2 * Bignum routines for RSA and DH and stuff.
14 * * Do not call the DIVMOD_WORD macro with expressions such as array
15 * subscripts, as some implementations object to this (see below).
16 * * Note that none of the division methods below will cope if the
17 * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
19 * If this condition occurs, in the case of the x86 DIV instruction,
20 * an overflow exception will occur, which (according to a correspondent)
21 * will manifest on Windows as something like
22 * 0xC0000095: Integer overflow
23 * The C variant won't give the right answer, either.
26 #if defined __GNUC__ && defined __i386__
27 typedef unsigned long BignumInt;
28 typedef unsigned long long BignumDblInt;
29 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
30 #define BIGNUM_TOP_BIT 0x80000000UL
31 #define BIGNUM_INT_BITS 32
32 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
33 #define DIVMOD_WORD(q, r, hi, lo, w) \
35 "=d" (r), "=a" (q) : \
36 "r" (w), "d" (hi), "a" (lo))
37 #elif defined _MSC_VER && defined _M_IX86
38 typedef unsigned __int32 BignumInt;
39 typedef unsigned __int64 BignumDblInt;
40 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
41 #define BIGNUM_TOP_BIT 0x80000000UL
42 #define BIGNUM_INT_BITS 32
43 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
44 /* Note: MASM interprets array subscripts in the macro arguments as
45 * assembler syntax, which gives the wrong answer. Don't supply them.
46 * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
47 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
55 /* 64-bit architectures can do 32x32->64 chunks at a time */
56 typedef unsigned int BignumInt;
57 typedef unsigned long BignumDblInt;
58 #define BIGNUM_INT_MASK 0xFFFFFFFFU
59 #define BIGNUM_TOP_BIT 0x80000000U
60 #define BIGNUM_INT_BITS 32
61 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
62 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
63 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
68 /* 64-bit architectures in which unsigned long is 32 bits, not 64 */
69 typedef unsigned long BignumInt;
70 typedef unsigned long long BignumDblInt;
71 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
72 #define BIGNUM_TOP_BIT 0x80000000UL
73 #define BIGNUM_INT_BITS 32
74 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
75 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
76 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
81 /* Fallback for all other cases */
82 typedef unsigned short BignumInt;
83 typedef unsigned long BignumDblInt;
84 #define BIGNUM_INT_MASK 0xFFFFU
85 #define BIGNUM_TOP_BIT 0x8000U
86 #define BIGNUM_INT_BITS 16
87 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
88 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
89 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
95 #define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
97 #define BIGNUM_INTERNAL
98 typedef BignumInt *Bignum;
102 BignumInt bnZero[1] = { 0 };
103 BignumInt bnOne[2] = { 1, 1 };
106 * The Bignum format is an array of `BignumInt'. The first
107 * element of the array counts the remaining elements. The
108 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
109 * significant digit first. (So it's trivial to extract the bit
110 * with value 2^n for any n.)
112 * All Bignums in this module are positive. Negative numbers must
113 * be dealt with outside it.
115 * INVARIANT: the most significant word of any Bignum must be
119 Bignum Zero = bnZero, One = bnOne;
121 static Bignum newbn(int length)
123 Bignum b = snewn(length + 1, BignumInt);
126 memset(b, 0, (length + 1) * sizeof(*b));
131 void bn_restore_invariant(Bignum b)
133 while (b[0] > 1 && b[b[0]] == 0)
137 Bignum copybn(Bignum orig)
139 Bignum b = snewn(orig[0] + 1, BignumInt);
142 memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
146 void freebn(Bignum b)
149 * Burn the evidence, just in case.
151 smemclr(b, sizeof(b[0]) * (b[0] + 1));
155 Bignum bn_power_2(int n)
157 Bignum ret = newbn(n / BIGNUM_INT_BITS + 1);
158 bignum_set_bit(ret, n, 1);
163 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
164 * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
167 static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
168 BignumInt *c, int len)
171 BignumDblInt carry = 0;
173 for (i = len-1; i >= 0; i--) {
174 carry += (BignumDblInt)a[i] + b[i];
175 c[i] = (BignumInt)carry;
176 carry >>= BIGNUM_INT_BITS;
179 return (BignumInt)carry;
183 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
184 * all big-endian arrays of 'len' BignumInts. Any borrow from the top
187 static void internal_sub(const BignumInt *a, const BignumInt *b,
188 BignumInt *c, int len)
191 BignumDblInt carry = 1;
193 for (i = len-1; i >= 0; i--) {
194 carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
195 c[i] = (BignumInt)carry;
196 carry >>= BIGNUM_INT_BITS;
202 * Input is in the first len words of a and b.
203 * Result is returned in the first 2*len words of c.
205 * 'scratch' must point to an array of BignumInt of size at least
206 * mul_compute_scratch(len). (This covers the needs of internal_mul
207 * and all its recursive calls to itself.)
209 #define KARATSUBA_THRESHOLD 50
210 static int mul_compute_scratch(int len)
213 while (len > KARATSUBA_THRESHOLD) {
214 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
215 int midlen = botlen + 1;
221 static void internal_mul(const BignumInt *a, const BignumInt *b,
222 BignumInt *c, int len, BignumInt *scratch)
224 if (len > KARATSUBA_THRESHOLD) {
228 * Karatsuba divide-and-conquer algorithm. Cut each input in
229 * half, so that it's expressed as two big 'digits' in a giant
235 * Then the product is of course
237 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
239 * and we compute the three coefficients by recursively
240 * calling ourself to do half-length multiplications.
