*/
#include <stdio.h>
+#include <assert.h>
#include <stdlib.h>
#include <string.h>
+#include <limits.h>
+#include <ctype.h>
#include "misc.h"
-#if defined __GNUC__ && defined __i386__
-typedef unsigned long BignumInt;
-typedef unsigned long long BignumDblInt;
-#define BIGNUM_INT_MASK 0xFFFFFFFFUL
-#define BIGNUM_TOP_BIT 0x80000000UL
-#define BIGNUM_INT_BITS 32
-#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
-#else
-typedef unsigned short BignumInt;
-typedef unsigned long BignumDblInt;
-#define BIGNUM_INT_MASK 0xFFFFU
-#define BIGNUM_TOP_BIT 0x8000U
-#define BIGNUM_INT_BITS 16
-#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
-#endif
-
-#define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
+#include "sshbn.h"
#define BIGNUM_INTERNAL
typedef BignumInt *Bignum;
BignumInt bnZero[1] = { 0 };
BignumInt bnOne[2] = { 1, 1 };
+BignumInt bnTen[2] = { 1, 10 };
/*
* The Bignum format is an array of `BignumInt'. The first
* nonzero.
*/
-Bignum Zero = bnZero, One = bnOne;
+Bignum Zero = bnZero, One = bnOne, Ten = bnTen;
static Bignum newbn(int length)
{
- Bignum b = snewn(length + 1, BignumInt);
- if (!b)
- abort(); /* FIXME */
+ Bignum b;
+
+ assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);
+
+ b = snewn(length + 1, BignumInt);
memset(b, 0, (length + 1) * sizeof(*b));
b[0] = length;
return b;
/*
* Burn the evidence, just in case.
*/
- memset(b, 0, sizeof(b[0]) * (b[0] + 1));
+ smemclr(b, sizeof(b[0]) * (b[0] + 1));
sfree(b);
}
Bignum bn_power_2(int n)
{
- Bignum ret = newbn(n / BIGNUM_INT_BITS + 1);
+ Bignum ret;
+
+ assert(n >= 0);
+
+ ret = newbn(n / BIGNUM_INT_BITS + 1);
bignum_set_bit(ret, n, 1);
return ret;
}
+/*
+ * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
+ * big-endian arrays of 'len' BignumInts. Returns the carry off the
+ * top.
+ */
+static BignumCarry internal_add(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len)
+{
+ int i;
+ BignumCarry carry = 0;
+
+ for (i = len-1; i >= 0; i--)
+ BignumADC(c[i], carry, a[i], b[i], carry);
+
+ return (BignumInt)carry;
+}
+
+/*
+ * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
+ * all big-endian arrays of 'len' BignumInts. Any borrow from the top
+ * is ignored.
+ */
+static void internal_sub(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len)
+{
+ int i;
+ BignumCarry carry = 1;
+
+ for (i = len-1; i >= 0; i--)
+ BignumADC(c[i], carry, a[i], ~b[i], carry);
+}
+
/*
* Compute c = a * b.
* Input is in the first len words of a and b.
* Result is returned in the first 2*len words of c.
+ *
+ * 'scratch' must point to an array of BignumInt of size at least
+ * mul_compute_scratch(len). (This covers the needs of internal_mul
+ * and all its recursive calls to itself.)
*/
-static void internal_mul(BignumInt *a, BignumInt *b,
- BignumInt *c, int len)
+#define KARATSUBA_THRESHOLD 50
+static int mul_compute_scratch(int len)
+{
+ int ret = 0;
+ while (len > KARATSUBA_THRESHOLD) {
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+ int midlen = botlen + 1;
+ ret += 4*midlen;
+ len = midlen;
+ }
+ return ret;
+}
+static void internal_mul(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len, BignumInt *scratch)
{
- int i, j;
- BignumDblInt t;
+ if (len > KARATSUBA_THRESHOLD) {
+ int i;
+
+ /*
+ * Karatsuba divide-and-conquer algorithm. Cut each input in
+ * half, so that it's expressed as two big 'digits' in a giant
+ * base D:
+ *
+ * a = a_1 D + a_0
+ * b = b_1 D + b_0
+ *
+ * Then the product is of course
+ *
+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
+ *
+ * and we compute the three coefficients by recursively
+ * calling ourself to do half-length multiplications.
+ *
+ * The clever bit that makes this worth doing is that we only
+ * need _one_ half-length multiplication for the central
+ * coefficient rather than the two that it obviouly looks
+ * like, because we can use a single multiplication to compute
+ *
+ * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
+ *
+ * and then we subtract the other two coefficients (a_1 b_1
+ * and a_0 b_0) which we were computing anyway.
+ *
+ * Hence we get to multiply two numbers of length N in about
+ * three times as much work as it takes to multiply numbers of
+ * length N/2, which is obviously better than the four times
+ * as much work it would take if we just did a long
+ * conventional multiply.
+ */
+
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+ int midlen = botlen + 1;
+ BignumCarry carry;
+#ifdef KARA_DEBUG
+ int i;
+#endif
- for (j = 0; j < 2 * len; j++)
- c[j] = 0;
+ /*
+ * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
+ * in the output array, so we can compute them immediately in
+ * place.
+ */
+
+#ifdef KARA_DEBUG
+ printf("a1,a0 = 0x");
+ for (i = 0; i < len; i++) {
+ if (i == toplen) printf(", 0x");
+ printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
+ }
+ printf("\n");
+ printf("b1,b0 = 0x");
+ for (i = 0; i < len; i++) {
+ if (i == toplen) printf(", 0x");
+ printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
+ }
+ printf("\n");
+#endif
- for (i = len - 1; i >= 0; i--) {
- t = 0;
- for (j = len - 1; j >= 0; j--) {
- t += MUL_WORD(a[i], (BignumDblInt) b[j]);
- t += (BignumDblInt) c[i + j + 1];
- c[i + j + 1] = (BignumInt) t;
- t = t >> BIGNUM_INT_BITS;
- }
- c[i] = (BignumInt) t;
+ /* a_1 b_1 */
+ internal_mul(a, b, c, toplen, scratch);
+#ifdef KARA_DEBUG
+ printf("a1b1 = 0x");
+ for (i = 0; i < 2*toplen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
+ }
+ printf("\n");
+#endif
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
+#ifdef KARA_DEBUG
+ printf("a0b0 = 0x");
+ for (i = 0; i < 2*botlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
+ }
+ printf("\n");
+#endif
+
+ /* Zero padding. midlen exceeds toplen by at most 2, so just
+ * zero the first two words of each input and the rest will be
+ * copied over. */
+ scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
+
+ for (i = 0; i < toplen; i++) {
+ scratch[midlen - toplen + i] = a[i]; /* a_1 */
+ scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
+ }
+
+ /* compute a_1 + a_0 */
+ scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
+#ifdef KARA_DEBUG
+ printf("a1plusa0 = 0x");
+ for (i = 0; i < midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
+ }
+ printf("\n");
+#endif
+ /* compute b_1 + b_0 */
+ scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
+ scratch+midlen+1, botlen);
+#ifdef KARA_DEBUG
+ printf("b1plusb0 = 0x");
+ for (i = 0; i < midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * Now we can do the third multiplication.
+ */
+ internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
+ scratch + 4*midlen);
+#ifdef KARA_DEBUG
+ printf("a1plusa0timesb1plusb0 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * Now we can reuse the first half of 'scratch' to compute the
+ * sum of the outer two coefficients, to subtract from that
+ * product to obtain the middle one.
+ */
+ scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
+ for (i = 0; i < 2*toplen; i++)
+ scratch[2*midlen - 2*toplen + i] = c[i];
+ scratch[1] = internal_add(scratch+2, c + 2*toplen,
+ scratch+2, 2*botlen);
+#ifdef KARA_DEBUG
+ printf("a1b1plusa0b0 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
+ }
+ printf("\n");
+#endif
+
+ internal_sub(scratch + 2*midlen, scratch,
+ scratch + 2*midlen, 2*midlen);
+#ifdef KARA_DEBUG
+ printf("a1b0plusa0b1 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * And now all we need to do is to add that middle coefficient
+ * back into the output. We may have to propagate a carry
+ * further up the output, but we can be sure it won't
+ * propagate right the way off the top.
