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 #define KARATSUBA_THRESHOLD 50
206 static void internal_mul(const BignumInt *a, const BignumInt *b,
207 BignumInt *c, int len)
212 if (len > KARATSUBA_THRESHOLD) {
215 * Karatsuba divide-and-conquer algorithm. Cut each input in
216 * half, so that it's expressed as two big 'digits' in a giant
222 * Then the product is of course
224 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
226 * and we compute the three coefficients by recursively
227 * calling ourself to do half-length multiplications.
229 * The clever bit that makes this worth doing is that we only
230 * need _one_ half-length multiplication for the central
231 * coefficient rather than the two that it obviouly looks
232 * like, because we can use a single multiplication to compute
234 * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
236 * and then we subtract the other two coefficients (a_1 b_1
237 * and a_0 b_0) which we were computing anyway.
239 * Hence we get to multiply two numbers of length N in about
240 * three times as much work as it takes to multiply numbers of
241 * length N/2, which is obviously better than the four times
242 * as much work it would take if we just did a long
243 * conventional multiply.
246 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
247 int midlen = botlen + 1;
255 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
256 * in the output array, so we can compute them immediately in
261 printf("a1,a0 = 0x");
262 for (i = 0; i < len; i++) {
263 if (i == toplen) printf(", 0x");
264 printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
267 printf("b1,b0 = 0x");
268 for (i = 0; i < len; i++) {
269 if (i == toplen) printf(", 0x");
270 printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
276 internal_mul(a, b, c, toplen);
279 for (i = 0; i < 2*toplen; i++) {
280 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
286 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);
289 for (i = 0; i < 2*botlen; i++) {
290 printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
296 * We must allocate scratch space for the central coefficient,
297 * and also for the two input values that we multiply when
298 * computing it. Since either or both may carry into the
299 * (botlen+1)th word, we must use a slightly longer length
302 scratch = snewn(4 * midlen, BignumInt);
304 /* Zero padding. midlen exceeds toplen by at most 2, so just
305 * zero the first two words of each input and the rest will be
307 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
309 for (j = 0; j < toplen; j++) {
310 scratch[midlen - toplen + j] = a[j]; /* a_1 */
311 scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
314 /* compute a_1 + a_0 */
315 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
317 printf("a1plusa0 = 0x");
318 for (i = 0; i < midlen; i++) {
319 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
323 /* compute b_1 + b_0 */
324 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
325 scratch+midlen+1, botlen);
327 printf("b1plusb0 = 0x");
328 for (i = 0; i < midlen; i++) {
329 printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
335 * Now we can do the third multiplication.
337 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);
339 printf("a1plusa0timesb1plusb0 = 0x");
340 for (i = 0; i < 2*midlen; i++) {
341 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
347 * Now we can reuse the first half of 'scratch' to compute the
348 * sum of the outer two coefficients, to subtract from that
349 * product to obtain the middle one.
351 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
352 for (j = 0; j < 2*toplen; j++)
353 scratch[2*midlen - 2*toplen + j] = c[j];
354 scratch[1] = internal_add(scratch+2, c + 2*toplen,
355 scratch+2, 2*botlen);
357 printf("a1b1plusa0b0 = 0x");
358 for (i = 0; i < 2*midlen; i++) {
359 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
364 internal_sub(scratch + 2*midlen, scratch,
365 scratch + 2*midlen, 2*midlen);
367 printf("a1b0plusa0b1 = 0x");
368 for (i = 0; i < 2*midlen; i++) {
369 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
375 * And now all we need to do is to add that middle coefficient
376 * back into the output. We may have to propagate a carry
377 * further up the output, but we can be sure it won't
378 * propagate right the way off the top.
380 carry = internal_add(c + 2*len - botlen - 2*midlen,
382 c + 2*len - botlen - 2*midlen, 2*midlen);
383 j = 2*len - botlen - 2*midlen - 1;
387 c[j] = (BignumInt)carry;
388 carry >>= BIGNUM_INT_BITS;
393 for (i = 0; i < 2*len; i++) {
394 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
400 for (j = 0; j < 4 * midlen; j++)
407 * Multiply in the ordinary O(N^2) way.
410 for (j = 0; j < 2 * len; j++)
413 for (i = len - 1; i >= 0; i--) {
415 for (j = len - 1; j >= 0; j--) {
416 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
417 t += (BignumDblInt) c[i + j + 1];
418 c[i + j + 1] = (BignumInt) t;
419 t = t >> BIGNUM_INT_BITS;
421 c[i] = (BignumInt) t;
427 * Variant form of internal_mul used for the initial step of
428 * Montgomery reduction. Only bothers outputting 'len' words
429 * (everything above that is thrown away).
