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 memset(b, 0, 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)
632 for (i = 0; i <= alen - mlen; i++) {
634 unsigned int q, r, c, ai1;
648 /* Find q = h:a[i] / m0 */
653 * To illustrate it, suppose a BignumInt is 8 bits, and
654 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
655 * our initial division will be 0xA123 / 0xA1, which
656 * will give a quotient of 0x100 and a divide overflow.
657 * However, the invariants in this division algorithm
658 * are not violated, since the full number A1:23:... is
659 * _less_ than the quotient prefix A1:B2:... and so the
660 * following correction loop would have sorted it out.
662 * In this situation we set q to be the largest
663 * quotient we _can_ stomach (0xFF, of course).
667 /* Macro doesn't want an array subscript expression passed
668 * into it (see definition), so use a temporary. */
669 BignumInt tmplo = a[i];
670 DIVMOD_WORD(q, r, h, tmplo, m0);
672 /* Refine our estimate of q by looking at
673 h:a[i]:a[i+1] / m0:m1 */
675 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
678 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
679 if (r >= (BignumDblInt) m0 &&
680 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
684 /* Subtract q * m from a[i...] */
686 for (k = mlen - 1; k >= 0; k--) {
687 t = MUL_WORD(q, m[k]);
689 c = (unsigned)(t >> BIGNUM_INT_BITS);
690 if ((BignumInt) t > a[i + k])
692 a[i + k] -= (BignumInt) t;
695 /* Add back m in case of borrow */
698 for (k = mlen - 1; k >= 0; k--) {
701 a[i + k] = (BignumInt) t;
702 t = t >> BIGNUM_INT_BITS;
707 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
712 * Compute (base ^ exp) % mod, the pedestrian way.
714 Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
716 BignumInt *a, *b, *n, *m, *scratch;
718 int mlen, scratchlen, i, j;
722 * The most significant word of mod needs to be non-zero. It
723 * should already be, but let's make sure.
725 assert(mod[mod[0]] != 0);
728 * Make sure the base is smaller than the modulus, by reducing
729 * it modulo the modulus if not.
731 base = bigmod(base_in, mod);
733 /* Allocate m of size mlen, copy mod to m */
734 /* We use big endian internally */
736 m = snewn(mlen, BignumInt);
737 for (j = 0; j < mlen; j++)
738 m[j] = mod[mod[0] - j];
740 /* Shift m left to make msb bit set */
741 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
742 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
745 for (i = 0; i < mlen - 1; i++)
746 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
747 m[mlen - 1] = m[mlen - 1] << mshift;
750 /* Allocate n of size mlen, copy base to n */
751 n = snewn(mlen, BignumInt);
753 for (j = 0; j < i; j++)
755 for (j = 0; j < (int)base[0]; j++)
756 n[i + j] = base[base[0] - j];
758 /* Allocate a and b of size 2*mlen. Set a = 1 */
759 a = snewn(2 * mlen, BignumInt);
760 b = snewn(2 * mlen, BignumInt);
761 for (i = 0; i < 2 * mlen; i++)
765 /* Scratch space for multiplies */
766 scratchlen = mul_compute_scratch(mlen);
767 scratch = snewn(scratchlen, BignumInt);
769 /* Skip leading zero bits of exp. */
771 j = BIGNUM_INT_BITS-1;
772 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
776 j = BIGNUM_INT_BITS-1;
780 /* Main computation */
781 while (i < (int)exp[0]) {
783 internal_mul(a + mlen, a + mlen, b, mlen, scratch);
784 internal_mod(b, mlen * 2, m, mlen, NULL, 0);
785 if ((exp[exp[0] - i] & (1 << j)) != 0) {
786 internal_mul(b + mlen, n, a, mlen, scratch);
787 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
797 j = BIGNUM_INT_BITS-1;
800 /* Fixup result in case the modulus was shifted */
802 for (i = mlen - 1; i < 2 * mlen - 1; i++)
803 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
804 a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
805 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
806 for (i = 2 * mlen - 1; i >= mlen; i--)
807 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
810 /* Copy result to buffer */
811 result = newbn(mod[0]);
812 for (i = 0; i < mlen; i++)
813 result[result[0] - i] = a[i + mlen];
814 while (result[0] > 1 && result[result[0]] == 0)
817 /* Free temporary arrays */
818 for (i = 0; i < 2 * mlen; i++)
821 for (i = 0; i < scratchlen; i++)
824 for (i = 0; i < 2 * mlen; i++)
827 for (i = 0; i < mlen; i++)
830 for (i = 0; i < mlen; i++)
840 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
841 * technique where possible, falling back to modpow_simple otherwise.