242 * The clever bit that makes this worth doing is that we only
243 * need _one_ half-length multiplication for the central
244 * coefficient rather than the two that it obviouly looks
245 * like, because we can use a single multiplication to compute
247 * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
249 * and then we subtract the other two coefficients (a_1 b_1
250 * and a_0 b_0) which we were computing anyway.
252 * Hence we get to multiply two numbers of length N in about
253 * three times as much work as it takes to multiply numbers of
254 * length N/2, which is obviously better than the four times
255 * as much work it would take if we just did a long
256 * conventional multiply.
259 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
260 int midlen = botlen + 1;
267 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
268 * in the output array, so we can compute them immediately in
273 printf("a1,a0 = 0x");
274 for (i = 0; i < len; i++) {
275 if (i == toplen) printf(", 0x");
276 printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
279 printf("b1,b0 = 0x");
280 for (i = 0; i < len; i++) {
281 if (i == toplen) printf(", 0x");
282 printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
288 internal_mul(a, b, c, toplen, scratch);
291 for (i = 0; i < 2*toplen; i++) {
292 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
298 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
301 for (i = 0; i < 2*botlen; i++) {
302 printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
307 /* Zero padding. midlen exceeds toplen by at most 2, so just
308 * zero the first two words of each input and the rest will be
310 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
312 for (i = 0; i < toplen; i++) {
313 scratch[midlen - toplen + i] = a[i]; /* a_1 */
314 scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
317 /* compute a_1 + a_0 */
318 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
320 printf("a1plusa0 = 0x");
321 for (i = 0; i < midlen; i++) {
322 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
326 /* compute b_1 + b_0 */
327 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
328 scratch+midlen+1, botlen);
330 printf("b1plusb0 = 0x");
331 for (i = 0; i < midlen; i++) {
332 printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
338 * Now we can do the third multiplication.
340 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
343 printf("a1plusa0timesb1plusb0 = 0x");
344 for (i = 0; i < 2*midlen; i++) {
345 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
351 * Now we can reuse the first half of 'scratch' to compute the
352 * sum of the outer two coefficients, to subtract from that
353 * product to obtain the middle one.
355 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
356 for (i = 0; i < 2*toplen; i++)
357 scratch[2*midlen - 2*toplen + i] = c[i];
358 scratch[1] = internal_add(scratch+2, c + 2*toplen,
359 scratch+2, 2*botlen);
361 printf("a1b1plusa0b0 = 0x");
362 for (i = 0; i < 2*midlen; i++) {
363 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
368 internal_sub(scratch + 2*midlen, scratch,
369 scratch + 2*midlen, 2*midlen);
371 printf("a1b0plusa0b1 = 0x");
372 for (i = 0; i < 2*midlen; i++) {
373 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
379 * And now all we need to do is to add that middle coefficient
380 * back into the output. We may have to propagate a carry
381 * further up the output, but we can be sure it won't
382 * propagate right the way off the top.
384 carry = internal_add(c + 2*len - botlen - 2*midlen,
386 c + 2*len - botlen - 2*midlen, 2*midlen);
387 i = 2*len - botlen - 2*midlen - 1;
391 c[i] = (BignumInt)carry;
392 carry >>= BIGNUM_INT_BITS;
397 for (i = 0; i < 2*len; i++) {
398 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
407 const BignumInt *ap, *bp;
411 * Multiply in the ordinary O(N^2) way.
414 for (i = 0; i < 2 * len; i++)
417 for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
419 for (cp = cps, bp = b + len; cp--, bp-- > b ;) {
420 t = (MUL_WORD(*ap, *bp) + carry) + *cp;
422 carry = (BignumInt)(t >> BIGNUM_INT_BITS);
430 * Variant form of internal_mul used for the initial step of
431 * Montgomery reduction. Only bothers outputting 'len' words
432 * (everything above that is thrown away).
434 static void internal_mul_low(const BignumInt *a, const BignumInt *b,
435 BignumInt *c, int len, BignumInt *scratch)
437 if (len > KARATSUBA_THRESHOLD) {
441 * Karatsuba-aware version of internal_mul_low. As before, we
442 * express each input value as a shifted combination of two
448 * Then the full product is, as before,
450 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
452 * Provided we choose D on the large side (so that a_0 and b_0
453 * are _at least_ as long as a_1 and b_1), we don't need the
454 * topmost term at all, and we only need half of the middle
455 * term. So there's no point in doing the proper Karatsuba
456 * optimisation which computes the middle term using the top
457 * one, because we'd take as long computing the top one as
458 * just computing the middle one directly.
460 * So instead, we do a much more obvious thing: we call the
461 * fully optimised internal_mul to compute a_0 b_0, and we
462 * recursively call ourself to compute the _bottom halves_ of
463 * a_1 b_0 and a_0 b_1, each of which we add into the result
464 * in the obvious way.
466 * In other words, there's no actual Karatsuba _optimisation_
467 * in this function; the only benefit in doing it this way is
468 * that we call internal_mul proper for a large part of the
469 * work, and _that_ can optimise its operation.