431 static void internal_mul_low(const BignumInt *a, const BignumInt *b,
432 BignumInt *c, int len)
437 if (len > KARATSUBA_THRESHOLD) {
440 * Karatsuba-aware version of internal_mul_low. As before, we
441 * express each input value as a shifted combination of two
447 * Then the full product is, as before,
449 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
451 * Provided we choose D on the large side (so that a_0 and b_0
452 * are _at least_ as long as a_1 and b_1), we don't need the
453 * topmost term at all, and we only need half of the middle
454 * term. So there's no point in doing the proper Karatsuba
455 * optimisation which computes the middle term using the top
456 * one, because we'd take as long computing the top one as
457 * just computing the middle one directly.
459 * So instead, we do a much more obvious thing: we call the
460 * fully optimised internal_mul to compute a_0 b_0, and we
461 * recursively call ourself to compute the _bottom halves_ of
462 * a_1 b_0 and a_0 b_1, each of which we add into the result
463 * in the obvious way.
465 * In other words, there's no actual Karatsuba _optimisation_
466 * in this function; the only benefit in doing it this way is
467 * that we call internal_mul proper for a large part of the
468 * work, and _that_ can optimise its operation.
471 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
475 * Allocate scratch space for the various bits and pieces
476 * we're going to be adding together. We need botlen*2 words
477 * for a_0 b_0 (though we may end up throwing away its topmost
478 * word), and toplen words for each of a_1 b_0 and a_0 b_1.
479 * That adds up to exactly 2*len.
481 scratch = snewn(len*2, BignumInt);
484 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);
487 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);
490 internal_mul_low(a + len - toplen, b, scratch, toplen);
492 /* Copy the bottom half of the big coefficient into place */
493 for (j = 0; j < botlen; j++)
494 c[toplen + j] = scratch[2*toplen + botlen + j];
496 /* Add the two small coefficients, throwing away the returned carry */
497 internal_add(scratch, scratch + toplen, scratch, toplen);
499 /* And add that to the large coefficient, leaving the result in c. */
500 internal_add(scratch, scratch + 2*toplen + botlen - toplen,
504 for (j = 0; j < len*2; j++)
510 for (j = 0; j < len; j++)
513 for (i = len - 1; i >= 0; i--) {
515 for (j = len - 1; j >= len - i - 1; j--) {
516 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
517 t += (BignumDblInt) c[i + j + 1 - len];
518 c[i + j + 1 - len] = (BignumInt) t;
519 t = t >> BIGNUM_INT_BITS;
527 * Montgomery reduction. Expects x to be a big-endian array of 2*len
528 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
529 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
530 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
533 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
534 * each, containing respectively n and the multiplicative inverse of
537 * 'tmp' is an array of at least '3*len' BignumInts used as scratch
540 static void monty_reduce(BignumInt *x, const BignumInt *n,
541 const BignumInt *mninv, BignumInt *tmp, int len)
547 * Multiply x by (-n)^{-1} mod r. This gives us a value m such
548 * that mn is congruent to -x mod r. Hence, mn+x is an exact
549 * multiple of r, and is also (obviously) congruent to x mod n.
551 internal_mul_low(x + len, mninv, tmp, len);
554 * Compute t = (mn+x)/r in ordinary, non-modular, integer
555 * arithmetic. By construction this is exact, and is congruent mod
556 * n to x * r^{-1}, i.e. the answer we want.
558 * The following multiply leaves that answer in the _most_
559 * significant half of the 'x' array, so then we must shift it
562 internal_mul(tmp, n, tmp+len, len);
563 carry = internal_add(x, tmp+len, x, 2*len);
564 for (i = 0; i < len; i++)
565 x[len + i] = x[i], x[i] = 0;
568 * Reduce t mod n. This doesn't require a full-on division by n,
569 * but merely a test and single optional subtraction, since we can
570 * show that 0 <= t < 2n.
573 * + we computed m mod r, so 0 <= m < r.
574 * + so 0 <= mn < rn, obviously
575 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
576 * + yielding 0 <= (mn+x)/r < 2n as required.
