*/
#include <stdio.h>
+#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include "misc.h"
+/*
+ * Usage notes:
+ * * Do not call the DIVMOD_WORD macro with expressions such as array
+ * subscripts, as some implementations object to this (see below).
+ * * Note that none of the division methods below will cope if the
+ * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
+ * to avoid this case.
+ * If this condition occurs, in the case of the x86 DIV instruction,
+ * an overflow exception will occur, which (according to a correspondent)
+ * will manifest on Windows as something like
+ * 0xC0000095: Integer overflow
+ * The C variant won't give the right answer, either.
+ */
+
#if defined __GNUC__ && defined __i386__
typedef unsigned long BignumInt;
typedef unsigned long long BignumDblInt;
#define BIGNUM_TOP_BIT 0x80000000UL
#define BIGNUM_INT_BITS 32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
+#define DIVMOD_WORD(q, r, hi, lo, w) \
+ __asm__("div %2" : \
+ "=d" (r), "=a" (q) : \
+ "r" (w), "d" (hi), "a" (lo))
+#elif defined _MSC_VER && defined _M_IX86
+typedef unsigned __int32 BignumInt;
+typedef unsigned __int64 BignumDblInt;
+#define BIGNUM_INT_MASK 0xFFFFFFFFUL
+#define BIGNUM_TOP_BIT 0x80000000UL
+#define BIGNUM_INT_BITS 32
+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
+/* Note: MASM interprets array subscripts in the macro arguments as
+ * assembler syntax, which gives the wrong answer. Don't supply them.
+ * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ __asm mov edx, hi \
+ __asm mov eax, lo \
+ __asm div w \
+ __asm mov r, edx \
+ __asm mov q, eax \
+} while(0)
+#elif defined _LP64
+/* 64-bit architectures can do 32x32->64 chunks at a time */
+typedef unsigned int BignumInt;
+typedef unsigned long BignumDblInt;
+#define BIGNUM_INT_MASK 0xFFFFFFFFU
+#define BIGNUM_TOP_BIT 0x80000000U
+#define BIGNUM_INT_BITS 32
+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
+ q = n / w; \
+ r = n % w; \
+} while (0)
+#elif defined _LLP64
+/* 64-bit architectures in which unsigned long is 32 bits, not 64 */
+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)
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
+ q = n / w; \
+ r = n % w; \
+} while (0)
#else
+/* Fallback for all other cases */
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)
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
+ q = n / w; \
+ r = n % w; \
+} while (0)
#endif
#define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
}
/*
+ * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
+ * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
+ * off the top.
+ */
+static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len)
+{
+ int i;
+ BignumDblInt carry = 0;
+
+ for (i = len-1; i >= 0; i--) {
+ carry += (BignumDblInt)a[i] + b[i];
+ c[i] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ }
+
+ 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;
+ BignumDblInt carry = 1;
+
+ for (i = len-1; i >= 0; i--) {
+ carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
+ c[i] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ }
+}
+
+/*
* 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.
*/
-static void internal_mul(BignumInt *a, BignumInt *b,
+#define KARATSUBA_THRESHOLD 50
+static void internal_mul(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i, j;
BignumDblInt t;
- for (j = 0; j < 2 * len; j++)
- c[j] = 0;
+ if (len > KARATSUBA_THRESHOLD) {
+
+ /*
+ * 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;
+ BignumInt *scratch;
+ BignumDblInt carry;
+
+ /*
+ * 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.
+ */
+
+ /* a_1 b_1 */
+ internal_mul(a, b, c, toplen);
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);
+
+ /*
+ * We must allocate scratch space for the central coefficient,
+ * and also for the two input values that we multiply when
+ * computing it. Since either or both may carry into the
+ * (botlen+1)th word, we must use a slightly longer length
+ * 'midlen'.
+ */
+ scratch = snewn(4 * midlen, BignumInt);
+
+ /* 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 (j = 0; j < toplen; j++) {
+ scratch[midlen - toplen + j] = a[j]; /* a_1 */
+ scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
+ }
+
+ /* compute a_1 + a_0 */
+ scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
+ /* compute b_1 + b_0 */
+ scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
+ scratch+midlen+1, botlen);
+
+ /*
+ * Now we can do the third multiplication.