472 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
475 * Scratch space for the various bits and pieces we're going
476 * to be adding together: we need botlen*2 words for a_0 b_0
477 * (though we may end up throwing away its topmost word), and
478 * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
483 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
487 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
491 internal_mul_low(a + len - toplen, b, scratch, toplen,
494 /* Copy the bottom half of the big coefficient into place */
495 for (i = 0; i < botlen; i++)
496 c[toplen + i] = scratch[2*toplen + botlen + i];
498 /* Add the two small coefficients, throwing away the returned carry */
499 internal_add(scratch, scratch + toplen, scratch, toplen);
501 /* And add that to the large coefficient, leaving the result in c. */
502 internal_add(scratch, scratch + 2*toplen + botlen - toplen,
509 const BignumInt *ap, *bp;
513 * Multiply in the ordinary O(N^2) way.
516 for (i = 0; i < len; i++)
519 for (cps = c + len, ap = a + len; ap-- > a; cps--) {
521 for (cp = cps, bp = b + len; bp--, cp-- > c ;) {
522 t = (MUL_WORD(*ap, *bp) + carry) + *cp;
524 carry = (BignumInt)(t >> BIGNUM_INT_BITS);
531 * Montgomery reduction. Expects x to be a big-endian array of 2*len
532 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
533 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
534 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
537 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
538 * each, containing respectively n and the multiplicative inverse of
541 * 'tmp' is an array of BignumInt used as scratch space, of length at
542 * least 3*len + mul_compute_scratch(len).
544 static void monty_reduce(BignumInt *x, const BignumInt *n,
545 const BignumInt *mninv, BignumInt *tmp, int len)
551 * Multiply x by (-n)^{-1} mod r. This gives us a value m such
552 * that mn is congruent to -x mod r. Hence, mn+x is an exact
553 * multiple of r, and is also (obviously) congruent to x mod n.
555 internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
558 * Compute t = (mn+x)/r in ordinary, non-modular, integer
559 * arithmetic. By construction this is exact, and is congruent mod
560 * n to x * r^{-1}, i.e. the answer we want.
562 * The following multiply leaves that answer in the _most_
563 * significant half of the 'x' array, so then we must shift it
566 internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
567 carry = internal_add(x, tmp+len, x, 2*len);
568 for (i = 0; i < len; i++)
569 x[len + i] = x[i], x[i] = 0;
572 * Reduce t mod n. This doesn't require a full-on division by n,
573 * but merely a test and single optional subtraction, since we can
574 * show that 0 <= t < 2n.
577 * + we computed m mod r, so 0 <= m < r.
578 * + so 0 <= mn < rn, obviously
579 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
580 * + yielding 0 <= (mn+x)/r < 2n as required.
583 for (i = 0; i < len; i++)
584 if (x[len + i] != n[i])
587 if (carry || i >= len || x[len + i] > n[i])
588 internal_sub(x+len, n, x+len, len);
591 static void internal_add_shifted(BignumInt *number,
592 unsigned n, int shift)
594 int word = 1 + (shift / BIGNUM_INT_BITS);
595 int bshift = shift % BIGNUM_INT_BITS;
598 addend = (BignumDblInt)n << bshift;
601 addend += number[word];
602 number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
603 addend >>= BIGNUM_INT_BITS;
610 * Input in first alen words of a and first mlen words of m.
611 * Output in first alen words of a
612 * (of which first alen-mlen words will be zero).
613 * The MSW of m MUST have its high bit set.
614 * Quotient is accumulated in the `quotient' array, which is a Bignum
615 * rather than the internal bigendian format. Quotient parts are shifted
616 * left by `qshift' before adding into quot.
618 static void internal_mod(BignumInt *a, int alen,
619 BignumInt *m, int mlen,
620 BignumInt *quot, int qshift)
627 assert(m0 >> (BIGNUM_INT_BITS-1) == 1);
633 for (i = 0; i <= alen - mlen; i++) {
635 unsigned int q, r, c, ai1;
649 /* Find q = h:a[i] / m0 */
654 * To illustrate it, suppose a BignumInt is 8 bits, and
655 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
656 * our initial division will be 0xA123 / 0xA1, which
657 * will give a quotient of 0x100 and a divide overflow.
658 * However, the invariants in this division algorithm
659 * are not violated, since the full number A1:23:... is
660 * _less_ than the quotient prefix A1:B2:... and so the
661 * following correction loop would have sorted it out.
663 * In this situation we set q to be the largest
664 * quotient we _can_ stomach (0xFF, of course).
668 /* Macro doesn't want an array subscript expression passed
669 * into it (see definition), so use a temporary. */
670 BignumInt tmplo = a[i];
671 DIVMOD_WORD(q, r, h, tmplo, m0);
673 /* Refine our estimate of q by looking at
674 h:a[i]:a[i+1] / m0:m1 */
676 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
679 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
680 if (r >= (BignumDblInt) m0 &&
681 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
685 /* Subtract q * m from a[i...] */
687 for (k = mlen - 1; k >= 0; k--) {
688 t = MUL_WORD(q, m[k]);
690 c = (unsigned)(t >> BIGNUM_INT_BITS);
691 if ((BignumInt) t > a[i + k])
693 a[i + k] -= (BignumInt) t;
696 /* Add back m in case of borrow */
699 for (k = mlen - 1; k >= 0; k--) {
702 a[i + k] = (BignumInt) t;
703 t = t >> BIGNUM_INT_BITS;
708 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
713 * Compute (base ^ exp) % mod, the pedestrian way.