579 for (i = 0; i < len; i++)
580 if (x[len + i] != n[i])
583 if (carry || i >= len || x[len + i] > n[i])
584 internal_sub(x+len, n, x+len, len);
587 static void internal_add_shifted(BignumInt *number,
588 unsigned n, int shift)
590 int word = 1 + (shift / BIGNUM_INT_BITS);
591 int bshift = shift % BIGNUM_INT_BITS;
594 addend = (BignumDblInt)n << bshift;
597 addend += number[word];
598 number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
599 addend >>= BIGNUM_INT_BITS;
606 * Input in first alen words of a and first mlen words of m.
607 * Output in first alen words of a
608 * (of which first alen-mlen words will be zero).
609 * The MSW of m MUST have its high bit set.
610 * Quotient is accumulated in the `quotient' array, which is a Bignum
611 * rather than the internal bigendian format. Quotient parts are shifted
612 * left by `qshift' before adding into quot.
614 static void internal_mod(BignumInt *a, int alen,
615 BignumInt *m, int mlen,
616 BignumInt *quot, int qshift)
628 for (i = 0; i <= alen - mlen; i++) {
630 unsigned int q, r, c, ai1;
644 /* Find q = h:a[i] / m0 */
649 * To illustrate it, suppose a BignumInt is 8 bits, and
650 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
651 * our initial division will be 0xA123 / 0xA1, which
652 * will give a quotient of 0x100 and a divide overflow.
653 * However, the invariants in this division algorithm
654 * are not violated, since the full number A1:23:... is
655 * _less_ than the quotient prefix A1:B2:... and so the
656 * following correction loop would have sorted it out.
658 * In this situation we set q to be the largest
659 * quotient we _can_ stomach (0xFF, of course).
663 /* Macro doesn't want an array subscript expression passed
664 * into it (see definition), so use a temporary. */
665 BignumInt tmplo = a[i];
666 DIVMOD_WORD(q, r, h, tmplo, m0);
668 /* Refine our estimate of q by looking at
669 h:a[i]:a[i+1] / m0:m1 */
671 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
674 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
675 if (r >= (BignumDblInt) m0 &&
676 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
680 /* Subtract q * m from a[i...] */
682 for (k = mlen - 1; k >= 0; k--) {
683 t = MUL_WORD(q, m[k]);
685 c = (unsigned)(t >> BIGNUM_INT_BITS);
686 if ((BignumInt) t > a[i + k])
688 a[i + k] -= (BignumInt) t;
691 /* Add back m in case of borrow */
694 for (k = mlen - 1; k >= 0; k--) {
697 a[i + k] = (BignumInt) t;
698 t = t >> BIGNUM_INT_BITS;
703 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
708 * Compute (base ^ exp) % mod, the pedestrian way.
710 Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
712 BignumInt *a, *b, *n, *m;
718 * The most significant word of mod needs to be non-zero. It
719 * should already be, but let's make sure.
721 assert(mod[mod[0]] != 0);
724 * Make sure the base is smaller than the modulus, by reducing
725 * it modulo the modulus if not.
727 base = bigmod(base_in, mod);
729 /* Allocate m of size mlen, copy mod to m */
730 /* We use big endian internally */
732 m = snewn(mlen, BignumInt);
733 for (j = 0; j < mlen; j++)
734 m[j] = mod[mod[0] - j];
736 /* Shift m left to make msb bit set */
737 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
738 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
741 for (i = 0; i < mlen - 1; i++)
742 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
743 m[mlen - 1] = m[mlen - 1] << mshift;
746 /* Allocate n of size mlen, copy base to n */
747 n = snewn(mlen, BignumInt);
749 for (j = 0; j < i; j++)
751 for (j = 0; j < (int)base[0]; j++)
752 n[i + j] = base[base[0] - j];
754 /* Allocate a and b of size 2*mlen. Set a = 1 */
755 a = snewn(2 * mlen, BignumInt);
756 b = snewn(2 * mlen, BignumInt);
757 for (i = 0; i < 2 * mlen; i++)
761 /* Skip leading zero bits of exp. */
763 j = BIGNUM_INT_BITS-1;
764 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
768 j = BIGNUM_INT_BITS-1;
772 /* Main computation */
773 while (i < (int)exp[0]) {
775 internal_mul(a + mlen, a + mlen, b, mlen);
776 internal_mod(b, mlen * 2, m, mlen, NULL, 0);
777 if ((exp[exp[0] - i] & (1 << j)) != 0) {
778 internal_mul(b + mlen, n, a, mlen);
779 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
789 j = BIGNUM_INT_BITS-1;
792 /* Fixup result in case the modulus was shifted */
794 for (i = mlen - 1; i < 2 * mlen - 1; i++)
795 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
796 a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
797 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
798 for (i = 2 * mlen - 1; i >= mlen; i--)
799 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
802 /* Copy result to buffer */
803 result = newbn(mod[0]);
804 for (i = 0; i < mlen; i++)
805 result[result[0] - i] = a[i + mlen];
806 while (result[0] > 1 && result[result[0]] == 0)
809 /* Free temporary arrays */
810 for (i = 0; i < 2 * mlen; i++)
813 for (i = 0; i < 2 * mlen; i++)
816 for (i = 0; i < mlen; i++)
819 for (i = 0; i < mlen; i++)
829 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
830 * technique where possible, falling back to modpow_simple otherwise.