+ */
+ carry = internal_add(c + 2*len - botlen - 2*midlen,
+ scratch + 2*midlen,
+ c + 2*len - botlen - 2*midlen, 2*midlen);
+ i = 2*len - botlen - 2*midlen - 1;
+ while (carry) {
+ assert(i >= 0);
+ BignumADC(c[i], carry, c[i], 0, carry);
+ i--;
+ }
+#ifdef KARA_DEBUG
+ printf("ab = 0x");
+ for (i = 0; i < 2*len; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
+ }
+ printf("\n");
+#endif
+
+ } else {
+ int i;
+ BignumInt carry;
+ const BignumInt *ap, *bp;
+ BignumInt *cp, *cps;
+
+ /*
+ * Multiply in the ordinary O(N^2) way.
+ */
+
+ for (i = 0; i < 2 * len; i++)
+ c[i] = 0;
+
+ for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
+ carry = 0;
+ for (cp = cps, bp = b + len; cp--, bp-- > b ;)
+ BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
+ *cp = carry;
+ }
}
}
+/*
+ * Variant form of internal_mul used for the initial step of
+ * Montgomery reduction. Only bothers outputting 'len' words
+ * (everything above that is thrown away).
+ */
+static void internal_mul_low(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len, BignumInt *scratch)
+{
+ if (len > KARATSUBA_THRESHOLD) {
+ int i;
+
+ /*
+ * Karatsuba-aware version of internal_mul_low. As before, we
+ * express each input value as a shifted combination of two
+ * halves:
+ *
+ * a = a_1 D + a_0
+ * b = b_1 D + b_0
+ *
+ * Then the full product is, as before,
+ *
+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
+ *
+ * Provided we choose D on the large side (so that a_0 and b_0
+ * are _at least_ as long as a_1 and b_1), we don't need the
+ * topmost term at all, and we only need half of the middle
+ * term. So there's no point in doing the proper Karatsuba
+ * optimisation which computes the middle term using the top
+ * one, because we'd take as long computing the top one as
+ * just computing the middle one directly.
+ *
+ * So instead, we do a much more obvious thing: we call the
+ * fully optimised internal_mul to compute a_0 b_0, and we
+ * recursively call ourself to compute the _bottom halves_ of
+ * a_1 b_0 and a_0 b_1, each of which we add into the result
+ * in the obvious way.
+ *
+ * In other words, there's no actual Karatsuba _optimisation_
+ * in this function; the only benefit in doing it this way is
+ * that we call internal_mul proper for a large part of the
+ * work, and _that_ can optimise its operation.
+ */
+
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+
+ /*
+ * Scratch space for the various bits and pieces we're going
+ * to be adding together: we need botlen*2 words for a_0 b_0
+ * (though we may end up throwing away its topmost word), and
+ * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
+ * to exactly 2*len.
+ */
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
+ scratch + 2*len);
+
+ /* a_1 b_0 */
+ internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
+ scratch + 2*len);
+
+ /* a_0 b_1 */
+ internal_mul_low(a + len - toplen, b, scratch, toplen,
+ scratch + 2*len);
+
+ /* Copy the bottom half of the big coefficient into place */
+ for (i = 0; i < botlen; i++)
+ c[toplen + i] = scratch[2*toplen + botlen + i];
+
+ /* Add the two small coefficients, throwing away the returned carry */
+ internal_add(scratch, scratch + toplen, scratch, toplen);
+
+ /* And add that to the large coefficient, leaving the result in c. */
+ internal_add(scratch, scratch + 2*toplen + botlen - toplen,
+ c, toplen);
+
+ } else {
+ int i;
+ BignumInt carry;
+ const BignumInt *ap, *bp;
+ BignumInt *cp, *cps;
+
+ /*
+ * Multiply in the ordinary O(N^2) way.
+ */
+
+ for (i = 0; i < len; i++)
+ c[i] = 0;
+
+ for (cps = c + len, ap = a + len; ap-- > a; cps--) {
+ carry = 0;
+ for (cp = cps, bp = b + len; bp--, cp-- > c ;)
+ BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
+ }
+ }
+}
+
+/*
+ * Montgomery reduction. Expects x to be a big-endian array of 2*len
+ * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
+ * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
+ * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
+ * x' < n.
+ *
+ * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
+ * each, containing respectively n and the multiplicative inverse of
+ * -n mod r.
+ *
+ * 'tmp' is an array of BignumInt used as scratch space, of length at
+ * least 3*len + mul_compute_scratch(len).
+ */
+static void monty_reduce(BignumInt *x, const BignumInt *n,
+ const BignumInt *mninv, BignumInt *tmp, int len)
+{
+ int i;
+ BignumInt carry;
+
+ /*
+ * Multiply x by (-n)^{-1} mod r. This gives us a value m such
+ * that mn is congruent to -x mod r. Hence, mn+x is an exact
+ * multiple of r, and is also (obviously) congruent to x mod n.
+ */
+ internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
+
+ /*
+ * Compute t = (mn+x)/r in ordinary, non-modular, integer
+ * arithmetic. By construction this is exact, and is congruent mod
+ * n to x * r^{-1}, i.e. the answer we want.
+ *
+ * The following multiply leaves that answer in the _most_
+ * significant half of the 'x' array, so then we must shift it
+ * down.
+ */
+ internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
+ carry = internal_add(x, tmp+len, x, 2*len);
+ for (i = 0; i < len; i++)
+ x[len + i] = x[i], x[i] = 0;
+
+ /*
+ * Reduce t mod n. This doesn't require a full-on division by n,
+ * but merely a test and single optional subtraction, since we can
+ * show that 0 <= t < 2n.
+ *
+ * Proof:
+ * + we computed m mod r, so 0 <= m < r.
+ * + so 0 <= mn < rn, obviously
+ * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
+ * + yielding 0 <= (mn+x)/r < 2n as required.
+ */
+ if (!carry) {
+ for (i = 0; i < len; i++)
+ if (x[len + i] != n[i])
+ break;
+ }
+ if (carry || i >= len || x[len + i] > n[i])
+ internal_sub(x+len, n, x+len, len);
+}
+
static void internal_add_shifted(BignumInt *number,
- unsigned n, int shift)
+ BignumInt n, int shift)
{
int word = 1 + (shift / BIGNUM_INT_BITS);
int bshift = shift % BIGNUM_INT_BITS;
- BignumDblInt addend;
+ BignumInt addendh, addendl;
+ BignumCarry carry;
+
+ addendl = n << bshift;
+ addendh = (bshift == 0 ? 0 : n >> (BIGNUM_INT_BITS - bshift));
+
+ assert(word <= number[0]);
+ BignumADC(number[word], carry, number[word], addendl, 0);
+ word++;
+ if (!addendh && !carry)
+ return;
+ assert(word <= number[0]);
+ BignumADC(number[word], carry, number[word], addendh, carry);
+ word++;
+ while (carry) {
+ assert(word <= number[0]);
+ BignumADC(number[word], carry, number[word], 0, carry);
+ word++;
+ }
+}
- addend = n << bshift;
+static int bn_clz(BignumInt x)
+{
+ /*
+ * Count the leading zero bits in x. Equivalently, how far left
+ * would we need to shift x to make its top bit set?
+ *
+ * Precondition: x != 0.
+ */
- while (addend) {
- addend += number[word];
- number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
- addend >>= BIGNUM_INT_BITS;
- word++;
+ /* FIXME: would be nice to put in some compiler intrinsics under
+ * ifdef here */
+ int i, ret = 0;
+ for (i = BIGNUM_INT_BITS / 2; i != 0; i >>= 1) {
+ if ((x >> (BIGNUM_INT_BITS-i)) == 0) {
+ x <<= i;
+ ret += i;
+ }
}
+ return ret;
+}
+
+static BignumInt reciprocal_word(BignumInt d)
+{
+ BignumInt dshort, recip, prodh, prodl;
+ int corrections;
+
+ /*
+ * Input: a BignumInt value d, with its top bit set.
+ */
+ assert(d >> (BIGNUM_INT_BITS-1) == 1);
+
+ /*
+ * Output: a value, shifted to fill a BignumInt, which is strictly
+ * less than 1/(d+1), i.e. is an *under*-estimate (but by as
+ * little as possible within the constraints) of the reciprocal of
+ * any number whose first BIGNUM_INT_BITS bits match d.
