* 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);