832 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
834 BignumInt *a, *b, *x, *n, *mninv, *tmp;
836 Bignum base, base2, r, rn, inv, result;
839 * The most significant word of mod needs to be non-zero. It
840 * should already be, but let's make sure.
842 assert(mod[mod[0]] != 0);
845 * mod had better be odd, or we can't do Montgomery multiplication
846 * using a power of two at all.
849 return modpow_simple(base_in, exp, mod);
852 * Make sure the base is smaller than the modulus, by reducing
853 * it modulo the modulus if not.
855 base = bigmod(base_in, mod);
858 * Compute the inverse of n mod r, for monty_reduce. (In fact we
859 * want the inverse of _minus_ n mod r, but we'll sort that out
863 r = bn_power_2(BIGNUM_INT_BITS * len);
864 inv = modinv(mod, r);
867 * Multiply the base by r mod n, to get it into Montgomery
870 base2 = modmul(base, r, mod);
874 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
876 freebn(r); /* won't need this any more */
879 * Set up internal arrays of the right lengths, in big-endian
880 * format, containing the base, the modulus, and the modulus's
883 n = snewn(len, BignumInt);
884 for (j = 0; j < len; j++)
885 n[len - 1 - j] = mod[j + 1];
887 mninv = snewn(len, BignumInt);
888 for (j = 0; j < len; j++)
889 mninv[len - 1 - j] = (j < inv[0] ? inv[j + 1] : 0);
890 freebn(inv); /* we don't need this copy of it any more */
891 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
892 x = snewn(len, BignumInt);
893 for (j = 0; j < len; j++)
895 internal_sub(x, mninv, mninv, len);
897 /* x = snewn(len, BignumInt); */ /* already done above */
898 for (j = 0; j < len; j++)
899 x[len - 1 - j] = (j < base[0] ? base[j + 1] : 0);
900 freebn(base); /* we don't need this copy of it any more */
902 a = snewn(2*len, BignumInt);
903 b = snewn(2*len, BignumInt);
904 for (j = 0; j < len; j++)
905 a[2*len - 1 - j] = (j < rn[0] ? rn[j + 1] : 0);
908 tmp = snewn(3*len, BignumInt);
910 /* Skip leading zero bits of exp. */
912 j = BIGNUM_INT_BITS-1;
913 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
917 j = BIGNUM_INT_BITS-1;
921 /* Main computation */
922 while (i < (int)exp[0]) {
924 internal_mul(a + len, a + len, b, len);
925 monty_reduce(b, n, mninv, tmp, len);
926 if ((exp[exp[0] - i] & (1 << j)) != 0) {
927 internal_mul(b + len, x, a, len);
928 monty_reduce(a, n, mninv, tmp, len);
938 j = BIGNUM_INT_BITS-1;
942 * Final monty_reduce to get back from the adjusted Montgomery
945 monty_reduce(a, n, mninv, tmp, len);
947 /* Copy result to buffer */
948 result = newbn(mod[0]);
949 for (i = 0; i < len; i++)
950 result[result[0] - i] = a[i + len];
951 while (result[0] > 1 && result[result[0]] == 0)
954 /* Free temporary arrays */
955 for (i = 0; i < 3 * len; i++)
958 for (i = 0; i < 2 * len; i++)
961 for (i = 0; i < 2 * len; i++)
964 for (i = 0; i < len; i++)
967 for (i = 0; i < len; i++)
970 for (i = 0; i < len; i++)
978 * Compute (p * q) % mod.
979 * The most significant word of mod MUST be non-zero.
980 * We assume that the result array is the same size as the mod array.