+ *
+ * Ideally we'd like to _totally_ fill BignumInt, i.e. always
+ * return a value with the top bit set. Unfortunately we can't
+ * quite guarantee that for all inputs and also return a fixed
+ * exponent. So instead we take our reciprocal to be
+ * 2^(BIGNUM_INT_BITS*2-1) / d, so that it has the top bit clear
+ * only in the exceptional case where d takes exactly the maximum
+ * value BIGNUM_INT_MASK; in that case, the top bit is clear and
+ * the next bit down is set.
+ */
+
+ /*
+ * Start by computing a half-length version of the answer, by
+ * straightforward division within a BignumInt.
+ */
+ dshort = (d >> (BIGNUM_INT_BITS/2)) + 1;
+ recip = (BIGNUM_TOP_BIT + dshort - 1) / dshort;
+ recip <<= BIGNUM_INT_BITS - BIGNUM_INT_BITS/2;
+
+ /*
+ * Newton-Raphson iteration to improve that starting reciprocal
+ * estimate: take f(x) = d - 1/x, and then the N-R formula gives
+ * x_new = x - f(x)/f'(x) = x - (d-1/x)/(1/x^2) = x(2-d*x). Or,
+ * taking our fixed-point representation into account, take f(x)
+ * to be d - K/x (where K = 2^(BIGNUM_INT_BITS*2-1) as discussed
+ * above) and then we get (2K - d*x) * x/K.
+ *
+ * Newton-Raphson doubles the number of correct bits at every
+ * iteration, and the initial division above already gave us half
+ * the output word, so it's only worth doing one iteration.
+ */
+ BignumMULADD(prodh, prodl, recip, d, recip);
+ prodl = ~prodl;
+ prodh = ~prodh;
+ {
+ BignumCarry c;
+ BignumADC(prodl, c, prodl, 1, 0);
+ prodh += c;
+ }
+ BignumMUL(prodh, prodl, prodh, recip);
+ recip = (prodh << 1) | (prodl >> (BIGNUM_INT_BITS-1));
+
+ /*
+ * Now make sure we have the best possible reciprocal estimate,
+ * before we return it. We might have been off by a handful either
+ * way - not enough to bother with any better-thought-out kind of
+ * correction loop.
+ */
+ BignumMULADD(prodh, prodl, recip, d, recip);
+ corrections = 0;
+ if (prodh >= BIGNUM_TOP_BIT) {
+ do {
+ BignumCarry c = 1;
+ BignumADC(prodl, c, prodl, ~d, c); prodh += BIGNUM_INT_MASK + c;
+ recip--;
+ corrections++;
+ } while (prodh >= ((BignumInt)1 << (BIGNUM_INT_BITS-1)));
+ } else {
+ while (1) {
+ BignumInt newprodh, newprodl;
+ BignumCarry c = 0;
+ BignumADC(newprodl, c, prodl, d, c); newprodh = prodh + c;
+ if (newprodh >= BIGNUM_TOP_BIT)
+ break;
+ prodh = newprodh;
+ prodl = newprodl;
+ recip++;
+ corrections++;
+ }
+ }
+
+ return recip;
}
/*
* Input in first alen words of a and first mlen words of m.
* Output in first alen words of a
* (of which first alen-mlen words will be zero).
- * The MSW of m MUST have its high bit set.
* Quotient is accumulated in the `quotient' array, which is a Bignum
- * rather than the internal bigendian format. Quotient parts are shifted
- * left by `qshift' before adding into quot.
+ * rather than the internal bigendian format.
+ *
+ * 'recip' must be the result of calling reciprocal_word() on the top
+ * BIGNUM_INT_BITS of the modulus (denoted m0 in comments below), with
+ * the topmost set bit normalised to the MSB of the input to
+ * reciprocal_word. 'rshift' is how far left the top nonzero word of
+ * the modulus had to be shifted to set that top bit.
*/
static void internal_mod(BignumInt *a, int alen,
BignumInt *m, int mlen,
- BignumInt *quot, int qshift)
+ BignumInt *quot, BignumInt recip, int rshift)
{
- BignumInt m0, m1;
- unsigned int h;
int i, k;
- m0 = m[0];
- if (mlen > 1)
- m1 = m[1];
- else
- m1 = 0;
+#ifdef DIVISION_DEBUG
+ {
+ int d;
+ printf("start division, m=0x");
+ for (d = 0; d < mlen; d++)
+ printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)m[d]);
+ printf(", recip=%#0*llx, rshift=%d\n",
+ BIGNUM_INT_BITS/4, (unsigned long long)recip, rshift);
+ }
+#endif
- for (i = 0; i <= alen - mlen; i++) {
- BignumDblInt t;
- unsigned int q, r, c, ai1;
+ /*
+ * Repeatedly use that reciprocal estimate to get a decent number
+ * of quotient bits, and subtract off the resulting multiple of m.
+ *
+ * Normally we expect to terminate this loop by means of finding
+ * out q=0 part way through, but one way in which we might not get
+ * that far in the first place is if the input a is actually zero,
+ * in which case we'll discard zero words from the front of a
+ * until we reach the termination condition in the for statement
+ * here.
+ */
+ for (i = 0; i <= alen - mlen ;) {
+ BignumInt product;
+ BignumInt aword, q;
+ int shift, full_bitoffset, bitoffset, wordoffset;
+
+#ifdef DIVISION_DEBUG
+ {
+ int d;
+ printf("main loop, a=0x");
+ for (d = 0; d < alen; d++)
+ printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
+ printf("\n");
+ }
+#endif
- if (i == 0) {
- h = 0;
- } else {
- h = a[i - 1];
- a[i - 1] = 0;
- }
+ if (a[i] == 0) {
+#ifdef DIVISION_DEBUG
+ printf("zero word at i=%d\n", i);
+#endif
+ i++;
+ continue;
+ }
+
+ aword = a[i];
+ shift = bn_clz(aword);
+ aword <<= shift;
+ if (shift > 0 && i+1 < alen)
+ aword |= a[i+1] >> (BIGNUM_INT_BITS - shift);
+
+ {
+ BignumInt unused;
+ BignumMUL(q, unused, recip, aword);
+ (void)unused;
+ }
+
+#ifdef DIVISION_DEBUG
+ printf("i=%d, aword=%#0*llx, shift=%d, q=%#0*llx\n",
+ i, BIGNUM_INT_BITS/4, (unsigned long long)aword,
+ shift, BIGNUM_INT_BITS/4, (unsigned long long)q);
+#endif
- if (i == alen - 1)
- ai1 = 0;
- else
- ai1 = a[i + 1];
-
- /* Find q = h:a[i] / m0 */
- t = ((BignumDblInt) h << BIGNUM_INT_BITS) + a[i];
- q = t / m0;
- r = t % m0;
-
- /* Refine our estimate of q by looking at
- h:a[i]:a[i+1] / m0:m1 */
- t = (BignumDblInt) m1 * (BignumDblInt) q;
- if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
- q--;
- t -= m1;
- r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
- if (r >= (BignumDblInt) m0 &&
- t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
- }
+ /*
+ * Work out the right bit and word offsets to use when
+ * subtracting q*m from a.
+ *
+ * aword was taken from a[i], which means its LSB was at bit
+ * position (alen-1-i) * BIGNUM_INT_BITS. But then we shifted
+ * it left by 'shift', so now the low bit of aword corresponds
+ * to bit position (alen-1-i) * BIGNUM_INT_BITS - shift, i.e.
+ * aword is approximately equal to a / 2^(that).
+ *
+ * m0 comes from the top word of mod, so its LSB is at bit
+ * position (mlen-1) * BIGNUM_INT_BITS - rshift, i.e. it can
+ * be considered to be m / 2^(that power). 'recip' is the
+ * reciprocal of m0, times 2^(BIGNUM_INT_BITS*2-1), i.e. it's
+ * about 2^((mlen+1) * BIGNUM_INT_BITS - rshift - 1) / m.
+ *
+ * Hence, recip * aword is approximately equal to the product
+ * of those, which simplifies to
+ *
+ * a/m * 2^((mlen+2+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
+ *
+ * But we've also shifted recip*aword down by BIGNUM_INT_BITS
+ * to form q, so we have
+ *
+ * q ~= a/m * 2^((mlen+1+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
+ *
+ * and hence, when we now compute q*m, it will be about
+ * a*2^(all that lot), i.e. the negation of that expression is
+ * how far left we have to shift the product q*m to make it
+ * approximately equal to a.