715 Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
717 BignumInt *a, *b, *n, *m, *scratch;
719 int mlen, scratchlen, i, j;
723 * The most significant word of mod needs to be non-zero. It
724 * should already be, but let's make sure.
726 assert(mod[mod[0]] != 0);
729 * Make sure the base is smaller than the modulus, by reducing
730 * it modulo the modulus if not.
732 base = bigmod(base_in, mod);
734 /* Allocate m of size mlen, copy mod to m */
735 /* We use big endian internally */
737 m = snewn(mlen, BignumInt);
738 for (j = 0; j < mlen; j++)
739 m[j] = mod[mod[0] - j];
741 /* Shift m left to make msb bit set */
742 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
743 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
746 for (i = 0; i < mlen - 1; i++)
747 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
748 m[mlen - 1] = m[mlen - 1] << mshift;
751 /* Allocate n of size mlen, copy base to n */
752 n = snewn(mlen, BignumInt);
754 for (j = 0; j < i; j++)
756 for (j = 0; j < (int)base[0]; j++)
757 n[i + j] = base[base[0] - j];
759 /* Allocate a and b of size 2*mlen. Set a = 1 */
760 a = snewn(2 * mlen, BignumInt);
761 b = snewn(2 * mlen, BignumInt);
762 for (i = 0; i < 2 * mlen; i++)
766 /* Scratch space for multiplies */
767 scratchlen = mul_compute_scratch(mlen);
768 scratch = snewn(scratchlen, BignumInt);
770 /* Skip leading zero bits of exp. */
772 j = BIGNUM_INT_BITS-1;
773 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
777 j = BIGNUM_INT_BITS-1;
781 /* Main computation */
782 while (i < (int)exp[0]) {
784 internal_mul(a + mlen, a + mlen, b, mlen, scratch);
785 internal_mod(b, mlen * 2, m, mlen, NULL, 0);
786 if ((exp[exp[0] - i] & (1 << j)) != 0) {
787 internal_mul(b + mlen, n, a, mlen, scratch);
788 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
798 j = BIGNUM_INT_BITS-1;
801 /* Fixup result in case the modulus was shifted */
803 for (i = mlen - 1; i < 2 * mlen - 1; i++)
804 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
805 a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
806 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
807 for (i = 2 * mlen - 1; i >= mlen; i--)
808 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
811 /* Copy result to buffer */
812 result = newbn(mod[0]);
813 for (i = 0; i < mlen; i++)
814 result[result[0] - i] = a[i + mlen];
815 while (result[0] > 1 && result[result[0]] == 0)
818 /* Free temporary arrays */
819 smemclr(a, 2 * mlen * sizeof(*a));
821 smemclr(scratch, scratchlen * sizeof(*scratch));
823 smemclr(b, 2 * mlen * sizeof(*b));
825 smemclr(m, mlen * sizeof(*m));
827 smemclr(n, mlen * sizeof(*n));
836 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
837 * technique where possible, falling back to modpow_simple otherwise.
839 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
841 BignumInt *a, *b, *x, *n, *mninv, *scratch;
842 int len, scratchlen, i, j;
843 Bignum base, base2, r, rn, inv, result;
846 * The most significant word of mod needs to be non-zero. It
847 * should already be, but let's make sure.
849 assert(mod[mod[0]] != 0);
852 * mod had better be odd, or we can't do Montgomery multiplication
853 * using a power of two at all.
856 return modpow_simple(base_in, exp, mod);
859 * Make sure the base is smaller than the modulus, by reducing
860 * it modulo the modulus if not.
862 base = bigmod(base_in, mod);
865 * Compute the inverse of n mod r, for monty_reduce. (In fact we
866 * want the inverse of _minus_ n mod r, but we'll sort that out
870 r = bn_power_2(BIGNUM_INT_BITS * len);
871 inv = modinv(mod, r);
872 assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */
875 * Multiply the base by r mod n, to get it into Montgomery
878 base2 = modmul(base, r, mod);
882 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
884 freebn(r); /* won't need this any more */
887 * Set up internal arrays of the right lengths, in big-endian
888 * format, containing the base, the modulus, and the modulus's
891 n = snewn(len, BignumInt);
892 for (j = 0; j < len; j++)
893 n[len - 1 - j] = mod[j + 1];
895 mninv = snewn(len, BignumInt);
896 for (j = 0; j < len; j++)
897 mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
898 freebn(inv); /* we don't need this copy of it any more */
899 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
900 x = snewn(len, BignumInt);
901 for (j = 0; j < len; j++)
903 internal_sub(x, mninv, mninv, len);
905 /* x = snewn(len, BignumInt); */ /* already done above */
906 for (j = 0; j < len; j++)
907 x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
908 freebn(base); /* we don't need this copy of it any more */
910 a = snewn(2*len, BignumInt);
911 b = snewn(2*len, BignumInt);
912 for (j = 0; j < len; j++)
913 a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
916 /* Scratch space for multiplies */
917 scratchlen = 3*len + mul_compute_scratch(len);
918 scratch = snewn(scratchlen, BignumInt);
920 /* Skip leading zero bits of exp. */
922 j = BIGNUM_INT_BITS-1;
923 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
927 j = BIGNUM_INT_BITS-1;
931 /* Main computation */
932 while (i < (int)exp[0]) {
934 internal_mul(a + len, a + len, b, len, scratch);
935 monty_reduce(b, n, mninv, scratch, len);
936 if ((exp[exp[0] - i] & (1 << j)) != 0) {
937 internal_mul(b + len, x, a, len, scratch);
938 monty_reduce(a, n, mninv, scratch, len);
948 j = BIGNUM_INT_BITS-1;
952 * Final monty_reduce to get back from the adjusted Montgomery
955 monty_reduce(a, n, mninv, scratch, len);
957 /* Copy result to buffer */
958 result = newbn(mod[0]);
959 for (i = 0; i < len; i++)
960 result[result[0] - i] = a[i + len];
961 while (result[0] > 1 && result[result[0]] == 0)
964 /* Free temporary arrays */
965 smemclr(scratch, scratchlen * sizeof(*scratch));
967 smemclr(a, 2 * len * sizeof(*a));
969 smemclr(b, 2 * len * sizeof(*b));
971 smemclr(mninv, len * sizeof(*mninv));
973 smemclr(n, len * sizeof(*n));
975 smemclr(x, len * sizeof(*x));
982 * Compute (p * q) % mod.