982 Bignum modmul(Bignum p, Bignum q, Bignum mod)
984 BignumInt *a, *n, *m, *o;
986 int pqlen, mlen, rlen, i, j;
989 /* Allocate m of size mlen, copy mod to m */
990 /* We use big endian internally */
992 m = snewn(mlen, BignumInt);
993 for (j = 0; j < mlen; j++)
994 m[j] = mod[mod[0] - j];
996 /* Shift m left to make msb bit set */
997 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
998 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1001 for (i = 0; i < mlen - 1; i++)
1002 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1003 m[mlen - 1] = m[mlen - 1] << mshift;
1006 pqlen = (p[0] > q[0] ? p[0] : q[0]);
1008 /* Allocate n of size pqlen, copy p to n */
1009 n = snewn(pqlen, BignumInt);
1011 for (j = 0; j < i; j++)
1013 for (j = 0; j < (int)p[0]; j++)
1014 n[i + j] = p[p[0] - j];
1016 /* Allocate o of size pqlen, copy q to o */
1017 o = snewn(pqlen, BignumInt);
1019 for (j = 0; j < i; j++)
1021 for (j = 0; j < (int)q[0]; j++)
1022 o[i + j] = q[q[0] - j];
1024 /* Allocate a of size 2*pqlen for result */
1025 a = snewn(2 * pqlen, BignumInt);
1027 /* Main computation */
1028 internal_mul(n, o, a, pqlen);
1029 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1031 /* Fixup result in case the modulus was shifted */
1033 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
1034 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
1035 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
1036 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
1037 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
1038 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
1041 /* Copy result to buffer */
1042 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
1043 result = newbn(rlen);
1044 for (i = 0; i < rlen; i++)
1045 result[result[0] - i] = a[i + 2 * pqlen - rlen];
1046 while (result[0] > 1 && result[result[0]] == 0)
1049 /* Free temporary arrays */
1050 for (i = 0; i < 2 * pqlen; i++)
1053 for (i = 0; i < mlen; i++)
1056 for (i = 0; i < pqlen; i++)
1059 for (i = 0; i < pqlen; i++)
1068 * The most significant word of mod MUST be non-zero.
1069 * We assume that the result array is the same size as the mod array.
1070 * We optionally write out a quotient if `quotient' is non-NULL.
1071 * We can avoid writing out the result if `result' is NULL.
1073 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
1077 int plen, mlen, i, j;
1079 /* Allocate m of size mlen, copy mod to m */
1080 /* We use big endian internally */
1082 m = snewn(mlen, BignumInt);
1083 for (j = 0; j < mlen; j++)
1084 m[j] = mod[mod[0] - j];
1086 /* Shift m left to make msb bit set */
1087 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
1088 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
1091 for (i = 0; i < mlen - 1; i++)
1092 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
1093 m[mlen - 1] = m[mlen - 1] << mshift;
1097 /* Ensure plen > mlen */
1101 /* Allocate n of size plen, copy p to n */
1102 n = snewn(plen, BignumInt);
1103 for (j = 0; j < plen; j++)
1105 for (j = 1; j <= (int)p[0]; j++)
1108 /* Main computation */
1109 internal_mod(n, plen, m, mlen, quotient, mshift);
1111 /* Fixup result in case the modulus was shifted */
1113 for (i = plen - mlen - 1; i < plen - 1; i++)
1114 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
1115 n[plen - 1] = n[plen - 1] << mshift;
1116 internal_mod(n, plen, m, mlen, quotient, 0);
1117 for (i = plen - 1; i >= plen - mlen; i--)
1118 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
1121 /* Copy result to buffer */
1123 for (i = 1; i <= (int)result[0]; i++) {
1125 result[i] = j >= 0 ? n[j] : 0;
1129 /* Free temporary arrays */
1130 for (i = 0; i < mlen; i++)
1133 for (i = 0; i < plen; i++)
1139 * Decrement a number.
1141 void decbn(Bignum bn)
1144 while (i < (int)bn[0] && bn[i] == 0)
1145 bn[i++] = BIGNUM_INT_MASK;
1149 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
1154 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
1157 for (i = 1; i <= w; i++)
1159 for (i = nbytes; i--;) {
1160 unsigned char byte = *data++;
1161 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
1164 while (result[0] > 1 && result[result[0]] == 0)
1170 * Read an SSH-1-format bignum from a data buffer. Return the number
1171 * of bytes consumed, or -1 if there wasn't enough data.
1173 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
1175 const unsigned char *p = data;
1183 for (i = 0; i < 2; i++)
1184 w = (w << 8) + *p++;
1185 b = (w + 7) / 8; /* bits -> bytes */
1190 if (!result) /* just return length */
1193 *result = bignum_from_bytes(p, b);
1195 return p + b - data;
1199 * Return the bit count of a bignum, for SSH-1 encoding.