+ */
+ full_bitoffset = -((mlen+1+i-alen)*BIGNUM_INT_BITS + shift-rshift-1);
+#ifdef DIVISION_DEBUG
+ printf("full_bitoffset=%d\n", full_bitoffset);
+#endif
- /* Subtract q * m from a[i...] */
- c = 0;
- for (k = mlen - 1; k >= 0; k--) {
- t = (BignumDblInt) q * (BignumDblInt) m[k];
- t += c;
- c = t >> BIGNUM_INT_BITS;
- if ((BignumInt) t > a[i + k])
- c++;
- a[i + k] -= (BignumInt) t;
- }
+ if (full_bitoffset < 0) {
+ /*
+ * If we find ourselves needing to shift q*m _right_, that
+ * means we've reached the bottom of the quotient. Clip q
+ * so that its right shift becomes zero, and if that means
+ * q becomes _actually_ zero, this loop is done.
+ */
+ if (full_bitoffset <= -BIGNUM_INT_BITS)
+ break;
+ q >>= -full_bitoffset;
+ full_bitoffset = 0;
+ if (!q)
+ break;
+#ifdef DIVISION_DEBUG
+ printf("now full_bitoffset=%d, q=%#0*llx\n",
+ full_bitoffset, BIGNUM_INT_BITS/4, (unsigned long long)q);
+#endif
+ }
- /* Add back m in case of borrow */
- if (c != h) {
- t = 0;
- for (k = mlen - 1; k >= 0; k--) {
- t += m[k];
- t += a[i + k];
- a[i + k] = (BignumInt) t;
- t = t >> BIGNUM_INT_BITS;
- }
- q--;
- }
- if (quot)
- internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
+ wordoffset = full_bitoffset / BIGNUM_INT_BITS;
+ bitoffset = full_bitoffset % BIGNUM_INT_BITS;
+#ifdef DIVISION_DEBUG
+ printf("wordoffset=%d, bitoffset=%d\n", wordoffset, bitoffset);
+#endif
+
+ /* wordoffset as computed above is the offset between the LSWs
+ * of m and a. But in fact m and a are stored MSW-first, so we
+ * need to adjust it to be the offset between the actual array
+ * indices, and flip the sign too. */
+ wordoffset = alen - mlen - wordoffset;
+
+ if (bitoffset == 0) {
+ BignumCarry c = 1;
+ BignumInt prev_hi_word = 0;
+ for (k = mlen - 1; wordoffset+k >= i; k--) {
+ BignumInt mword = k<0 ? 0 : m[k];
+ BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
+#ifdef DIVISION_DEBUG
+ printf(" aligned sub: product word for m[%d] = %#0*llx\n",
+ k, BIGNUM_INT_BITS/4,
+ (unsigned long long)product);
+#endif
+#ifdef DIVISION_DEBUG
+ printf(" aligned sub: subtrahend for a[%d] = %#0*llx\n",
+ wordoffset+k, BIGNUM_INT_BITS/4,
+ (unsigned long long)product);
+#endif
+ BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~product, c);
+ }
+ } else {
+ BignumInt add_word = 0;
+ BignumInt c = 1;
+ BignumInt prev_hi_word = 0;
+ for (k = mlen - 1; wordoffset+k >= i; k--) {
+ BignumInt mword = k<0 ? 0 : m[k];
+ BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
+#ifdef DIVISION_DEBUG
+ printf(" unaligned sub: product word for m[%d] = %#0*llx\n",
+ k, BIGNUM_INT_BITS/4,
+ (unsigned long long)product);
+#endif
+
+ add_word |= product << bitoffset;
+
+#ifdef DIVISION_DEBUG
+ printf(" unaligned sub: subtrahend for a[%d] = %#0*llx\n",
+ wordoffset+k,
+ BIGNUM_INT_BITS/4, (unsigned long long)add_word);
+#endif
+ BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~add_word, c);
+
+ add_word = product >> (BIGNUM_INT_BITS - bitoffset);
+ }
+ }
+
+ if (quot) {
+#ifdef DIVISION_DEBUG
+ printf("adding quotient word %#0*llx << %d\n",
+ BIGNUM_INT_BITS/4, (unsigned long long)q, full_bitoffset);
+#endif
+ internal_add_shifted(quot, q, full_bitoffset);
+#ifdef DIVISION_DEBUG
+ {
+ int d;
+ printf("now quot=0x");
+ for (d = quot[0]; d > 0; d--)
+ printf("%0*llx", BIGNUM_INT_BITS/4,
+ (unsigned long long)quot[d]);
+ printf("\n");
+ }
+#endif
+ }
+ }
+
+#ifdef DIVISION_DEBUG
+ {
+ int d;
+ printf("end main loop, a=0x");
+ for (d = 0; d < alen; d++)
+ printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
+ if (quot) {
+ printf(", quot=0x");
+ for (d = quot[0]; d > 0; d--)
+ printf("%0*llx", BIGNUM_INT_BITS/4,
+ (unsigned long long)quot[d]);
+ }
+ printf("\n");
+ }
+#endif
+
+ /*
+ * The above loop should terminate with the remaining value in a
+ * being strictly less than 2*m (if a >= 2*m then we should always
+ * have managed to get a nonzero q word), but we can't guarantee
+ * that it will be strictly less than m: consider a case where the
+ * remainder is 1, and another where the remainder is m-1. By the
+ * time a contains a value that's _about m_, you clearly can't
+ * distinguish those cases by looking at only the top word of a -
+ * you have to go all the way down to the bottom before you find
+ * out whether it's just less or just more than m.
+ *
+ * Hence, we now do a final fixup in which we subtract one last
+ * copy of m, or don't, accordingly. We should never have to
+ * subtract more than one copy of m here.
+ */
+ for (i = 0; i < alen; i++) {
+ /* Compare a with m, word by word, from the MSW down. As soon
+ * as we encounter a difference, we know whether we need the
+ * fixup. */
+ int mindex = mlen-alen+i;
+ BignumInt mword = mindex < 0 ? 0 : m[mindex];
+ if (a[i] < mword) {
+#ifdef DIVISION_DEBUG
+ printf("final fixup not needed, a < m\n");
+#endif
+ return;
+ } else if (a[i] > mword) {
+#ifdef DIVISION_DEBUG
+ printf("final fixup is needed, a > m\n");
+#endif
+ break;
+ }
+ /* If neither of those cases happened, the words are the same,
+ * so keep going and look at the next one. */
+ }
+#ifdef DIVISION_DEBUG
+ if (i == mlen) /* if we printed neither of the above diagnostics */
+ printf("final fixup is needed, a == m\n");
+#endif
+
+ /*
+ * If we got here without returning, then a >= m, so we must
+ * subtract m, and increment the quotient.
+ */
+ {
+ BignumCarry c = 1;
+ for (i = alen - 1; i >= 0; i--) {
+ int mindex = mlen-alen+i;
+ BignumInt mword = mindex < 0 ? 0 : m[mindex];
+ BignumADC(a[i], c, a[i], ~mword, c);
+ }
+ }
+ if (quot)
+ internal_add_shifted(quot, 1, 0);
+
+#ifdef DIVISION_DEBUG
+ {
+ int d;
+ printf("after final fixup, a=0x");
+ for (d = 0; d < alen; d++)
+ printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
+ if (quot) {
+ printf(", quot=0x");
+ for (d = quot[0]; d > 0; d--)
+ printf("%0*llx", BIGNUM_INT_BITS/4,
+ (unsigned long long)quot[d]);
+ }
+ printf("\n");
}
+#endif
}
/*
- * Compute (base ^ exp) % mod.
- * The base MUST be smaller than the modulus.
- * The most significant word of mod MUST be non-zero.
- * We assume that the result array is the same size as the mod array.
+ * Compute (base ^ exp) % mod, the pedestrian way.
*/
-Bignum modpow(Bignum base, Bignum exp, Bignum mod)
+Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
- BignumInt *a, *b, *n, *m;
- int mshift;
- int mlen, i, j;
- Bignum result;
+ BignumInt *a, *b, *n, *m, *scratch;
+ BignumInt recip;
+ int rshift;
+ int mlen, scratchlen, i, j;
+ Bignum base, result;
+
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
+ /*
+ * Make sure the base is smaller than the modulus, by reducing
+ * it modulo the modulus if not.