843 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
845 BignumInt *a, *b, *x, *n, *mninv, *scratch;
846 int len, scratchlen, i, j;
847 Bignum base, base2, r, rn, inv, result;
850 * The most significant word of mod needs to be non-zero. It
851 * should already be, but let's make sure.
853 assert(mod[mod[0]] != 0);
856 * mod had better be odd, or we can't do Montgomery multiplication
857 * using a power of two at all.
860 return modpow_simple(base_in, exp, mod);
863 * Make sure the base is smaller than the modulus, by reducing
864 * it modulo the modulus if not.
866 base = bigmod(base_in, mod);
869 * Compute the inverse of n mod r, for monty_reduce. (In fact we
870 * want the inverse of _minus_ n mod r, but we'll sort that out
874 r = bn_power_2(BIGNUM_INT_BITS * len);
875 inv = modinv(mod, r);
878 * Multiply the base by r mod n, to get it into Montgomery
881 base2 = modmul(base, r, mod);
885 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
887 freebn(r); /* won't need this any more */
890 * Set up internal arrays of the right lengths, in big-endian
891 * format, containing the base, the modulus, and the modulus's
894 n = snewn(len, BignumInt);
895 for (j = 0; j < len; j++)
896 n[len - 1 - j] = mod[j + 1];
898 mninv = snewn(len, BignumInt);
899 for (j = 0; j < len; j++)
900 mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
901 freebn(inv); /* we don't need this copy of it any more */
902 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
903 x = snewn(len, BignumInt);
904 for (j = 0; j < len; j++)
906 internal_sub(x, mninv, mninv, len);
908 /* x = snewn(len, BignumInt); */ /* already done above */
909 for (j = 0; j < len; j++)
910 x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
911 freebn(base); /* we don't need this copy of it any more */
913 a = snewn(2*len, BignumInt);
914 b = snewn(2*len, BignumInt);
915 for (j = 0; j < len; j++)
916 a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
919 /* Scratch space for multiplies */
920 scratchlen = 3*len + mul_compute_scratch(len);
921 scratch = snewn(scratchlen, BignumInt);
923 /* Skip leading zero bits of exp. */
925 j = BIGNUM_INT_BITS-1;
926 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
930 j = BIGNUM_INT_BITS-1;
934 /* Main computation */
935 while (i < (int)exp[0]) {
937 internal_mul(a + len, a + len, b, len, scratch);
938 monty_reduce(b, n, mninv, scratch, len);
939 if ((exp[exp[0] - i] & (1 << j)) != 0) {
940 internal_mul(b + len, x, a, len, scratch);
941 monty_reduce(a, n, mninv, scratch, len);
951 j = BIGNUM_INT_BITS-1;
955 * Final monty_reduce to get back from the adjusted Montgomery
958 monty_reduce(a, n, mninv, scratch, len);
960 /* Copy result to buffer */
961 result = newbn(mod[0]);
962 for (i = 0; i < len; i++)
963 result[result[0] - i] = a[i + len];
964 while (result[0] > 1 && result[result[0]] == 0)
967 /* Free temporary arrays */
968 for (i = 0; i < scratchlen; i++)
971 for (i = 0; i < 2 * len; i++)
974 for (i = 0; i < 2 * len; i++)
977 for (i = 0; i < len; i++)
980 for (i = 0; i < len; i++)
983 for (i = 0; i < len; i++)
991 * Compute (p * q) % mod.
992 * The most significant word of mod MUST be non-zero.
993 * We assume that the result array is the same size as the mod array.