1201 int bignum_bitcount(Bignum bn)
1203 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
1204 while (bitcount >= 0
1205 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
1206 return bitcount + 1;
1210 * Return the byte length of a bignum when SSH-1 encoded.
1212 int ssh1_bignum_length(Bignum bn)
1214 return 2 + (bignum_bitcount(bn) + 7) / 8;
1218 * Return the byte length of a bignum when SSH-2 encoded.
1220 int ssh2_bignum_length(Bignum bn)
1222 return 4 + (bignum_bitcount(bn) + 8) / 8;
1226 * Return a byte from a bignum; 0 is least significant, etc.
1228 int bignum_byte(Bignum bn, int i)
1230 if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
1231 return 0; /* beyond the end */
1233 return (bn[i / BIGNUM_INT_BYTES + 1] >>
1234 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
1238 * Return a bit from a bignum; 0 is least significant, etc.
1240 int bignum_bit(Bignum bn, int i)
1242 if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
1243 return 0; /* beyond the end */
1245 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
1249 * Set a bit in a bignum; 0 is least significant, etc.
1251 void bignum_set_bit(Bignum bn, int bitnum, int value)
1253 if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
1254 abort(); /* beyond the end */
1256 int v = bitnum / BIGNUM_INT_BITS + 1;
1257 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
1266 * Write a SSH-1-format bignum into a buffer. It is assumed the
1267 * buffer is big enough. Returns the number of bytes used.
1269 int ssh1_write_bignum(void *data, Bignum bn)
1271 unsigned char *p = data;
1272 int len = ssh1_bignum_length(bn);
1274 int bitc = bignum_bitcount(bn);
1276 *p++ = (bitc >> 8) & 0xFF;
1277 *p++ = (bitc) & 0xFF;
1278 for (i = len - 2; i--;)
1279 *p++ = bignum_byte(bn, i);
1284 * Compare two bignums. Returns like strcmp.
1286 int bignum_cmp(Bignum a, Bignum b)
1288 int amax = a[0], bmax = b[0];
1289 int i = (amax > bmax ? amax : bmax);
1291 BignumInt aval = (i > amax ? 0 : a[i]);
1292 BignumInt bval = (i > bmax ? 0 : b[i]);
1303 * Right-shift one bignum to form another.
1305 Bignum bignum_rshift(Bignum a, int shift)
1308 int i, shiftw, shiftb, shiftbb, bits;
1311 bits = bignum_bitcount(a) - shift;
1312 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
1315 shiftw = shift / BIGNUM_INT_BITS;
1316 shiftb = shift % BIGNUM_INT_BITS;
1317 shiftbb = BIGNUM_INT_BITS - shiftb;
1319 ai1 = a[shiftw + 1];
1320 for (i = 1; i <= (int)ret[0]; i++) {
1322 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
1323 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
1331 * Non-modular multiplication and addition.
1333 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
1335 int alen = a[0], blen = b[0];
1336 int mlen = (alen > blen ? alen : blen);
1337 int rlen, i, maxspot;
1338 BignumInt *workspace;
1341 /* mlen space for a, mlen space for b, 2*mlen for result */
1342 workspace = snewn(mlen * 4, BignumInt);
1343 for (i = 0; i < mlen; i++) {
1344 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
1345 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
1348 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
1349 workspace + 2 * mlen, mlen);
1351 /* now just copy the result back */
1352 rlen = alen + blen + 1;
1353 if (addend && rlen <= (int)addend[0])
1354 rlen = addend[0] + 1;
1357 for (i = 1; i <= (int)ret[0]; i++) {
1358 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
1364 /* now add in the addend, if any */
1366 BignumDblInt carry = 0;
1367 for (i = 1; i <= rlen; i++) {
1368 carry += (i <= (int)ret[0] ? ret[i] : 0);
1369 carry += (i <= (int)addend[0] ? addend[i] : 0);
1370 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1371 carry >>= BIGNUM_INT_BITS;
1372 if (ret[i] != 0 && i > maxspot)
1383 * Non-modular multiplication.