+ */
+ base = bigmod(base_in, mod);
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
- /* Shift m left to make msb bit set */
- for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
- if ((m[0] << mshift) & BIGNUM_TOP_BIT)
- break;
- if (mshift) {
- for (i = 0; i < mlen - 1; i++)
- m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
- m[mlen - 1] = m[mlen - 1] << mshift;
- }
-
/* Allocate n of size mlen, copy base to n */
n = snewn(mlen, BignumInt);
i = mlen - base[0];
for (j = 0; j < i; j++)
n[j] = 0;
- for (j = 0; j < base[0]; j++)
+ for (j = 0; j < (int)base[0]; j++)
n[i + j] = base[base[0] - j];
/* Allocate a and b of size 2*mlen. Set a = 1 */
a[i] = 0;
a[2 * mlen - 1] = 1;
+ /* Scratch space for multiplies */
+ scratchlen = mul_compute_scratch(mlen);
+ scratch = snewn(scratchlen, BignumInt);
+
/* Skip leading zero bits of exp. */
i = 0;
j = BIGNUM_INT_BITS-1;
- while (i < exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
+ while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
j--;
if (j < 0) {
i++;
}
}
+ /* Compute reciprocal of the top full word of the modulus */
+ {
+ BignumInt m0 = m[0];
+ rshift = bn_clz(m0);
+ if (rshift) {
+ m0 <<= rshift;
+ if (mlen > 1)
+ m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
+ }
+ recip = reciprocal_word(m0);
+ }
+
/* Main computation */
- while (i < exp[0]) {
+ while (i < (int)exp[0]) {
while (j >= 0) {
- internal_mul(a + mlen, a + mlen, b, mlen);
- internal_mod(b, mlen * 2, m, mlen, NULL, 0);
- if ((exp[exp[0] - i] & (1 << j)) != 0) {
- internal_mul(b + mlen, n, a, mlen);
- internal_mod(a, mlen * 2, m, mlen, NULL, 0);
+ internal_mul(a + mlen, a + mlen, b, mlen, scratch);
+ internal_mod(b, mlen * 2, m, mlen, NULL, recip, rshift);
+ if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
+ internal_mul(b + mlen, n, a, mlen, scratch);
+ internal_mod(a, mlen * 2, m, mlen, NULL, recip, rshift);
} else {
BignumInt *t;
t = a;
j = BIGNUM_INT_BITS-1;
}
- /* Fixup result in case the modulus was shifted */
- if (mshift) {
- for (i = mlen - 1; i < 2 * mlen - 1; i++)
- a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
- a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
- internal_mod(a, mlen * 2, m, mlen, NULL, 0);
- for (i = 2 * mlen - 1; i >= mlen; i--)
- a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
- }
-
/* Copy result to buffer */
result = newbn(mod[0]);
for (i = 0; i < mlen; i++)
result[0]--;
/* Free temporary arrays */
- for (i = 0; i < 2 * mlen; i++)
- a[i] = 0;
+ smemclr(a, 2 * mlen * sizeof(*a));
sfree(a);
- for (i = 0; i < 2 * mlen; i++)
- b[i] = 0;
+ smemclr(scratch, scratchlen * sizeof(*scratch));
+ sfree(scratch);
+ smemclr(b, 2 * mlen * sizeof(*b));
sfree(b);
- for (i = 0; i < mlen; i++)
- m[i] = 0;
+ smemclr(m, mlen * sizeof(*m));
sfree(m);
- for (i = 0; i < mlen; i++)
- n[i] = 0;
+ smemclr(n, mlen * sizeof(*n));
+ sfree(n);
+
+ freebn(base);
+
+ return result;
+}
+
+/*
+ * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
+ * technique where possible, falling back to modpow_simple otherwise.
+ */
+Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
+{
+ BignumInt *a, *b, *x, *n, *mninv, *scratch;
+ int len, scratchlen, i, j;
+ Bignum base, base2, r, rn, inv, result;
+
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
+ /*
+ * mod had better be odd, or we can't do Montgomery multiplication
+ * using a power of two at all.
+ */
+ if (!(mod[1] & 1))
+ return modpow_simple(base_in, exp, mod);
+
+ /*
+ * Make sure the base is smaller than the modulus, by reducing
+ * it modulo the modulus if not.
+ */
+ base = bigmod(base_in, mod);
+
+ /*
+ * Compute the inverse of n mod r, for monty_reduce. (In fact we
+ * want the inverse of _minus_ n mod r, but we'll sort that out
+ * below.)
+ */
+ len = mod[0];
+ r = bn_power_2(BIGNUM_INT_BITS * len);
+ inv = modinv(mod, r);
+ assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */
+
+ /*
+ * Multiply the base by r mod n, to get it into Montgomery
+ * representation.
+ */
+ base2 = modmul(base, r, mod);
+ freebn(base);
+ base = base2;
+
+ rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
+
+ freebn(r); /* won't need this any more */
+
+ /*
+ * Set up internal arrays of the right lengths, in big-endian
+ * format, containing the base, the modulus, and the modulus's
+ * inverse.
+ */
+ n = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ n[len - 1 - j] = mod[j + 1];
+
+ mninv = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
+ freebn(inv); /* we don't need this copy of it any more */
+ /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
+ x = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ x[j] = 0;
+ internal_sub(x, mninv, mninv, len);
+
+ /* x = snewn(len, BignumInt); */ /* already done above */
+ for (j = 0; j < len; j++)
+ x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
+ freebn(base); /* we don't need this copy of it any more */
+
+ a = snewn(2*len, BignumInt);
+ b = snewn(2*len, BignumInt);
+ for (j = 0; j < len; j++)
+ a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
+ freebn(rn);
+
+ /* Scratch space for multiplies */
+ scratchlen = 3*len + mul_compute_scratch(len);
+ scratch = snewn(scratchlen, BignumInt);
+
+ /* Skip leading zero bits of exp. */
+ i = 0;
+ j = BIGNUM_INT_BITS-1;
+ while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
+ j--;
+ if (j < 0) {
+ i++;
+ j = BIGNUM_INT_BITS-1;
+ }
+ }
+
+ /* Main computation */
+ while (i < (int)exp[0]) {
+ while (j >= 0) {
+ internal_mul(a + len, a + len, b, len, scratch);
+ monty_reduce(b, n, mninv, scratch, len);
+ if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
+ internal_mul(b + len, x, a, len, scratch);
+ monty_reduce(a, n, mninv, scratch, len);
+ } else {
+ BignumInt *t;
+ t = a;
+ a = b;
+ b = t;
+ }
+ j--;
+ }
+ i++;
+ j = BIGNUM_INT_BITS-1;
+ }
+
+ /*
+ * Final monty_reduce to get back from the adjusted Montgomery
+ * representation.
+ */
+ monty_reduce(a, n, mninv, scratch, len);
+
+ /* Copy result to buffer */
+ result = newbn(mod[0]);
+ for (i = 0; i < len; i++)
+ result[result[0] - i] = a[i + len];
+ while (result[0] > 1 && result[result[0]] == 0)
+ result[0]--;
+
+ /* Free temporary arrays */
+ smemclr(scratch, scratchlen * sizeof(*scratch));
+ sfree(scratch);
+ smemclr(a, 2 * len * sizeof(*a));
+ sfree(a);
+ smemclr(b, 2 * len * sizeof(*b));
+ sfree(b);
+ smemclr(mninv, len * sizeof(*mninv));
+ sfree(mninv);
+ smemclr(n, len * sizeof(*n));
sfree(n);
+ smemclr(x, len * sizeof(*x));
+ sfree(x);
return result;
}
*/
Bignum modmul(Bignum p, Bignum q, Bignum mod)
{
- BignumInt *a, *n, *m, *o;
- int mshift;
+ BignumInt *a, *n, *m, *o, *scratch;
+ BignumInt recip;
+ int rshift, scratchlen;
int pqlen, mlen, rlen, i, j;
Bignum result;
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
- /* Shift m left to make msb bit set */
- for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
- if ((m[0] << mshift) & BIGNUM_TOP_BIT)
- break;
- if (mshift) {
- for (i = 0; i < mlen - 1; i++)
- m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
- m[mlen - 1] = m[mlen - 1] << mshift;
- }
-
pqlen = (p[0] > q[0] ? p[0] : q[0]);
+ /*
+ * Make sure that we're allowing enough space. The shifting below
+ * will underflow the vectors we allocate if pqlen is too small.
