⭐ 欢迎来到虫虫下载站! | 📦 资源下载 📁 资源专辑 ℹ️ 关于我们
⭐ 虫虫下载站

📄 sshbn.c

📁 大名鼎鼎的远程登录软件putty的Symbian版源码
💻 C
📖 第 1 页 / 共 2 页
字号:
    bn[i]--;}Bignum bignum_from_bytes(const unsigned char *data, int nbytes){    Bignum result;    int w, i;    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 = nbytes; i--;) {	unsigned char byte = *data++;	result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);    }    while (result[0] > 1 && result[result[0]] == 0)	result[0]--;    return result;}/* * Read an ssh1-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, 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;    *result = bignum_from_bytes(p, b);    return p + b - data;}/* * Return the bit count of a bignum, for ssh1 encoding. */int bignum_bitcount(Bignum bn){    int bitcount = bn[0] * BIGNUM_INT_BITS - 1;    while (bitcount >= 0	   && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;    return bitcount + 1;}/* * Return the byte length of a bignum when ssh1 encoded. */int ssh1_bignum_length(Bignum bn){    return 2 + (bignum_bitcount(bn) + 7) / 8;}/* * Return the byte length of a bignum when ssh2 encoded. */int ssh2_bignum_length(Bignum bn){    return 4 + (bignum_bitcount(bn) + 8) / 8;}/* * Return a byte from a bignum; 0 is least significant, etc. */int bignum_byte(Bignum bn, int i){    if (i >= BIGNUM_INT_BYTES * bn[0])	return 0;		       /* beyond the end */    else	return (bn[i / BIGNUM_INT_BYTES + 1] >>		((i % BIGNUM_INT_BYTES)*8)) & 0xFF;}/* * Return a bit from a bignum; 0 is least significant, etc. */int bignum_bit(Bignum bn, int i){    if (i >= BIGNUM_INT_BITS * bn[0])	return 0;		       /* beyond the end */    else	return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;}/* * Set a bit in a bignum; 0 is least significant, etc. */void bignum_set_bit(Bignum bn, int bitnum, int value){    if (bitnum >= BIGNUM_INT_BITS * bn[0])	abort();		       /* beyond the end */    else {	int v = bitnum / BIGNUM_INT_BITS + 1;	int mask = 1 << (bitnum % BIGNUM_INT_BITS);	if (value)	    bn[v] |= mask;	else	    bn[v] &= ~mask;    }}/* * Write a ssh1-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){    unsigned char *p = data;    int len = ssh1_bignum_length(bn);    int i;    int bitc = bignum_bitcount(bn);    *p++ = (bitc >> 8) & 0xFF;    *p++ = (bitc) & 0xFF;    for (i = len - 2; i--;)	*p++ = bignum_byte(bn, i);    return len;}/* * Compare two bignums. Returns like strcmp. */int bignum_cmp(Bignum a, Bignum b){    int amax = a[0], bmax = b[0];    int i = (amax > bmax ? amax : bmax);    while (i) {	BignumInt aval = (i > amax ? 0 : a[i]);	BignumInt bval = (i > bmax ? 0 : b[i]);	if (aval < bval)	    return -1;	if (aval > bval)	    return +1;	i--;    }    return 0;}/* * Right-shift one bignum to form another. */Bignum bignum_rshift(Bignum a, int shift){    Bignum ret;    int i, shiftw, shiftb, shiftbb, bits;    BignumInt ai, ai1;    bits = bignum_bitcount(a) - shift;    ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);    if (ret) {	shiftw = shift / BIGNUM_INT_BITS;	shiftb = shift % BIGNUM_INT_BITS;	shiftbb = BIGNUM_INT_BITS - shiftb;	ai1 = a[shiftw + 1];	for (i = 1; i <= ret[0]; i++) {	    ai = ai1;	    ai1 = (i + shiftw + 1 <= a[0] ? a[i + shiftw + 1] : 0);	    ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;	}    }    return ret;}/* * Non-modular multiplication and addition. */Bignum bigmuladd(Bignum a, Bignum b, Bignum addend){    int alen = a[0], blen = b[0];    int mlen = (alen > blen ? alen : blen);    int rlen, i, maxspot;    BignumInt *workspace;    Bignum ret;    /* 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);    }    internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,		 workspace + 2 * mlen, mlen);    /* now just copy the result back */    rlen = alen + blen + 1;    if (addend && rlen <= addend[0])	rlen = addend[0] + 1;    ret = newbn(rlen);    maxspot = 0;    for (i = 1; i <= ret[0]; i++) {	ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);	if (ret[i] != 0)	    maxspot = i;    }    ret[0] = maxspot;    /* now add in the addend, if any */    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);	    ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;	    carry >>= BIGNUM_INT_BITS;	    if (ret[i] != 0 && i > maxspot)		maxspot = i;	}    }    ret[0] = maxspot;    sfree(workspace);    return ret;}/* * Non-modular multiplication. */Bignum bigmul(Bignum a, Bignum b){    return bigmuladd(a, b, NULL);}/* * Create a bignum which is the bitmask covering another one. That * is, the smallest integer which is >= N and is also one less than * a power of two. */Bignum bignum_bitmask(Bignum n){    Bignum ret = copybn(n);    int i;    BignumInt j;    i = ret[0];    while (n[i] == 0 && i > 0)	i--;    if (i <= 0)	return ret;		       /* input was zero */    j = 1;    while (j < n[i])	j = 2 * j + 1;    ret[i] = j;    while (--i > 0)	ret[i] = BIGNUM_INT_MASK;    return ret;}/* * Convert a (max 32-bit) long into a bignum. */Bignum bignum_from_long(unsigned long nn){    Bignum ret;    BignumDblInt n = nn;    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 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;    }    ret[0] = maxspot;    return ret;}/* * Compute the residue of a bignum, modulo a (max 16-bit) short. */unsigned short bignum_mod_short(Bignum number, unsigned short modulus){    BignumDblInt mod, r;    int i;    r = 0;    mod = modulus;    for (i = number[0]; i > 0; i--)	r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;    return (unsigned short) r;}#ifdef DEBUGvoid diagbn(char *prefix, Bignum md){    int i, nibbles, morenibbles;    static const char hex[] = "0123456789ABCDEF";    debug(("%s0x", prefix ? prefix : ""));    nibbles = (3 + bignum_bitcount(md)) / 4;    if (nibbles < 1)	nibbles = 1;    morenibbles = 4 * md[0] - nibbles;    for (i = 0; i < morenibbles; i++)	debug(("-"));    for (i = nibbles; i--;)	debug(("%c",	       hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));    if (prefix)	debug(("\n"));}#endif/* * Simple division. */Bignum bigdiv(Bignum a, Bignum b){    Bignum q = newbn(a[0]);    bigdivmod(a, b, NULL, q);    return q;}/* * Simple remainder. */Bignum bigmod(Bignum a, Bignum b){    Bignum r = newbn(b[0]);    bigdivmod(a, b, r, NULL);    return r;}/* * Greatest common divisor. */Bignum biggcd(Bignum av, Bignum bv){    Bignum a = copybn(av);    Bignum b = copybn(bv);    while (bignum_cmp(b, Zero) != 0) {	Bignum t = newbn(b[0]);	bigdivmod(a, b, t, NULL);	while (t[0] > 1 && t[t[0]] == 0)	    t[0]--;	freebn(a);	a = b;	b = t;    }    freebn(b);    return a;}/* * Modular inverse, using Euclid's extended algorithm. */Bignum modinv(Bignum number, Bignum modulus){    Bignum a = copybn(modulus);    Bignum b = copybn(number);    Bignum xp = copybn(Zero);    Bignum x = copybn(One);    int sign = +1;    while (bignum_cmp(b, One) != 0) {	Bignum t = newbn(b[0]);	Bignum q = newbn(a[0]);	bigdivmod(a, b, t, q);	while (t[0] > 1 && t[t[0]] == 0)	    t[0]--;	freebn(a);	a = b;	b = t;	t = xp;	xp = x;	x = bigmuladd(q, xp, t);	sign = -sign;	freebn(t);	freebn(q);    }    freebn(b);    freebn(a);    freebn(xp);    /* now we know that sign * x == 1, and that x < modulus */    if (sign < 0) {	/* set a new x to be modulus - x */	Bignum newx = newbn(modulus[0]);	BignumInt carry = 0;	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);	    newx[i] = aword - bword - carry;	    bword = ~bword;	    carry = carry ? (newx[i] >= bword) : (newx[i] > bword);	    if (newx[i] != 0)		maxspot = i;	}	newx[0] = maxspot;	freebn(x);	x = newx;    }    /* and return. */    return x;}/* * Render a bignum into decimal. Return a malloced string holding * the decimal representation. */char *bignum_decimal(Bignum x){    int ndigits, ndigit;    int i, iszero;    BignumDblInt carry;    char *ret;    BignumInt *workspace;    /*     * First, estimate the number of digits. Since log(10)/log(2)     * is just greater than 93/28 (the joys of continued fraction     * approximations...) we know that for every 93 bits, we need     * at most 28 digits. This will tell us how much to malloc.     *     * Formally: if x has i bits, that means x is strictly less     * than 2^i. Since 2 is less than 10^(28/93), this is less than     * 10^(28i/93). We need an integer power of ten, so we must     * 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 = bignum_bitcount(x);    ndigits = (28 * i + 92) / 93;      /* multiply by 28/93 and round up */    ndigits++;			       /* allow for trailing \0 */    ret = snewn(ndigits, char);    /*     * Now allocate some workspace to hold the binary form as we     * repeatedly divide it by ten. Initialise this to the     * big-endian form of the number.     */    workspace = snewn(x[0], BignumInt);    for (i = 0; i < x[0]; i++)	workspace[i] = x[x[0] - i];    /*     * Next, write the decimal number starting with the last digit.     * We use ordinary short division, dividing 10 into the     * workspace.     */    ndigit = ndigits - 1;    ret[ndigit] = '\0';    do {	iszero = 1;	carry = 0;	for (i = 0; i < x[0]; i++) {	    carry = (carry << BIGNUM_INT_BITS) + workspace[i];	    workspace[i] = (BignumInt) (carry / 10);	    if (workspace[i])		iszero = 0;	    carry %= 10;	}	ret[--ndigit] = (char) (carry + '0');    } while (!iszero);    /*     * There's a chance we've fallen short of the start of the     * string. Correct if so.     */    if (ndigit > 0)	memmove(ret, ret + ndigit, ndigits - ndigit);    /*     * Done.     */    sfree(workspace);    return ret;}

⌨️ 快捷键说明

复制代码 Ctrl + C
搜索代码 Ctrl + F
全屏模式 F11
切换主题 Ctrl + Shift + D
显示快捷键 ?
增大字号 Ctrl + =
减小字号 Ctrl + -