983 * The most significant word of mod MUST be non-zero.
984 * We assume that the result array is the same size as the mod array.
986 Bignum modmul(Bignum p, Bignum q, Bignum mod)
988 BignumInt *a, *n, *m, *o, *scratch;
989 int mshift, scratchlen;
990 int pqlen, mlen, rlen, i, j;
994 * The most significant word of mod needs to be non-zero. It
995 * should already be, but let's make sure.
997 assert(mod[mod[0]] != 0);
999 /* Allocate m of size mlen, copy mod to m */
1000 /* We use big endian internally */
1002 m = snewn(mlen, BignumInt);
1003 for (j = 0; j < mlen; j++)
1004 m[j] = mod[mod[0] - j];
1006 /* Shift m left to make msb bit set */
1007 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
1008 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1011 for (i = 0; i < mlen - 1; i++)
1012 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1013 m[mlen - 1] = m[mlen - 1] << mshift;
1016 pqlen = (p[0] > q[0] ? p[0] : q[0]);
1019 * Make sure that we're allowing enough space. The shifting below
1020 * will underflow the vectors we allocate if pqlen is too small.
1022 if (2*pqlen <= mlen)
1025 /* Allocate n of size pqlen, copy p to n */
1026 n = snewn(pqlen, BignumInt);
1028 for (j = 0; j < i; j++)
1030 for (j = 0; j < (int)p[0]; j++)
1031 n[i + j] = p[p[0] - j];
1033 /* Allocate o of size pqlen, copy q to o */
1034 o = snewn(pqlen, BignumInt);
1036 for (j = 0; j < i; j++)
1038 for (j = 0; j < (int)q[0]; j++)
1039 o[i + j] = q[q[0] - j];
1041 /* Allocate a of size 2*pqlen for result */
1042 a = snewn(2 * pqlen, BignumInt);
1044 /* Scratch space for multiplies */
1045 scratchlen = mul_compute_scratch(pqlen);
1046 scratch = snewn(scratchlen, BignumInt);
1048 /* Main computation */
1049 internal_mul(n, o, a, pqlen, scratch);
1050 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1052 /* Fixup result in case the modulus was shifted */
1054 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
1055 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
1056 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
1057 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1058 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
1059 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
1062 /* Copy result to buffer */
1063 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
1064 result = newbn(rlen);
1065 for (i = 0; i < rlen; i++)
1066 result[result[0] - i] = a[i + 2 * pqlen - rlen];
1067 while (result[0] > 1 && result[result[0]] == 0)
1070 /* Free temporary arrays */
1071 smemclr(scratch, scratchlen * sizeof(*scratch));
1073 smemclr(a, 2 * pqlen * sizeof(*a));
1075 smemclr(m, mlen * sizeof(*m));
1077 smemclr(n, pqlen * sizeof(*n));
1079 smemclr(o, pqlen * sizeof(*o));
1087 * The most significant word of mod MUST be non-zero.
1088 * We assume that the result array is the same size as the mod array.
1089 * We optionally write out a quotient if `quotient' is non-NULL.
1090 * We can avoid writing out the result if `result' is NULL.
1092 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
1096 int plen, mlen, i, j;
1099 * The most significant word of mod needs to be non-zero. It
1100 * should already be, but let's make sure.
1102 assert(mod[mod[0]] != 0);
1104 /* Allocate m of size mlen, copy mod to m */
1105 /* We use big endian internally */
1107 m = snewn(mlen, BignumInt);
1108 for (j = 0; j < mlen; j++)
1109 m[j] = mod[mod[0] - j];
1111 /* Shift m left to make msb bit set */
1112 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
1113 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1116 for (i = 0; i < mlen - 1; i++)
1117 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1118 m[mlen - 1] = m[mlen - 1] << mshift;
1122 /* Ensure plen > mlen */
1126 /* Allocate n of size plen, copy p to n */
1127 n = snewn(plen, BignumInt);
1128 for (j = 0; j < plen; j++)
1130 for (j = 1; j <= (int)p[0]; j++)
1133 /* Main computation */
1134 internal_mod(n, plen, m, mlen, quotient, mshift);
1136 /* Fixup result in case the modulus was shifted */
1138 for (i = plen - mlen - 1; i < plen - 1; i++)
1139 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
1140 n[plen - 1] = n[plen - 1] << mshift;
1141 internal_mod(n, plen, m, mlen, quotient, 0);
1142 for (i = plen - 1; i >= plen - mlen; i--)
1143 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
1146 /* Copy result to buffer */
1148 for (i = 1; i <= (int)result[0]; i++) {
1150 result[i] = j >= 0 ? n[j] : 0;
1154 /* Free temporary arrays */
1155 smemclr(m, mlen * sizeof(*m));
1157 smemclr(n, plen * sizeof(*n));
1162 * Decrement a number.