+ */
+ if (2*pqlen <= mlen)
+ pqlen = mlen/2 + 1;
+
/* Allocate n of size pqlen, copy p to n */
n = snewn(pqlen, BignumInt);
i = pqlen - p[0];
for (j = 0; j < i; j++)
n[j] = 0;
- for (j = 0; j < p[0]; j++)
+ for (j = 0; j < (int)p[0]; j++)
n[i + j] = p[p[0] - j];
/* Allocate o of size pqlen, copy q to o */
i = pqlen - q[0];
for (j = 0; j < i; j++)
o[j] = 0;
- for (j = 0; j < q[0]; j++)
+ for (j = 0; j < (int)q[0]; j++)
o[i + j] = q[q[0] - j];
/* Allocate a of size 2*pqlen for result */
a = snewn(2 * pqlen, BignumInt);
- /* Main computation */
- internal_mul(n, o, a, pqlen);
- internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
-
- /* Fixup result in case the modulus was shifted */
- if (mshift) {
- for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
- a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
- a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
- internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
- for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
- a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
+ /* Scratch space for multiplies */
+ scratchlen = mul_compute_scratch(pqlen);
+ scratch = snewn(scratchlen, BignumInt);
+
+ /* Compute reciprocal of the top full word of the modulus */
+ {
+ BignumInt m0 = m[0];
+ rshift = bn_clz(m0);
+ if (rshift) {
+ m0 <<= rshift;
+ if (mlen > 1)
+ m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
+ }
+ recip = reciprocal_word(m0);
}
+ /* Main computation */
+ internal_mul(n, o, a, pqlen, scratch);
+ internal_mod(a, pqlen * 2, m, mlen, NULL, recip, rshift);
+
/* Copy result to buffer */
rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
result = newbn(rlen);
result[0]--;
/* Free temporary arrays */
- for (i = 0; i < 2 * pqlen; i++)
- a[i] = 0;
+ smemclr(scratch, scratchlen * sizeof(*scratch));
+ sfree(scratch);
+ smemclr(a, 2 * pqlen * sizeof(*a));
sfree(a);
- for (i = 0; i < mlen; i++)
- m[i] = 0;
+ smemclr(m, mlen * sizeof(*m));
sfree(m);
- for (i = 0; i < pqlen; i++)
- n[i] = 0;
+ smemclr(n, pqlen * sizeof(*n));
sfree(n);
- for (i = 0; i < pqlen; i++)
- o[i] = 0;
+ smemclr(o, pqlen * sizeof(*o));
sfree(o);
return result;
}
+Bignum modsub(const Bignum a, const Bignum b, const Bignum n)
+{
+ Bignum a1, b1, ret;
+
+ if (bignum_cmp(a, n) >= 0) a1 = bigmod(a, n);
+ else a1 = a;
+ if (bignum_cmp(b, n) >= 0) b1 = bigmod(b, n);
+ else b1 = b;
+
+ if (bignum_cmp(a1, b1) >= 0) /* a >= b */
+ {
+ ret = bigsub(a1, b1);
+ }
+ else
+ {
+ /* Handle going round the corner of the modulus without having
+ * negative support in Bignum */
+ Bignum tmp = bigsub(n, b1);
+ assert(tmp);
+ ret = bigadd(tmp, a1);
+ freebn(tmp);
+ }
+
+ if (a != a1) freebn(a1);
+ if (b != b1) freebn(b1);
+
+ return ret;
+}
+
/*
* Compute p % mod.
* The most significant word of mod MUST be non-zero.
static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
{
BignumInt *n, *m;
- int mshift;
+ BignumInt recip;
+ int rshift;
int plen, mlen, i, j;
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
- /* Shift m left to make msb bit set */
- for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
- if ((m[0] << mshift) & BIGNUM_TOP_BIT)
- break;
- if (mshift) {
- for (i = 0; i < mlen - 1; i++)
- m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
- m[mlen - 1] = m[mlen - 1] << mshift;
- }
-
plen = p[0];
/* Ensure plen > mlen */
if (plen <= mlen)
n = snewn(plen, BignumInt);
for (j = 0; j < plen; j++)
n[j] = 0;
- for (j = 1; j <= p[0]; j++)
+ for (j = 1; j <= (int)p[0]; j++)
n[plen - j] = p[j];
- /* Main computation */
- internal_mod(n, plen, m, mlen, quotient, mshift);
-
- /* Fixup result in case the modulus was shifted */
- if (mshift) {
- for (i = plen - mlen - 1; i < plen - 1; i++)
- n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
- n[plen - 1] = n[plen - 1] << mshift;
- internal_mod(n, plen, m, mlen, quotient, 0);
- for (i = plen - 1; i >= plen - mlen; i--)
- n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
+ /* Compute reciprocal of the top full word of the modulus */
+ {
+ BignumInt m0 = m[0];
+ rshift = bn_clz(m0);
+ if (rshift) {
+ m0 <<= rshift;
+ if (mlen > 1)
+ m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
+ }
+ recip = reciprocal_word(m0);
}
+ /* Main computation */
+ internal_mod(n, plen, m, mlen, quotient, recip, rshift);
+
/* Copy result to buffer */
if (result) {
- for (i = 1; i <= result[0]; i++) {
+ for (i = 1; i <= (int)result[0]; i++) {
int j = plen - i;
result[i] = j >= 0 ? n[j] : 0;
}
}
/* Free temporary arrays */
- for (i = 0; i < mlen; i++)
- m[i] = 0;
+ smemclr(m, mlen * sizeof(*m));
sfree(m);
- for (i = 0; i < plen; i++)
- n[i] = 0;
+ smemclr(n, plen * sizeof(*n));
sfree(n);
}
void decbn(Bignum bn)
{
int i = 1;
- while (i < bn[0] && bn[i] == 0)
+ while (i < (int)bn[0] && bn[i] == 0)
bn[i++] = BIGNUM_INT_MASK;
bn[i]--;
}
Bignum result;
int w, i;
+ assert(nbytes >= 0 && nbytes < INT_MAX/8);
+
w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
result = newbn(w);
result[i] = 0;
for (i = nbytes; i--;) {
unsigned char byte = *data++;
- result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
+ result[1 + i / BIGNUM_INT_BYTES] |=
+ (BignumInt)byte << (8*i % BIGNUM_INT_BITS);
}
- while (result[0] > 1 && result[result[0]] == 0)
- result[0]--;
+ bn_restore_invariant(result);
return result;
}
+Bignum bignum_from_bytes_le(const unsigned char *data, int nbytes)
+{
+ Bignum result;
+ int w, i;
+
+ assert(nbytes >= 0 && nbytes < INT_MAX/8);
+
+ w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
+
+ result = newbn(w);
+ for (i = 1; i <= w; i++)
+ result[i] = 0;
+ for (i = 0; i < nbytes; ++i) {
+ unsigned char byte = *data++;
+ result[1 + i / BIGNUM_INT_BYTES] |=
+ (BignumInt)byte << (8*i % BIGNUM_INT_BITS);
+ }
+
+ bn_restore_invariant(result);
+ return result;
+}
+
+Bignum bignum_from_decimal(const char *decimal)
+{
+ Bignum result = copybn(Zero);
+
+ while (*decimal) {
+ Bignum tmp, tmp2;
+
+ if (!isdigit((unsigned char)*decimal)) {
+ freebn(result);
+ return 0;
+ }
+
+ tmp = bigmul(result, Ten);
+ tmp2 = bignum_from_long(*decimal - '0');
+ freebn(result);
+ result = bigadd(tmp, tmp2);
+ freebn(tmp);
+ freebn(tmp2);
+
+ decimal++;
+ }
+
+ return result;
+}
+
+Bignum bignum_random_in_range(const Bignum lower, const Bignum upper)
+{
+ Bignum ret = NULL;
+ unsigned char *bytes;
+ int upper_len = bignum_bitcount(upper);
+ int upper_bytes = upper_len / 8;
+ int upper_bits = upper_len % 8;
+ if (upper_bits) ++upper_bytes;
+
+ bytes = snewn(upper_bytes, unsigned char);
+ do {
+ int i;
+
+ if (ret) freebn(ret);
+
+ for (i = 0; i < upper_bytes; ++i)
+ {
+ bytes[i] = (unsigned char)random_byte();
+ }
+ /* Mask the top to reduce failure rate to 50/50 */
+ if (upper_bits)
+ {
+ bytes[i - 1] &= 0xFF >> (8 - upper_bits);
+ }
+
+ ret = bignum_from_bytes(bytes, upper_bytes);
+ } while (bignum_cmp(ret, lower) < 0 || bignum_cmp(ret, upper) > 0);
+ smemclr(bytes, upper_bytes);
+ sfree(bytes);
+
+ return ret;
+}
+
/*
- * Read an ssh1-format bignum from a data buffer. Return the number
- * of bytes consumed.