995 Bignum modmul(Bignum p, Bignum q, Bignum mod)
997 BignumInt *a, *n, *m, *o, *scratch;
998 int mshift, scratchlen;
999 int pqlen, mlen, rlen, i, j;
1002 /* Allocate m of size mlen, copy mod to m */
1003 /* We use big endian internally */
1005 m = snewn(mlen, BignumInt);
1006 for (j = 0; j < mlen; j++)
1007 m[j] = mod[mod[0] - j];
1009 /* Shift m left to make msb bit set */
1010 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
1011 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1014 for (i = 0; i < mlen - 1; i++)
1015 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1016 m[mlen - 1] = m[mlen - 1] << mshift;
1019 pqlen = (p[0] > q[0] ? p[0] : q[0]);
1021 /* Allocate n of size pqlen, copy p to n */
1022 n = snewn(pqlen, BignumInt);
1024 for (j = 0; j < i; j++)
1026 for (j = 0; j < (int)p[0]; j++)
1027 n[i + j] = p[p[0] - j];
1029 /* Allocate o of size pqlen, copy q to o */
1030 o = snewn(pqlen, BignumInt);
1032 for (j = 0; j < i; j++)
1034 for (j = 0; j < (int)q[0]; j++)
1035 o[i + j] = q[q[0] - j];
1037 /* Allocate a of size 2*pqlen for result */
1038 a = snewn(2 * pqlen, BignumInt);
1040 /* Scratch space for multiplies */
1041 scratchlen = mul_compute_scratch(pqlen);
1042 scratch = snewn(scratchlen, BignumInt);
1044 /* Main computation */
1045 internal_mul(n, o, a, pqlen, scratch);
1046 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1048 /* Fixup result in case the modulus was shifted */
1050 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
1051 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
1052 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
1053 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1054 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
1055 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
1058 /* Copy result to buffer */
1059 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
1060 result = newbn(rlen);
1061 for (i = 0; i < rlen; i++)
1062 result[result[0] - i] = a[i + 2 * pqlen - rlen];
1063 while (result[0] > 1 && result[result[0]] == 0)
1066 /* Free temporary arrays */
1067 for (i = 0; i < scratchlen; i++)
1070 for (i = 0; i < 2 * pqlen; i++)
1073 for (i = 0; i < mlen; i++)
1076 for (i = 0; i < pqlen; i++)
1079 for (i = 0; i < pqlen; i++)
1088 * The most significant word of mod MUST be non-zero.
1089 * We assume that the result array is the same size as the mod array.
1090 * We optionally write out a quotient if `quotient' is non-NULL.
1091 * We can avoid writing out the result if `result' is NULL.
1093 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
1097 int plen, mlen, i, j;
1099 /* Allocate m of size mlen, copy mod to m */
1100 /* We use big endian internally */
1102 m = snewn(mlen, BignumInt);
1103 for (j = 0; j < mlen; j++)
1104 m[j] = mod[mod[0] - j];
1106 /* Shift m left to make msb bit set */
1107 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
1108 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1111 for (i = 0; i < mlen - 1; i++)
1112 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1113 m[mlen - 1] = m[mlen - 1] << mshift;
1117 /* Ensure plen > mlen */
1121 /* Allocate n of size plen, copy p to n */
1122 n = snewn(plen, BignumInt);
1123 for (j = 0; j < plen; j++)
1125 for (j = 1; j <= (int)p[0]; j++)
1128 /* Main computation */
1129 internal_mod(n, plen, m, mlen, quotient, mshift);
1131 /* Fixup result in case the modulus was shifted */
1133 for (i = plen - mlen - 1; i < plen - 1; i++)
1134 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
1135 n[plen - 1] = n[plen - 1] << mshift;
1136 internal_mod(n, plen, m, mlen, quotient, 0);
1137 for (i = plen - 1; i >= plen - mlen; i--)
1138 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
1141 /* Copy result to buffer */
1143 for (i = 1; i <= (int)result[0]; i++) {
1145 result[i] = j >= 0 ? n[j] : 0;
1149 /* Free temporary arrays */
1150 for (i = 0; i < mlen; i++)
1153 for (i = 0; i < plen; i++)
1159 * Decrement a number.
1161 void decbn(Bignum bn)
1164 while (i < (int)bn[0] && bn[i] == 0)
1165 bn[i++] = BIGNUM_INT_MASK;
1169 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
1174 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
1177 for (i = 1; i <= w; i++)
1179 for (i = nbytes; i--;) {
1180 unsigned char byte = *data++;
1181 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
1184 while (result[0] > 1 && result[result[0]] == 0)
1190 * Read an SSH-1-format bignum from a data buffer. Return the number
1191 * of bytes consumed, or -1 if there wasn't enough data.
1193 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
1195 const unsigned char *p = data;
1203 for (i = 0; i < 2; i++)
1204 w = (w << 8) + *p++;
1205 b = (w + 7) / 8; /* bits -> bytes */
1210 if (!result) /* just return length */
1213 *result = bignum_from_bytes(p, b);
1215 return p + b - data;
1219 * Return the bit count of a bignum, for SSH-1 encoding.