1385 Bignum bigmul(Bignum a, Bignum b)
1387 return bigmuladd(a, b, NULL);
1393 Bignum bigadd(Bignum a, Bignum b)
1395 int alen = a[0], blen = b[0];
1396 int rlen = (alen > blen ? alen : blen) + 1;
1405 for (i = 1; i <= rlen; i++) {
1406 carry += (i <= (int)a[0] ? a[i] : 0);
1407 carry += (i <= (int)b[0] ? b[i] : 0);
1408 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1409 carry >>= BIGNUM_INT_BITS;
1410 if (ret[i] != 0 && i > maxspot)
1419 * Subtraction. Returns a-b, or NULL if the result would come out
1420 * negative (recall that this entire bignum module only handles
1421 * positive numbers).
1423 Bignum bigsub(Bignum a, Bignum b)
1425 int alen = a[0], blen = b[0];
1426 int rlen = (alen > blen ? alen : blen);
1435 for (i = 1; i <= rlen; i++) {
1436 carry += (i <= (int)a[0] ? a[i] : 0);
1437 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
1438 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1439 carry >>= BIGNUM_INT_BITS;
1440 if (ret[i] != 0 && i > maxspot)
1454 * Create a bignum which is the bitmask covering another one. That
1455 * is, the smallest integer which is >= N and is also one less than
1458 Bignum bignum_bitmask(Bignum n)
1460 Bignum ret = copybn(n);
1465 while (n[i] == 0 && i > 0)
1468 return ret; /* input was zero */
1474 ret[i] = BIGNUM_INT_MASK;
1479 * Convert a (max 32-bit) long into a bignum.
1481 Bignum bignum_from_long(unsigned long nn)
1484 BignumDblInt n = nn;
1487 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
1488 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
1490 ret[0] = (ret[2] ? 2 : 1);
1495 * Add a long to a bignum.
1497 Bignum bignum_add_long(Bignum number, unsigned long addendx)
1499 Bignum ret = newbn(number[0] + 1);
1501 BignumDblInt carry = 0, addend = addendx;
1503 for (i = 1; i <= (int)ret[0]; i++) {
1504 carry += addend & BIGNUM_INT_MASK;
1505 carry += (i <= (int)number[0] ? number[i] : 0);
1506 addend >>= BIGNUM_INT_BITS;
1507 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1508 carry >>= BIGNUM_INT_BITS;
1517 * Compute the residue of a bignum, modulo a (max 16-bit) short.
1519 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
1521 BignumDblInt mod, r;
1526 for (i = number[0]; i > 0; i--)
1527 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
1528 return (unsigned short) r;
1532 void diagbn(char *prefix, Bignum md)
1534 int i, nibbles, morenibbles;
1535 static const char hex[] = "0123456789ABCDEF";
1537 debug(("%s0x", prefix ? prefix : ""));
1539 nibbles = (3 + bignum_bitcount(md)) / 4;
1542 morenibbles = 4 * md[0] - nibbles;
1543 for (i = 0; i < morenibbles; i++)
1545 for (i = nibbles; i--;)
1547 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
1557 Bignum bigdiv(Bignum a, Bignum b)
1559 Bignum q = newbn(a[0]);
1560 bigdivmod(a, b, NULL, q);
1567 Bignum bigmod(Bignum a, Bignum b)
1569 Bignum r = newbn(b[0]);
1570 bigdivmod(a, b, r, NULL);
1575 * Greatest common divisor.
1577 Bignum biggcd(Bignum av, Bignum bv)
1579 Bignum a = copybn(av);
1580 Bignum b = copybn(bv);
1582 while (bignum_cmp(b, Zero) != 0) {
1583 Bignum t = newbn(b[0]);
1584 bigdivmod(a, b, t, NULL);
1585 while (t[0] > 1 && t[t[0]] == 0)
1597 * Modular inverse, using Euclid's extended algorithm.
1599 Bignum modinv(Bignum number, Bignum modulus)
1601 Bignum a = copybn(modulus);
1602 Bignum b = copybn(number);
1603 Bignum xp = copybn(Zero);
1604 Bignum x = copybn(One);
1607 while (bignum_cmp(b, One) != 0) {
1608 Bignum t = newbn(b[0]);
1609 Bignum q = newbn(a[0]);
1610 bigdivmod(a, b, t, q);
1611 while (t[0] > 1 && t[t[0]] == 0)
1618 x = bigmuladd(q, xp, t);
1628 /* now we know that sign * x == 1, and that x < modulus */
1630 /* set a new x to be modulus - x */
1631 Bignum newx = newbn(modulus[0]);
1632 BignumInt carry = 0;
1636 for (i = 1; i <= (int)newx[0]; i++) {
1637 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
1638 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
1639 newx[i] = aword - bword - carry;
1641 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
1655 * Render a bignum into decimal. Return a malloced string holding
1656 * the decimal representation.