+ * Read an SSH-1-format bignum from a data buffer. Return the number
+ * of bytes consumed, or -1 if there wasn't enough data.
*/
-int ssh1_read_bignum(const unsigned char *data, Bignum * result)
+int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
{
const unsigned char *p = data;
int i;
int w, b;
+ if (len < 2)
+ return -1;
+
w = 0;
for (i = 0; i < 2; i++)
w = (w << 8) + *p++;
b = (w + 7) / 8; /* bits -> bytes */
+ if (len < b+2)
+ return -1;
+
if (!result) /* just return length */
return b + 2;
}
/*
- * Return the bit count of a bignum, for ssh1 encoding.
+ * Return the bit count of a bignum, for SSH-1 encoding.
*/
int bignum_bitcount(Bignum bn)
{
}
/*
- * Return the byte length of a bignum when ssh1 encoded.
+ * Return the byte length of a bignum when SSH-1 encoded.
*/
int ssh1_bignum_length(Bignum bn)
{
}
/*
- * Return the byte length of a bignum when ssh2 encoded.
+ * Return the byte length of a bignum when SSH-2 encoded.
*/
int ssh2_bignum_length(Bignum bn)
{
*/
int bignum_byte(Bignum bn, int i)
{
- if (i >= BIGNUM_INT_BYTES * bn[0])
+ if (i < 0 || i >= (int)(BIGNUM_INT_BYTES * bn[0]))
return 0; /* beyond the end */
else
return (bn[i / BIGNUM_INT_BYTES + 1] >>
*/
int bignum_bit(Bignum bn, int i)
{
- if (i >= BIGNUM_INT_BITS * bn[0])
+ if (i < 0 || i >= (int)(BIGNUM_INT_BITS * bn[0]))
return 0; /* beyond the end */
else
return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
*/
void bignum_set_bit(Bignum bn, int bitnum, int value)
{
- if (bitnum >= BIGNUM_INT_BITS * bn[0])
- abort(); /* beyond the end */
- else {
+ if (bitnum < 0 || bitnum >= (int)(BIGNUM_INT_BITS * bn[0])) {
+ if (value) abort(); /* beyond the end */
+ } else {
int v = bitnum / BIGNUM_INT_BITS + 1;
- int mask = 1 << (bitnum % BIGNUM_INT_BITS);
+ BignumInt mask = (BignumInt)1 << (bitnum % BIGNUM_INT_BITS);
if (value)
bn[v] |= mask;
else
}
/*
- * Write a ssh1-format bignum into a buffer. It is assumed the
+ * Write a SSH-1-format bignum into a buffer. It is assumed the
* buffer is big enough. Returns the number of bytes used.
*/
int ssh1_write_bignum(void *data, Bignum bn)
int bignum_cmp(Bignum a, Bignum b)
{
int amax = a[0], bmax = b[0];
- int i = (amax > bmax ? amax : bmax);
+ int i;
+
+ /* Annoyingly we have two representations of zero */
+ if (amax == 1 && a[amax] == 0)
+ amax = 0;
+ if (bmax == 1 && b[bmax] == 0)
+ bmax = 0;
+
+ assert(amax == 0 || a[amax] != 0);
+ assert(bmax == 0 || b[bmax] != 0);
+
+ i = (amax > bmax ? amax : bmax);
while (i) {
BignumInt aval = (i > amax ? 0 : a[i]);
BignumInt bval = (i > bmax ? 0 : b[i]);
int i, shiftw, shiftb, shiftbb, bits;
BignumInt ai, ai1;
+ assert(shift >= 0);
+
bits = bignum_bitcount(a) - shift;
ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
shiftbb = BIGNUM_INT_BITS - shiftb;
ai1 = a[shiftw + 1];
- for (i = 1; i <= ret[0]; i++) {
+ for (i = 1; i <= (int)ret[0]; i++) {
ai = ai1;
- ai1 = (i + shiftw + 1 <= a[0] ? a[i + shiftw + 1] : 0);
+ ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
}
}
return ret;
}
+/*
+ * Left-shift one bignum to form another.
+ */
+Bignum bignum_lshift(Bignum a, int shift)
+{
+ Bignum ret;
+ int bits, shiftWords, shiftBits;
+
+ assert(shift >= 0);
+
+ bits = bignum_bitcount(a) + shift;
+ ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
+
+ shiftWords = shift / BIGNUM_INT_BITS;
+ shiftBits = shift % BIGNUM_INT_BITS;
+
+ if (shiftBits == 0)
+ {
+ memcpy(&ret[1 + shiftWords], &a[1], sizeof(BignumInt) * a[0]);
+ }
+ else
+ {
+ int i;
+ BignumInt carry = 0;
+
+ /* Remember that Bignum[0] is length, so add 1 */
+ for (i = shiftWords + 1; i < ((int)a[0]) + shiftWords + 1; ++i)
+ {
+ BignumInt from = a[i - shiftWords];
+ ret[i] = (from << shiftBits) | carry;
+ carry = from >> (BIGNUM_INT_BITS - shiftBits);
+ }
+ if (carry) ret[i] = carry;
+ }
+
+ return ret;
+}
+
/*
* Non-modular multiplication and addition.
*/
int alen = a[0], blen = b[0];
int mlen = (alen > blen ? alen : blen);
int rlen, i, maxspot;
+ int wslen;
BignumInt *workspace;
Bignum ret;
- /* mlen space for a, mlen space for b, 2*mlen for result */
- workspace = snewn(mlen * 4, BignumInt);
+ /* mlen space for a, mlen space for b, 2*mlen for result,
+ * plus scratch space for multiplication */
+ wslen = mlen * 4 + mul_compute_scratch(mlen);
+ workspace = snewn(wslen, BignumInt);
for (i = 0; i < mlen; i++) {
- workspace[0 * mlen + i] = (mlen - i <= a[0] ? a[mlen - i] : 0);
- workspace[1 * mlen + i] = (mlen - i <= b[0] ? b[mlen - i] : 0);
+ workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
+ workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
}
internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
- workspace + 2 * mlen, mlen);
+ workspace + 2 * mlen, mlen, workspace + 4 * mlen);
/* now just copy the result back */
rlen = alen + blen + 1;
- if (addend && rlen <= addend[0])
+ if (addend && rlen <= (int)addend[0])
rlen = addend[0] + 1;
ret = newbn(rlen);
maxspot = 0;
- for (i = 1; i <= ret[0]; i++) {
+ for (i = 1; i <= (int)ret[0]; i++) {
ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
if (ret[i] != 0)
maxspot = i;
/* now add in the addend, if any */
if (addend) {
- BignumDblInt carry = 0;
+ BignumCarry carry = 0;
for (i = 1; i <= rlen; i++) {
- carry += (i <= ret[0] ? ret[i] : 0);
- carry += (i <= addend[0] ? addend[i] : 0);
- ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
- carry >>= BIGNUM_INT_BITS;
+ BignumInt retword = (i <= (int)ret[0] ? ret[i] : 0);
+ BignumInt addword = (i <= (int)addend[0] ? addend[i] : 0);
+ BignumADC(ret[i], carry, retword, addword, carry);
if (ret[i] != 0 && i > maxspot)
maxspot = i;
}
}
ret[0] = maxspot;
+ smemclr(workspace, wslen * sizeof(*workspace));
+ sfree(workspace);
return ret;
}
return bigmuladd(a, b, NULL);
}
+/*
+ * Simple addition.