1164 void decbn(Bignum bn)
1167 while (i < (int)bn[0] && bn[i] == 0)
1168 bn[i++] = BIGNUM_INT_MASK;
1172 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
1177 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
1180 for (i = 1; i <= w; i++)
1182 for (i = nbytes; i--;) {
1183 unsigned char byte = *data++;
1184 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
1187 while (result[0] > 1 && result[result[0]] == 0)
1193 * Read an SSH-1-format bignum from a data buffer. Return the number
1194 * of bytes consumed, or -1 if there wasn't enough data.
1196 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
1198 const unsigned char *p = data;
1206 for (i = 0; i < 2; i++)
1207 w = (w << 8) + *p++;
1208 b = (w + 7) / 8; /* bits -> bytes */
1213 if (!result) /* just return length */
1216 *result = bignum_from_bytes(p, b);
1218 return p + b - data;
1222 * Return the bit count of a bignum, for SSH-1 encoding.
1224 int bignum_bitcount(Bignum bn)
1226 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
1227 while (bitcount >= 0
1228 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
1229 return bitcount + 1;
1233 * Return the byte length of a bignum when SSH-1 encoded.
1235 int ssh1_bignum_length(Bignum bn)
1237 return 2 + (bignum_bitcount(bn) + 7) / 8;
1241 * Return the byte length of a bignum when SSH-2 encoded.
1243 int ssh2_bignum_length(Bignum bn)
1245 return 4 + (bignum_bitcount(bn) + 8) / 8;
1249 * Return a byte from a bignum; 0 is least significant, etc.
1251 int bignum_byte(Bignum bn, int i)
1253 if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
1254 return 0; /* beyond the end */
1256 return (bn[i / BIGNUM_INT_BYTES + 1] >>
1257 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
1261 * Return a bit from a bignum; 0 is least significant, etc.
1263 int bignum_bit(Bignum bn, int i)
1265 if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
1266 return 0; /* beyond the end */
1268 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
1272 * Set a bit in a bignum; 0 is least significant, etc.
1274 void bignum_set_bit(Bignum bn, int bitnum, int value)
1276 if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
1277 abort(); /* beyond the end */
1279 int v = bitnum / BIGNUM_INT_BITS + 1;
1280 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
1289 * Write a SSH-1-format bignum into a buffer. It is assumed the
1290 * buffer is big enough. Returns the number of bytes used.
1292 int ssh1_write_bignum(void *data, Bignum bn)
1294 unsigned char *p = data;
1295 int len = ssh1_bignum_length(bn);
1297 int bitc = bignum_bitcount(bn);
1299 *p++ = (bitc >> 8) & 0xFF;
1300 *p++ = (bitc) & 0xFF;
1301 for (i = len - 2; i--;)
1302 *p++ = bignum_byte(bn, i);
1307 * Compare two bignums. Returns like strcmp.
1309 int bignum_cmp(Bignum a, Bignum b)
1311 int amax = a[0], bmax = b[0];
1312 int i = (amax > bmax ? amax : bmax);
1314 BignumInt aval = (i > amax ? 0 : a[i]);
1315 BignumInt bval = (i > bmax ? 0 : b[i]);
1326 * Right-shift one bignum to form another.
1328 Bignum bignum_rshift(Bignum a, int shift)
1331 int i, shiftw, shiftb, shiftbb, bits;
1334 bits = bignum_bitcount(a) - shift;
1335 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
1338 shiftw = shift / BIGNUM_INT_BITS;
1339 shiftb = shift % BIGNUM_INT_BITS;
1340 shiftbb = BIGNUM_INT_BITS - shiftb;
1342 ai1 = a[shiftw + 1];
1343 for (i = 1; i <= (int)ret[0]; i++) {
1345 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
1346 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
1354 * Non-modular multiplication and addition.
1356 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
1358 int alen = a[0], blen = b[0];
1359 int mlen = (alen > blen ? alen : blen);
1360 int rlen, i, maxspot;
1362 BignumInt *workspace;
1365 /* mlen space for a, mlen space for b, 2*mlen for result,
1366 * plus scratch space for multiplication */
1367 wslen = mlen * 4 + mul_compute_scratch(mlen);
1368 workspace = snewn(wslen, BignumInt);
1369 for (i = 0; i < mlen; i++) {
1370 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
1371 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
1374 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
1375 workspace + 2 * mlen, mlen, workspace + 4 * mlen);
1377 /* now just copy the result back */
1378 rlen = alen + blen + 1;
1379 if (addend && rlen <= (int)addend[0])
1380 rlen = addend[0] + 1;
1383 for (i = 1; i <= (int)ret[0]; i++) {
1384 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
1390 /* now add in the addend, if any */
1392 BignumDblInt carry = 0;
1393 for (i = 1; i <= rlen; i++) {
1394 carry += (i <= (int)ret[0] ? ret[i] : 0);
1395 carry += (i <= (int)addend[0] ? addend[i] : 0);
1396 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1397 carry >>= BIGNUM_INT_BITS;
1398 if (ret[i] != 0 && i > maxspot)
1404 smemclr(workspace, wslen * sizeof(*workspace));
1410 * Non-modular multiplication.