1221 int bignum_bitcount(Bignum bn)
1223 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
1224 while (bitcount >= 0
1225 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
1226 return bitcount + 1;
1230 * Return the byte length of a bignum when SSH-1 encoded.
1232 int ssh1_bignum_length(Bignum bn)
1234 return 2 + (bignum_bitcount(bn) + 7) / 8;
1238 * Return the byte length of a bignum when SSH-2 encoded.
1240 int ssh2_bignum_length(Bignum bn)
1242 return 4 + (bignum_bitcount(bn) + 8) / 8;
1246 * Return a byte from a bignum; 0 is least significant, etc.
1248 int bignum_byte(Bignum bn, int i)
1250 if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
1251 return 0; /* beyond the end */
1253 return (bn[i / BIGNUM_INT_BYTES + 1] >>
1254 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
1258 * Return a bit from a bignum; 0 is least significant, etc.
1260 int bignum_bit(Bignum bn, int i)
1262 if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
1263 return 0; /* beyond the end */
1265 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
1269 * Set a bit in a bignum; 0 is least significant, etc.
1271 void bignum_set_bit(Bignum bn, int bitnum, int value)
1273 if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
1274 abort(); /* beyond the end */
1276 int v = bitnum / BIGNUM_INT_BITS + 1;
1277 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
1286 * Write a SSH-1-format bignum into a buffer. It is assumed the
1287 * buffer is big enough. Returns the number of bytes used.
1289 int ssh1_write_bignum(void *data, Bignum bn)
1291 unsigned char *p = data;
1292 int len = ssh1_bignum_length(bn);
1294 int bitc = bignum_bitcount(bn);
1296 *p++ = (bitc >> 8) & 0xFF;
1297 *p++ = (bitc) & 0xFF;
1298 for (i = len - 2; i--;)
1299 *p++ = bignum_byte(bn, i);
1304 * Compare two bignums. Returns like strcmp.
1306 int bignum_cmp(Bignum a, Bignum b)
1308 int amax = a[0], bmax = b[0];
1309 int i = (amax > bmax ? amax : bmax);
1311 BignumInt aval = (i > amax ? 0 : a[i]);
1312 BignumInt bval = (i > bmax ? 0 : b[i]);
1323 * Right-shift one bignum to form another.
1325 Bignum bignum_rshift(Bignum a, int shift)
1328 int i, shiftw, shiftb, shiftbb, bits;
1331 bits = bignum_bitcount(a) - shift;
1332 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
1335 shiftw = shift / BIGNUM_INT_BITS;
1336 shiftb = shift % BIGNUM_INT_BITS;
1337 shiftbb = BIGNUM_INT_BITS - shiftb;
1339 ai1 = a[shiftw + 1];
1340 for (i = 1; i <= (int)ret[0]; i++) {
1342 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
1343 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
1351 * Non-modular multiplication and addition.
1353 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
1355 int alen = a[0], blen = b[0];
1356 int mlen = (alen > blen ? alen : blen);
1357 int rlen, i, maxspot;
1359 BignumInt *workspace;
1362 /* mlen space for a, mlen space for b, 2*mlen for result,
1363 * plus scratch space for multiplication */
1364 wslen = mlen * 4 + mul_compute_scratch(mlen);
1365 workspace = snewn(wslen, BignumInt);
1366 for (i = 0; i < mlen; i++) {
1367 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
1368 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
1371 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
1372 workspace + 2 * mlen, mlen, workspace + 4 * mlen);
1374 /* now just copy the result back */
1375 rlen = alen + blen + 1;
1376 if (addend && rlen <= (int)addend[0])
1377 rlen = addend[0] + 1;
1380 for (i = 1; i <= (int)ret[0]; i++) {
1381 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
1387 /* now add in the addend, if any */
1389 BignumDblInt carry = 0;
1390 for (i = 1; i <= rlen; i++) {
1391 carry += (i <= (int)ret[0] ? ret[i] : 0);
1392 carry += (i <= (int)addend[0] ? addend[i] : 0);
1393 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1394 carry >>= BIGNUM_INT_BITS;
1395 if (ret[i] != 0 && i > maxspot)
1401 for (i = 0; i < wslen; i++)
1408 * Non-modular multiplication.
