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 {
+ 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--;
+ }
}
/* Subtract q * m from a[i...] */
/*
* Read an ssh1-format bignum from a data buffer. Return the number
- * of bytes consumed.
+ * 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;