+ */
+ internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);
+
+ /*
+ * 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 (j = 0; j < 2*toplen; j++)
+ scratch[2*midlen - 2*toplen + j] = c[j];
+ scratch[1] = internal_add(scratch+2, c + 2*toplen,
+ scratch+2, 2*botlen);
+
+ internal_sub(scratch + 2*midlen, scratch,
+ scratch + 2*midlen, 2*midlen);
+
+ /*
+ * 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);
+ j = 2*len - botlen - 2*midlen - 1;
+ while (carry) {
+ assert(j >= 0);
+ carry += c[j];
+ c[j] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ }
+
+ /* Free scratch. */
+ for (j = 0; j < 4 * midlen; j++)
+ scratch[j] = 0;
+ sfree(scratch);
+
+ } else {
+
+ /*
+ * Multiply in the ordinary O(N^2) way.
+ */
+
+ for (j = 0; j < 2 * len; j++)
+ c[j] = 0;
+
+ 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;
+ }
+ }
+}
- 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;
+/*
+ * 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)
+{
+ int i, j;
+ BignumDblInt t;
+
+ if (len > KARATSUBA_THRESHOLD) {
+
+ /*
+ * 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 */
+ BignumInt *scratch;
+
+ /*
+ * Allocate 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.
+ */
+ scratch = snewn(len*2, BignumInt);
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);
+
+ /* a_1 b_0 */
+ internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);
+
+ /* a_0 b_1 */
+ internal_mul_low(a + len - toplen, b, scratch, toplen);
+
+ /* Copy the bottom half of the big coefficient into place */
+ for (j = 0; j < botlen; j++)
+ c[toplen + j] = scratch[2*toplen + botlen + j];
+
+ /* 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);
+
+ /* Free scratch. */
+ for (j = 0; j < len*2; j++)
+ scratch[j] = 0;
+ sfree(scratch);
+
+ } else {
+
+ for (j = 0; j < len; j++)
+ c[j] = 0;
+
+ for (i = len - 1; i >= 0; i--) {
+ t = 0;
+ for (j = len - 1; j >= len - i - 1; j--) {
+ t += MUL_WORD(a[i], (BignumDblInt) b[j]);
+ t += (BignumDblInt) c[i + j + 1 - len];
+ c[i + j + 1 - len] = (BignumInt) t;
+ t = t >> BIGNUM_INT_BITS;
+ }
+ }
+
+ }
+}
+
+/*
+ * 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 at least '3*len' BignumInts used as scratch
+ * space.
+ */
+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);
+
+ /*
+ * 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);
+ 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,
int bshift = shift % BIGNUM_INT_BITS;
BignumDblInt addend;
- addend = n << bshift;
+ addend = (BignumDblInt)n << bshift;
while (addend) {
addend += number[word];
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--;
+ if (h >= m0) {
+ /*
+ * Special case.
+ *
+ * To illustrate it, suppose a BignumInt is 8 bits, and
+ * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
+ * our initial division will be 0xA123 / 0xA1, which
+ * will give a quotient of 0x100 and a divide overflow.
+ * However, the invariants in this division algorithm
+ * are not violated, since the full number A1:23:... is
+ * _less_ than the quotient prefix A1:B2:... and so the
+ * following correction loop would have sorted it out.
+ *
+ * In this situation we set q to be the largest
+ * quotient we _can_ stomach (0xFF, of course).
+ */
+ q = BIGNUM_INT_MASK;
+ } else {
+ /* Macro doesn't want an array subscript expression passed
+ * into it (see definition), so use a temporary. */
+ BignumInt tmplo = a[i];
+ DIVMOD_WORD(q, r, h, tmplo, m0);
+
+ /* Refine our estimate of q by looking at
+ h:a[i]:a[i+1] / m0:m1 */
+ t = MUL_WORD(m1, 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--;
+ }
}
/* Subtract q * m from a[i...] */
c = 0;
for (k = mlen - 1; k >= 0; k--) {
- t = (BignumDblInt) q * (BignumDblInt) m[k];
+ t = MUL_WORD(q, m[k]);
t += c;
- c = t >> BIGNUM_INT_BITS;
+ c = (unsigned)(t >> BIGNUM_INT_BITS);
if ((BignumInt) t > a[i + k])
c++;
a[i + k] -= (BignumInt) t;
}
/*
- * 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. Uses the Montgomery multiplication
+ * technique.
*/
-Bignum modpow(Bignum base, Bignum exp, Bignum mod)
+Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
{
- BignumInt *a, *b, *n, *m;
- int mshift;
- int mlen, i, j;
- Bignum result;
+ BignumInt *a, *b, *x, *n, *mninv, *tmp;
+ int len, i, j;
+ Bignum base, base2, r, rn, inv, result;
- /* Allocate m of size mlen, copy mod to m */
- /* We use big endian internally */
- mlen = mod[0];
- m = snewn(mlen, BignumInt);
- for (j = 0; j < mlen; j++)
- m[j] = mod[mod[0] - 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);
- /* 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;
- }
+ /*
+ * Make sure the base is smaller than the modulus, by reducing
+ * it modulo the modulus if not.
