X-Git-Url: https://git.distorted.org.uk/u/mdw/putty/blobdiff_plain/a47e8bba76c3b1826d62c000b707b285e70e1e78..132c534fe341ccfea930d270e3e01ed9957f4f5a:/sshbn.c diff --git a/sshbn.c b/sshbn.c index 37422271..5620bb02 100644 --- a/sshbn.c +++ b/sshbn.c @@ -3,11 +3,26 @@ */ #include +#include #include #include #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; @@ -19,7 +34,51 @@ typedef unsigned long long BignumDblInt; __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. + * */ +#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 @@ -101,29 +160,353 @@ Bignum bn_power_2(int n) } /* + * 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, @@ -133,7 +516,7 @@ 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]; @@ -184,17 +567,39 @@ static void internal_mod(BignumInt *a, int alen, ai1 = a[i + 1]; /* Find q = h:a[i] / m0 */ - DIVMOD_WORD(q, r, h, a[i], 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--; + 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...] */ @@ -202,7 +607,7 @@ static void internal_mod(BignumInt *a, int alen, for (k = mlen - 1; k >= 0; 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; @@ -225,54 +630,90 @@ static void internal_mod(BignumInt *a, int alen, } /* - * 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++; @@ -281,13 +722,13 @@ Bignum modpow(Bignum base, Bignum exp, Bignum mod) } /* 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; @@ -300,36 +741,38 @@ Bignum modpow(Bignum base, Bignum exp, Bignum mod) 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; } @@ -370,7 +813,7 @@ Bignum modmul(Bignum p, Bignum q, Bignum mod) 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 */ @@ -378,7 +821,7 @@ Bignum modmul(Bignum p, Bignum q, Bignum mod) 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 */ @@ -462,7 +905,7 @@ static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient) 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 */ @@ -480,7 +923,7 @@ static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient) /* 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; } @@ -501,7 +944,7 @@ static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient) 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]--; } @@ -527,20 +970,26 @@ Bignum bignum_from_bytes(const unsigned char *data, int nbytes) } /* - * 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; @@ -550,7 +999,7 @@ int ssh1_read_bignum(const unsigned char *data, Bignum * result) } /* - * 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) { @@ -561,7 +1010,7 @@ 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) { @@ -569,7 +1018,7 @@ 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) { @@ -581,7 +1030,7 @@ 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] >> @@ -593,7 +1042,7 @@ int bignum_byte(Bignum bn, int i) */ 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; @@ -604,7 +1053,7 @@ int bignum_bit(Bignum bn, int i) */ 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; @@ -617,7 +1066,7 @@ void bignum_set_bit(Bignum bn, int bitnum, int value) } /* - * 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) @@ -671,9 +1120,9 @@ Bignum bignum_rshift(Bignum a, int shift) 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; } } @@ -695,8 +1144,8 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) /* 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, @@ -704,11 +1153,11 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) /* 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; @@ -719,8 +1168,8 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) 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) @@ -729,6 +1178,7 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) } ret[0] = maxspot; + sfree(workspace); return ret; } @@ -790,9 +1240,9 @@ Bignum bignum_add_long(Bignum number, unsigned long addendx) 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; @@ -814,7 +1264,7 @@ unsigned short bignum_mod_short(Bignum number, unsigned short modulus) 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; } @@ -908,6 +1358,7 @@ Bignum modinv(Bignum number, Bignum modulus) x = bigmuladd(q, xp, t); sign = -sign; freebn(t); + freebn(q); } freebn(b); @@ -922,9 +1373,9 @@ Bignum modinv(Bignum number, Bignum modulus) 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); @@ -964,9 +1415,14 @@ char *bignum_decimal(Bignum x) * 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); @@ -976,7 +1432,7 @@ char *bignum_decimal(Bignum x) * 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]; /* @@ -989,7 +1445,7 @@ char *bignum_decimal(Bignum x) 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]) @@ -1009,5 +1465,6 @@ char *bignum_decimal(Bignum x) /* * Done. */ + sfree(workspace); return ret; }