1412 Bignum bigmul(Bignum a, Bignum b)
1414 return bigmuladd(a, b, NULL);
1420 Bignum bigadd(Bignum a, Bignum b)
1422 int alen = a[0], blen = b[0];
1423 int rlen = (alen > blen ? alen : blen) + 1;
1432 for (i = 1; i <= rlen; i++) {
1433 carry += (i <= (int)a[0] ? a[i] : 0);
1434 carry += (i <= (int)b[0] ? b[i] : 0);
1435 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1436 carry >>= BIGNUM_INT_BITS;
1437 if (ret[i] != 0 && i > maxspot)
1446 * Subtraction. Returns a-b, or NULL if the result would come out
1447 * negative (recall that this entire bignum module only handles
1448 * positive numbers).
1450 Bignum bigsub(Bignum a, Bignum b)
1452 int alen = a[0], blen = b[0];
1453 int rlen = (alen > blen ? alen : blen);
1462 for (i = 1; i <= rlen; i++) {
1463 carry += (i <= (int)a[0] ? a[i] : 0);
1464 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
1465 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1466 carry >>= BIGNUM_INT_BITS;
1467 if (ret[i] != 0 && i > maxspot)
1481 * Create a bignum which is the bitmask covering another one. That
1482 * is, the smallest integer which is >= N and is also one less than
1485 Bignum bignum_bitmask(Bignum n)
1487 Bignum ret = copybn(n);
1492 while (n[i] == 0 && i > 0)
1495 return ret; /* input was zero */
1501 ret[i] = BIGNUM_INT_MASK;
1506 * Convert a (max 32-bit) long into a bignum.
1508 Bignum bignum_from_long(unsigned long nn)
1511 BignumDblInt n = nn;
1514 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
1515 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
1517 ret[0] = (ret[2] ? 2 : 1);
1522 * Add a long to a bignum.
1524 Bignum bignum_add_long(Bignum number, unsigned long addendx)
1526 Bignum ret = newbn(number[0] + 1);
1528 BignumDblInt carry = 0, addend = addendx;
1530 for (i = 1; i <= (int)ret[0]; i++) {
1531 carry += addend & BIGNUM_INT_MASK;
1532 carry += (i <= (int)number[0] ? number[i] : 0);
1533 addend >>= BIGNUM_INT_BITS;
1534 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1535 carry >>= BIGNUM_INT_BITS;
1544 * Compute the residue of a bignum, modulo a (max 16-bit) short.
1546 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
1548 BignumDblInt mod, r;
1553 for (i = number[0]; i > 0; i--)
1554 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
1555 return (unsigned short) r;
1559 void diagbn(char *prefix, Bignum md)
1561 int i, nibbles, morenibbles;
1562 static const char hex[] = "0123456789ABCDEF";
1564 debug(("%s0x", prefix ? prefix : ""));
1566 nibbles = (3 + bignum_bitcount(md)) / 4;
1569 morenibbles = 4 * md[0] - nibbles;
1570 for (i = 0; i < morenibbles; i++)
1572 for (i = nibbles; i--;)
1574 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
1584 Bignum bigdiv(Bignum a, Bignum b)
1586 Bignum q = newbn(a[0]);
1587 bigdivmod(a, b, NULL, q);
1594 Bignum bigmod(Bignum a, Bignum b)
1596 Bignum r = newbn(b[0]);
1597 bigdivmod(a, b, r, NULL);
1602 * Greatest common divisor.
1604 Bignum biggcd(Bignum av, Bignum bv)
1606 Bignum a = copybn(av);
1607 Bignum b = copybn(bv);
1609 while (bignum_cmp(b, Zero) != 0) {
1610 Bignum t = newbn(b[0]);
1611 bigdivmod(a, b, t, NULL);
1612 while (t[0] > 1 && t[t[0]] == 0)
1624 * Modular inverse, using Euclid's extended algorithm.
1626 Bignum modinv(Bignum number, Bignum modulus)
1628 Bignum a = copybn(modulus);
1629 Bignum b = copybn(number);
1630 Bignum xp = copybn(Zero);
1631 Bignum x = copybn(One);
1634 assert(number[number[0]] != 0);
1635 assert(modulus[modulus[0]] != 0);
1637 while (bignum_cmp(b, One) != 0) {
1640 if (bignum_cmp(b, Zero) == 0) {
1642 * Found a common factor between the inputs, so we cannot
1643 * return a modular inverse at all.
1650 bigdivmod(a, b, t, q);
1651 while (t[0] > 1 && t[t[0]] == 0)
1658 x = bigmuladd(q, xp, t);
1668 /* now we know that sign * x == 1, and that x < modulus */
1670 /* set a new x to be modulus - x */
1671 Bignum newx = newbn(modulus[0]);
1672 BignumInt carry = 0;
1676 for (i = 1; i <= (int)newx[0]; i++) {
1677 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
1678 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
1679 newx[i] = aword - bword - carry;
1681 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
1695 * Render a bignum into decimal. Return a malloced string holding
1696 * the decimal representation.
