+
+/*
+ * 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;
+}