1410 Bignum bigmul(Bignum a, Bignum b)
1412 return bigmuladd(a, b, NULL);
1418 Bignum bigadd(Bignum a, Bignum b)
1420 int alen = a[0], blen = b[0];
1421 int rlen = (alen > blen ? alen : blen) + 1;
1430 for (i = 1; i <= rlen; i++) {
1431 carry += (i <= (int)a[0] ? a[i] : 0);
1432 carry += (i <= (int)b[0] ? b[i] : 0);
1433 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1434 carry >>= BIGNUM_INT_BITS;
1435 if (ret[i] != 0 && i > maxspot)
1444 * Subtraction. Returns a-b, or NULL if the result would come out
1445 * negative (recall that this entire bignum module only handles
1446 * positive numbers).
1448 Bignum bigsub(Bignum a, Bignum b)
1450 int alen = a[0], blen = b[0];
1451 int rlen = (alen > blen ? alen : blen);
1460 for (i = 1; i <= rlen; i++) {
1461 carry += (i <= (int)a[0] ? a[i] : 0);
1462 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
1463 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1464 carry >>= BIGNUM_INT_BITS;
1465 if (ret[i] != 0 && i > maxspot)
1479 * Create a bignum which is the bitmask covering another one. That
1480 * is, the smallest integer which is >= N and is also one less than
1483 Bignum bignum_bitmask(Bignum n)
1485 Bignum ret = copybn(n);
1490 while (n[i] == 0 && i > 0)
1493 return ret; /* input was zero */
1499 ret[i] = BIGNUM_INT_MASK;
1504 * Convert a (max 32-bit) long into a bignum.
1506 Bignum bignum_from_long(unsigned long nn)
1509 BignumDblInt n = nn;
1512 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
1513 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
1515 ret[0] = (ret[2] ? 2 : 1);
1520 * Add a long to a bignum.
1522 Bignum bignum_add_long(Bignum number, unsigned long addendx)
1524 Bignum ret = newbn(number[0] + 1);
1526 BignumDblInt carry = 0, addend = addendx;
1528 for (i = 1; i <= (int)ret[0]; i++) {
1529 carry += addend & BIGNUM_INT_MASK;
1530 carry += (i <= (int)number[0] ? number[i] : 0);
1531 addend >>= BIGNUM_INT_BITS;
1532 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1533 carry >>= BIGNUM_INT_BITS;
1542 * Compute the residue of a bignum, modulo a (max 16-bit) short.
1544 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
1546 BignumDblInt mod, r;
1551 for (i = number[0]; i > 0; i--)
1552 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
1553 return (unsigned short) r;
1557 void diagbn(char *prefix, Bignum md)
1559 int i, nibbles, morenibbles;
1560 static const char hex[] = "0123456789ABCDEF";
1562 debug(("%s0x", prefix ? prefix : ""));
1564 nibbles = (3 + bignum_bitcount(md)) / 4;
1567 morenibbles = 4 * md[0] - nibbles;
1568 for (i = 0; i < morenibbles; i++)
1570 for (i = nibbles; i--;)
1572 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
1582 Bignum bigdiv(Bignum a, Bignum b)
1584 Bignum q = newbn(a[0]);
1585 bigdivmod(a, b, NULL, q);
1592 Bignum bigmod(Bignum a, Bignum b)
1594 Bignum r = newbn(b[0]);
1595 bigdivmod(a, b, r, NULL);
1600 * Greatest common divisor.
1602 Bignum biggcd(Bignum av, Bignum bv)
1604 Bignum a = copybn(av);
1605 Bignum b = copybn(bv);
1607 while (bignum_cmp(b, Zero) != 0) {
1608 Bignum t = newbn(b[0]);
1609 bigdivmod(a, b, t, NULL);
1610 while (t[0] > 1 && t[t[0]] == 0)
1622 * Modular inverse, using Euclid's extended algorithm.
1624 Bignum modinv(Bignum number, Bignum modulus)
1626 Bignum a = copybn(modulus);
1627 Bignum b = copybn(number);
1628 Bignum xp = copybn(Zero);
1629 Bignum x = copybn(One);
1632 while (bignum_cmp(b, One) != 0) {
1633 Bignum t = newbn(b[0]);
1634 Bignum q = newbn(a[0]);
1635 bigdivmod(a, b, t, q);
1636 while (t[0] > 1 && t[t[0]] == 0)
1643 x = bigmuladd(q, xp, t);
1653 /* now we know that sign * x == 1, and that x < modulus */
1655 /* set a new x to be modulus - x */
1656 Bignum newx = newbn(modulus[0]);
1657 BignumInt carry = 0;
1661 for (i = 1; i <= (int)newx[0]; i++) {
1662 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
1663 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
1664 newx[i] = aword - bword - carry;
1666 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
1680 * Render a bignum into decimal. Return a malloced string holding
1681 * the decimal representation.
