 @@ d=12 ;(AP) d by value (input) bicl3 #^X00000003,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 +45$: + bgeq 5$ ; while r < 0 + decl r2 ; q = q - 1 + addl r7,r3 ; [r',r] = [r',r] + d + adwc #0,r6 + brb 45$ + 5$: tstl r6 bneq 6$ cmpl r3,r7 blssu 42$ ; while [r',r] >= d' 6$: - subl r7,r3 ; r = r - d + subl r7,r3 ; [r',r] = [r',r] - d sbwc #0,r6 incl r2 ; q = q + 1 brb 5$ |