1698 char *bignum_decimal(Bignum x)
1700 int ndigits, ndigit;
1704 BignumInt *workspace;
1707 * First, estimate the number of digits. Since log(10)/log(2)
1708 * is just greater than 93/28 (the joys of continued fraction
1709 * approximations...) we know that for every 93 bits, we need
1710 * at most 28 digits. This will tell us how much to malloc.
1712 * Formally: if x has i bits, that means x is strictly less
1713 * than 2^i. Since 2 is less than 10^(28/93), this is less than
1714 * 10^(28i/93). We need an integer power of ten, so we must
1715 * round up (rounding down might make it less than x again).
1716 * Therefore if we multiply the bit count by 28/93, rounding
1717 * up, we will have enough digits.
1719 * i=0 (i.e., x=0) is an irritating special case.
1721 i = bignum_bitcount(x);
1723 ndigits = 1; /* x = 0 */
1725 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
1726 ndigits++; /* allow for trailing \0 */
1727 ret = snewn(ndigits, char);
1730 * Now allocate some workspace to hold the binary form as we
1731 * repeatedly divide it by ten. Initialise this to the
1732 * big-endian form of the number.
1734 workspace = snewn(x[0], BignumInt);
1735 for (i = 0; i < (int)x[0]; i++)
1736 workspace[i] = x[x[0] - i];
1739 * Next, write the decimal number starting with the last digit.
1740 * We use ordinary short division, dividing 10 into the
1743 ndigit = ndigits - 1;
1748 for (i = 0; i < (int)x[0]; i++) {
1749 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
1750 workspace[i] = (BignumInt) (carry / 10);
1755 ret[--ndigit] = (char) (carry + '0');
1759 * There's a chance we've fallen short of the start of the
1760 * string. Correct if so.
1763 memmove(ret, ret + ndigit, ndigits - ndigit);
1768 smemclr(workspace, x[0] * sizeof(*workspace));
1780 * gcc -Wall -g -O0 -DTESTBN -o testbn sshbn.c misc.c conf.c tree234.c unix/uxmisc.c -I. -I unix -I charset
1782 * Then feed to this program's standard input the output of
1783 * testdata/bignum.py .
1786 void modalfatalbox(char *p, ...)
1789 fprintf(stderr, "FATAL ERROR: ");
1791 vfprintf(stderr, p, ap);
1793 fputc('\n', stderr);
1797 #define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
1799 int main(int argc, char **argv)
1803 int passes = 0, fails = 0;
1805 while ((buf = fgetline(stdin)) != NULL) {
1806 int maxlen = strlen(buf);
1807 unsigned char *data = snewn(maxlen, unsigned char);
1808 unsigned char *ptrs[5], *q;
1817 while (*bufp && !isspace((unsigned char)*bufp))
1826 while (*bufp && !isxdigit((unsigned char)*bufp))
1833 while (*bufp && isxdigit((unsigned char)*bufp))
1837 if (ptrnum >= lenof(ptrs))
1841 for (i = -((end - start) & 1); i < end-start; i += 2) {
1842 unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
1843 val = val * 16 + fromxdigit(start[i+1]);
1850 if (!strcmp(buf, "mul")) {
1854 printf("%d: mul with %d parameters, expected 3\n", line, ptrnum);
1857 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1858 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1859 c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1862 if (bignum_cmp(c, p) == 0) {
1865 char *as = bignum_decimal(a);
1866 char *bs = bignum_decimal(b);
1867 char *cs = bignum_decimal(c);
1868 char *ps = bignum_decimal(p);
1870 printf("%d: fail: %s * %s gave %s expected %s\n",
1871 line, as, bs, ps, cs);
1883 } else if (!strcmp(buf, "modmul")) {
1884 Bignum a, b, m, c, p;
1887 printf("%d: modmul with %d parameters, expected 4\n",
1891 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1892 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1893 m = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1894 c = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
1895 p = modmul(a, b, m);
1897 if (bignum_cmp(c, p) == 0) {
1900 char *as = bignum_decimal(a);
1901 char *bs = bignum_decimal(b);
1902 char *ms = bignum_decimal(m);
1903 char *cs = bignum_decimal(c);
1904 char *ps = bignum_decimal(p);
1906 printf("%d: fail: %s * %s mod %s gave %s expected %s\n",
1907 line, as, bs, ms, ps, cs);
1921 } else if (!strcmp(buf, "pow")) {
1922 Bignum base, expt, modulus, expected, answer;
1925 printf("%d: mul with %d parameters, expected 4\n", line, ptrnum);
1929 base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1930 expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1931 modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1932 expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
1933 answer = modpow(base, expt, modulus);
1935 if (bignum_cmp(expected, answer) == 0) {
1938 char *as = bignum_decimal(base);
1939 char *bs = bignum_decimal(expt);
1940 char *cs = bignum_decimal(modulus);
1941 char *ds = bignum_decimal(answer);
1942 char *ps = bignum_decimal(expected);
1944 printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
1945 line, as, bs, cs, ds, ps);
1960 printf("%d: unrecognised test keyword: '%s'\n", line, buf);
1968 printf("passed %d failed %d total %d\n", passes, fails, passes+fails);