1683 char *bignum_decimal(Bignum x)
1685 int ndigits, ndigit;
1689 BignumInt *workspace;
1692 * First, estimate the number of digits. Since log(10)/log(2)
1693 * is just greater than 93/28 (the joys of continued fraction
1694 * approximations...) we know that for every 93 bits, we need
1695 * at most 28 digits. This will tell us how much to malloc.
1697 * Formally: if x has i bits, that means x is strictly less
1698 * than 2^i. Since 2 is less than 10^(28/93), this is less than
1699 * 10^(28i/93). We need an integer power of ten, so we must
1700 * round up (rounding down might make it less than x again).
1701 * Therefore if we multiply the bit count by 28/93, rounding
1702 * up, we will have enough digits.
1704 * i=0 (i.e., x=0) is an irritating special case.
1706 i = bignum_bitcount(x);
1708 ndigits = 1; /* x = 0 */
1710 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
1711 ndigits++; /* allow for trailing \0 */
1712 ret = snewn(ndigits, char);
1715 * Now allocate some workspace to hold the binary form as we
1716 * repeatedly divide it by ten. Initialise this to the
1717 * big-endian form of the number.
1719 workspace = snewn(x[0], BignumInt);
1720 for (i = 0; i < (int)x[0]; i++)
1721 workspace[i] = x[x[0] - i];
1724 * Next, write the decimal number starting with the last digit.
1725 * We use ordinary short division, dividing 10 into the
1728 ndigit = ndigits - 1;
1733 for (i = 0; i < (int)x[0]; i++) {
1734 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
1735 workspace[i] = (BignumInt) (carry / 10);
1740 ret[--ndigit] = (char) (carry + '0');
1744 * There's a chance we've fallen short of the start of the
1745 * string. Correct if so.
1748 memmove(ret, ret + ndigit, ndigits - ndigit);
1764 * gcc -g -O0 -DTESTBN -o testbn sshbn.c misc.c -I unix -I charset
1766 * Then feed to this program's standard input the output of
1767 * testdata/bignum.py .
1770 void modalfatalbox(char *p, ...)
1773 fprintf(stderr, "FATAL ERROR: ");
1775 vfprintf(stderr, p, ap);
1777 fputc('\n', stderr);
1781 #define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
1783 int main(int argc, char **argv)
1787 int passes = 0, fails = 0;
1789 while ((buf = fgetline(stdin)) != NULL) {
1790 int maxlen = strlen(buf);
1791 unsigned char *data = snewn(maxlen, unsigned char);
1792 unsigned char *ptrs[5], *q;
1801 while (*bufp && !isspace((unsigned char)*bufp))
1810 while (*bufp && !isxdigit((unsigned char)*bufp))
1817 while (*bufp && isxdigit((unsigned char)*bufp))
1821 if (ptrnum >= lenof(ptrs))
1825 for (i = -((end - start) & 1); i < end-start; i += 2) {
1826 unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
1827 val = val * 16 + fromxdigit(start[i+1]);
1834 if (!strcmp(buf, "mul")) {
1838 printf("%d: mul with %d parameters, expected 3\n", line);
1841 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1842 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1843 c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1846 if (bignum_cmp(c, p) == 0) {
1849 char *as = bignum_decimal(a);
1850 char *bs = bignum_decimal(b);
1851 char *cs = bignum_decimal(c);
1852 char *ps = bignum_decimal(p);
1854 printf("%d: fail: %s * %s gave %s expected %s\n",
1855 line, as, bs, ps, cs);
1867 } else if (!strcmp(buf, "pow")) {
1868 Bignum base, expt, modulus, expected, answer;
1871 printf("%d: mul with %d parameters, expected 3\n", line);
1875 base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1876 expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1877 modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1878 expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
1879 answer = modpow(base, expt, modulus);
1881 if (bignum_cmp(expected, answer) == 0) {
1884 char *as = bignum_decimal(base);
1885 char *bs = bignum_decimal(expt);
1886 char *cs = bignum_decimal(modulus);
1887 char *ds = bignum_decimal(answer);
1888 char *ps = bignum_decimal(expected);
1890 printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
1891 line, as, bs, cs, ds, ps);
1906 printf("%d: unrecognised test keyword: '%s'\n", line, buf);
1914 printf("passed %d failed %d total %d\n", passes, fails, passes+fails);