+ */
+ base = bigmod(base_in, mod);
- /* 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++)
- n[i + j] = base[base[0] - j];
+ /*
+ * mod had better be odd, or we can't do Montgomery multiplication
+ * using a power of two at all.
+ */
+ assert(mod[1] & 1);
- /* Allocate a and b of size 2*mlen. Set a = 1 */
- a = snewn(2 * mlen, BignumInt);
- b = snewn(2 * mlen, BignumInt);
- for (i = 0; i < 2 * mlen; i++)
- a[i] = 0;
- a[2 * mlen - 1] = 1;
+ /*
+ * 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);
+
+ /*
+ * 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 < 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 < 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 < rn[0] ? rn[j + 1] : 0);
+ freebn(rn);
+
+ tmp = snewn(3*len, 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] & (1 << j)) == 0) {
j--;
if (j < 0) {
i++;
}
/* 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);
+ internal_mul(a + len, a + len, b, len);
+ monty_reduce(b, n, mninv, tmp, len);
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(b + len, x, a, len);
+ monty_reduce(a, n, mninv, tmp, len);
} 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));
- }
+ /*
+ * Final monty_reduce to get back from the adjusted Montgomery
+ * representation.
+ */
+ monty_reduce(a, n, mninv, tmp, len);
/* Copy result to buffer */
result = newbn(mod[0]);
- for (i = 0; i < mlen; i++)
- result[result[0] - i] = a[i + mlen];
+ 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 */
- for (i = 0; i < 2 * mlen; i++)
+ for (i = 0; i < 3 * len; i++)
+ tmp[i] = 0;
+ sfree(tmp);
+ for (i = 0; i < 2 * len; i++)
a[i] = 0;
sfree(a);
- for (i = 0; i < 2 * mlen; i++)
+ for (i = 0; i < 2 * len; i++)
b[i] = 0;
sfree(b);
- for (i = 0; i < mlen; i++)
- m[i] = 0;
- sfree(m);
- for (i = 0; i < mlen; i++)
+ for (i = 0; i < len; i++)
+ mninv[i] = 0;
+ sfree(mninv);
+ for (i = 0; i < len; i++)
n[i] = 0;
sfree(n);
+ for (i = 0; i < len; i++)
+ x[i] = 0;
+ sfree(x);
return result;
}
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 */
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 */
/* 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;
}
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]--;
}
}
/*
- * 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 >= (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 >= (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])
+ if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
abort(); /* beyond the end */
else {
int v = bitnum / BIGNUM_INT_BITS + 1;
}
/*
- * 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)
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;
}
}
/* mlen space for a, mlen space for b, 2*mlen for result */
workspace = snewn(mlen * 4, 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,
/* 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;
if (addend) {
BignumDblInt carry = 0;
for (i = 1; i <= rlen; i++) {
- carry += (i <= ret[0] ? ret[i] : 0);
- carry += (i <= addend[0] ? addend[i] : 0);
+ carry += (i <= (int)ret[0] ? ret[i] : 0);
+ carry += (i <= (int)addend[0] ? addend[i] : 0);
ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
carry >>= BIGNUM_INT_BITS;
if (ret[i] != 0 && i > maxspot)
}
ret[0] = maxspot;
+ sfree(workspace);
return ret;
}
int i, maxspot = 0;
BignumDblInt carry = 0, addend = addendx;
- for (i = 1; i <= ret[0]; i++) {
+ for (i = 1; i <= (int)ret[0]; i++) {
carry += addend & BIGNUM_INT_MASK;
- carry += (i <= number[0] ? number[i] : 0);
+ carry += (i <= (int)number[0] ? number[i] : 0);
addend >>= BIGNUM_INT_BITS;
ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
carry >>= BIGNUM_INT_BITS;
r = 0;
mod = modulus;
for (i = number[0]; i > 0; i--)
- r = (r * 65536 + number[i]) % mod;
+ r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
return (unsigned short) r;
}
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);
* 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++) {
+ for (i = 0; i < (int)x[0]; i++) {
carry = (carry << BIGNUM_INT_BITS) + workspace[i];
workspace[i] = (BignumInt) (carry / 10);
if (workspace[i])
/*
* Done.
*/
+ sfree(workspace);
return ret;
}