1658 char *bignum_decimal(Bignum x)
1660 int ndigits, ndigit;
1664 BignumInt *workspace;
1667 * First, estimate the number of digits. Since log(10)/log(2)
1668 * is just greater than 93/28 (the joys of continued fraction
1669 * approximations...) we know that for every 93 bits, we need
1670 * at most 28 digits. This will tell us how much to malloc.
1672 * Formally: if x has i bits, that means x is strictly less
1673 * than 2^i. Since 2 is less than 10^(28/93), this is less than
1674 * 10^(28i/93). We need an integer power of ten, so we must
1675 * round up (rounding down might make it less than x again).
1676 * Therefore if we multiply the bit count by 28/93, rounding
1677 * up, we will have enough digits.
1679 * i=0 (i.e., x=0) is an irritating special case.
1681 i = bignum_bitcount(x);
1683 ndigits = 1; /* x = 0 */
1685 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
1686 ndigits++; /* allow for trailing \0 */
1687 ret = snewn(ndigits, char);
1690 * Now allocate some workspace to hold the binary form as we
1691 * repeatedly divide it by ten. Initialise this to the
1692 * big-endian form of the number.
1694 workspace = snewn(x[0], BignumInt);
1695 for (i = 0; i < (int)x[0]; i++)
1696 workspace[i] = x[x[0] - i];
1699 * Next, write the decimal number starting with the last digit.
1700 * We use ordinary short division, dividing 10 into the
1703 ndigit = ndigits - 1;
1708 for (i = 0; i < (int)x[0]; i++) {
1709 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
1710 workspace[i] = (BignumInt) (carry / 10);
1715 ret[--ndigit] = (char) (carry + '0');
1719 * There's a chance we've fallen short of the start of the
1720 * string. Correct if so.
1723 memmove(ret, ret + ndigit, ndigits - ndigit);
1739 * gcc -g -O0 -DTESTBN -o testbn sshbn.c misc.c -I unix -I charset
1741 * Then feed to this program's standard input the output of
1742 * testdata/bignum.py .
1745 void modalfatalbox(char *p, ...)
1748 fprintf(stderr, "FATAL ERROR: ");
1750 vfprintf(stderr, p, ap);
1752 fputc('\n', stderr);
1756 #define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
1758 int main(int argc, char **argv)
1762 int passes = 0, fails = 0;
1764 while ((buf = fgetline(stdin)) != NULL) {
1765 int maxlen = strlen(buf);
1766 unsigned char *data = snewn(maxlen, unsigned char);
1767 unsigned char *ptrs[5], *q;
1776 while (*bufp && !isspace((unsigned char)*bufp))
1785 while (*bufp && !isxdigit((unsigned char)*bufp))
1792 while (*bufp && isxdigit((unsigned char)*bufp))
1796 if (ptrnum >= lenof(ptrs))
1800 for (i = -((end - start) & 1); i < end-start; i += 2) {
1801 unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
1802 val = val * 16 + fromxdigit(start[i+1]);
1809 if (!strcmp(buf, "mul")) {
1813 printf("%d: mul with %d parameters, expected 3\n", line);
1816 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1817 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1818 c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1821 if (bignum_cmp(c, p) == 0) {
1824 char *as = bignum_decimal(a);
1825 char *bs = bignum_decimal(b);
1826 char *cs = bignum_decimal(c);
1827 char *ps = bignum_decimal(p);
1829 printf("%d: fail: %s * %s gave %s expected %s\n",
1830 line, as, bs, ps, cs);
1842 } else if (!strcmp(buf, "pow")) {
1843 Bignum base, expt, modulus, expected, answer;
1846 printf("%d: mul with %d parameters, expected 3\n", line);
1850 base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1851 expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1852 modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1853 expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
1854 answer = modpow(base, expt, modulus);
1856 if (bignum_cmp(expected, answer) == 0) {
1859 char *as = bignum_decimal(base);
1860 char *bs = bignum_decimal(expt);
1861 char *cs = bignum_decimal(modulus);
1862 char *ds = bignum_decimal(answer);
1863 char *ps = bignum_decimal(expected);
1865 printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
1866 line, as, bs, cs, ds, ps);
1881 printf("%d: unrecognised test keyword: '%s'\n", line, buf);
1889 printf("passed %d failed %d total %d\n", passes, fails, passes+fails);