+ */
+Bignum bigadd(Bignum a, Bignum b)
+{
+ int alen = a[0], blen = b[0];
+ int rlen = (alen > blen ? alen : blen) + 1;
+ int i, maxspot;
+ Bignum ret;
+ BignumCarry carry;
+
+ ret = newbn(rlen);
+
+ carry = 0;
+ maxspot = 0;
+ for (i = 1; i <= rlen; i++) {
+ BignumInt aword = (i <= (int)a[0] ? a[i] : 0);
+ BignumInt bword = (i <= (int)b[0] ? b[i] : 0);
+ BignumADC(ret[i], carry, aword, bword, carry);
+ if (ret[i] != 0 && i > maxspot)
+ maxspot = i;
+ }
+ ret[0] = maxspot;
+
+ return ret;
+}
+
+/*
+ * Subtraction. Returns a-b, or NULL if the result would come out
+ * negative (recall that this entire bignum module only handles
+ * positive numbers).
+ */
+Bignum bigsub(Bignum a, Bignum b)
+{
+ int alen = a[0], blen = b[0];
+ int rlen = (alen > blen ? alen : blen);
+ int i, maxspot;
+ Bignum ret;
+ BignumCarry carry;
+
+ ret = newbn(rlen);
+
+ carry = 1;
+ maxspot = 0;
+ for (i = 1; i <= rlen; i++) {
+ BignumInt aword = (i <= (int)a[0] ? a[i] : 0);
+ BignumInt bword = (i <= (int)b[0] ? b[i] : 0);
+ BignumADC(ret[i], carry, aword, ~bword, carry);
+ if (ret[i] != 0 && i > maxspot)
+ maxspot = i;
+ }
+ ret[0] = maxspot;
+
+ if (!carry) {
+ freebn(ret);
+ return NULL;
+ }
+
+ return ret;
+}
+
/*
* Create a bignum which is the bitmask covering another one. That
* is, the smallest integer which is >= N and is also one less than
}
/*
- * Convert a (max 32-bit) long into a bignum.
+ * Convert an unsigned long into a bignum.
*/
-Bignum bignum_from_long(unsigned long nn)
+Bignum bignum_from_long(unsigned long n)
{
+ const int maxwords =
+ (sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);
Bignum ret;
- BignumDblInt n = nn;
+ int i;
+
+ ret = newbn(maxwords);
+ ret[0] = 0;
+ for (i = 0; i < maxwords; i++) {
+ ret[i+1] = n >> (i * BIGNUM_INT_BITS);
+ if (ret[i+1] != 0)
+ ret[0] = i+1;
+ }
- ret = newbn(3);
- ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
- ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
- ret[3] = 0;
- ret[0] = (ret[2] ? 2 : 1);
return ret;
}
/*
* Add a long to a bignum.
*/
-Bignum bignum_add_long(Bignum number, unsigned long addendx)
+Bignum bignum_add_long(Bignum number, unsigned long n)
{
- Bignum ret = newbn(number[0] + 1);
- int i, maxspot = 0;
- BignumDblInt carry = 0, addend = addendx;
-
- for (i = 1; i <= ret[0]; i++) {
- carry += addend & BIGNUM_INT_MASK;
- carry += (i <= number[0] ? number[i] : 0);
- addend >>= BIGNUM_INT_BITS;
- ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
- carry >>= BIGNUM_INT_BITS;
- if (ret[i] != 0)
- maxspot = i;
+ const int maxwords =
+ (sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);
+ Bignum ret;
+ int words, i;
+ BignumCarry carry;
+
+ words = number[0];
+ if (words < maxwords)
+ words = maxwords;
+ words++;
+ ret = newbn(words);
+
+ carry = 0;
+ ret[0] = 0;
+ for (i = 0; i < words; i++) {
+ BignumInt nword = (i < maxwords ? n >> (i * BIGNUM_INT_BITS) : 0);
+ BignumInt numword = (i < number[0] ? number[i+1] : 0);
+ BignumADC(ret[i+1], carry, numword, nword, carry);
+ if (ret[i+1] != 0)
+ ret[0] = i+1;
}
- ret[0] = maxspot;
return ret;
}
*/
unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
{
- BignumDblInt mod, r;
+ unsigned long mod = modulus, r = 0;
+ /* Precompute (BIGNUM_INT_MASK+1) % mod */
+ unsigned long base_r = (BIGNUM_INT_MASK - modulus + 1) % mod;
int i;
- r = 0;
- mod = modulus;
- for (i = number[0]; i > 0; i--)
- r = (r * 65536 + number[i]) % mod;
+ for (i = number[0]; i > 0; i--) {
+ /*
+ * Conceptually, ((r << BIGNUM_INT_BITS) + number[i]) % mod
+ */
+ r = ((r * base_r) + (number[i] % mod)) % mod;
+ }
return (unsigned short) r;
}
{
Bignum q = newbn(a[0]);
bigdivmod(a, b, NULL, q);
+ while (q[0] > 1 && q[q[0]] == 0)
+ q[0]--;
return q;
}
{
Bignum r = newbn(b[0]);
bigdivmod(a, b, r, NULL);
+ while (r[0] > 1 && r[r[0]] == 0)
+ r[0]--;
return r;
}
Bignum x = copybn(One);
int sign = +1;
+ assert(number[number[0]] != 0);
+ assert(modulus[modulus[0]] != 0);
+
while (bignum_cmp(b, One) != 0) {
- Bignum t = newbn(b[0]);
- Bignum q = newbn(a[0]);
+ Bignum t, q;
+
+ if (bignum_cmp(b, Zero) == 0) {
+ /*
+ * Found a common factor between the inputs, so we cannot
+ * return a modular inverse at all.
+ */
+ freebn(b);
+ freebn(a);
+ freebn(xp);
+ freebn(x);
+ return NULL;
+ }
+
+ t = newbn(b[0]);
+ q = newbn(a[0]);
bigdivmod(a, b, t, q);
while (t[0] > 1 && t[t[0]] == 0)
t[0]--;
+ while (q[0] > 1 && q[q[0]] == 0)
+ q[0]--;
freebn(a);
a = b;
b = t;
x = bigmuladd(q, xp, t);
sign = -sign;
freebn(t);
+ freebn(q);
}
freebn(b);
int maxspot = 1;
int i;
- for (i = 1; i <= newx[0]; i++) {
- BignumInt aword = (i <= modulus[0] ? modulus[i] : 0);
- BignumInt bword = (i <= x[0] ? x[i] : 0);
+ for (i = 1; i <= (int)newx[0]; i++) {
+ BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
+ BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
newx[i] = aword - bword - carry;
bword = ~bword;
carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
{
int ndigits, ndigit;
int i, iszero;
- BignumDblInt carry;
+ BignumInt carry;
char *ret;
BignumInt *workspace;
* round up (rounding down might make it less than x again).
* Therefore if we multiply the bit count by 28/93, rounding
* up, we will have enough digits.
+ *
+ * i=0 (i.e., x=0) is an irritating special case.
*/
i = bignum_bitcount(x);
- ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
+ if (!i)
+ ndigits = 1; /* x = 0 */
+ else
+ ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
ndigits++; /* allow for trailing \0 */
ret = snewn(ndigits, char);
* big-endian form of the number.
*/
workspace = snewn(x[0], BignumInt);
- for (i = 0; i < x[0]; i++)
+ for (i = 0; i < (int)x[0]; i++)
workspace[i] = x[x[0] - i];
/*
do {
iszero = 1;
carry = 0;
- for (i = 0; i < x[0]; i++) {
- carry = (carry << BIGNUM_INT_BITS) + workspace[i];
- workspace[i] = (BignumInt) (carry / 10);
+ for (i = 0; i < (int)x[0]; i++) {
+ /*
+ * Conceptually, we want to compute
+ *
+ * (carry << BIGNUM_INT_BITS) + workspace[i]
+ * -----------------------------------------
+ * 10
+ *
+ * but we don't have an integer type longer than BignumInt
+ * to work with. So we have to do it in pieces.
+ */
+
+ BignumInt q, r;
+ q = workspace[i] / 10;
+ r = workspace[i] % 10;
+
+ /* I want (BIGNUM_INT_MASK+1)/10 but can't say so directly! */
+ q += carry * ((BIGNUM_INT_MASK-9) / 10 + 1);
+ r += carry * ((BIGNUM_INT_MASK-9) % 10);
+
+ q += r / 10;
+ r %= 10;
+
+ workspace[i] = q;
+ carry = r;
+
if (workspace[i])
iszero = 0;
- carry %= 10;
}
ret[--ndigit] = (char) (carry + '0');
} while (!iszero);
/*
* Done.
*/
+ smemclr(workspace, x[0] * sizeof(*workspace));
+ sfree(workspace);
return ret;
}
projects / PuTTY.git / blobdiff