diff options
author | Richard Levitte <levitte@openssl.org> | 2002-12-23 11:25:51 +0000 |
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committer | Richard Levitte <levitte@openssl.org> | 2002-12-23 11:25:51 +0000 |
commit | e9883d285dd6d4fe172eddf6588c3387d11b8c47 (patch) | |
tree | eb57f5f82d03c4e31192c698e5197e42e82544f8 /crypto | |
parent | 8d6ad9e39d3e104427c133ac45ce05f30da9f9a3 (diff) |
Finally, a bn_div_words() in VAX assembler that goes through all tests.
PR: 413
Diffstat (limited to 'crypto')
-rw-r--r-- | crypto/bn/asm/vms.mar | 136 |
1 files changed, 81 insertions, 55 deletions
diff --git a/crypto/bn/asm/vms.mar b/crypto/bn/asm/vms.mar index 8278adffd3..aefab15cdb 100644 --- a/crypto/bn/asm/vms.mar +++ b/crypto/bn/asm/vms.mar @@ -172,7 +172,7 @@ n=12 ;(AP) n by value (input) ; } ; ; 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). ; @@ -180,21 +180,26 @@ n=12 ;(AP) n by value (input) ; 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 @@ -204,15 +209,19 @@ n=12 ;(AP) n by value (input) ; 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) ; { @@ -220,14 +229,14 @@ n=12 ;(AP) n by value (input) ; 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 @@ -236,31 +245,37 @@ h=4 ;(AP) h by value (input) 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$ @@ -277,44 +292,55 @@ d=12 ;(AP) d by value (input) 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 .title vax_bn_add_words unsigned add of two arrays ; |