#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;
__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)
#else
typedef unsigned short BignumInt;
typedef unsigned long BignumDblInt;
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...] */
}
/*
- * Read an ssh1-format bignum from a data buffer. Return the number
+ * 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, int len, 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)
{
}
/*
- * 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)
{
}
/*
- * 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)
* 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);