- ep = &(*bufp)[pos];
- for ( ; NULL != (ep = strchr(ep, '\\')); ep++) {
- ep++;
- if ('}' != *ep)
- continue;
- *ep = '&';
- roff_ccond(r, ROFF_ccond, bufp, szp,
- ln, pos, pos + 2, offs);
+ ep = *bufp + pos;
+ while (NULL != (ep = strchr(ep, '\\'))) {
+ if ('}' == *(++ep)) {
+ *ep = '&';
+ roff_ccond(r, ln, ep - *bufp - 1);
+ }
+ ++ep;
+ }
+ return(rr ? ROFF_CONT : ROFF_IGN);
+}
+
+/*
+ * Parse a single signed integer number. Stop at the first non-digit.
+ * If there is at least one digit, return success and advance the
+ * parse point, else return failure and let the parse point unchanged.
+ * Ignore overflows, treat them just like the C language.
+ */
+static int
+roff_getnum(const char *v, int *pos, int *res)
+{
+ int myres, n, p;
+
+ if (NULL == res)
+ res = &myres;
+
+ p = *pos;
+ n = v[p] == '-';
+ if (n)
+ p++;
+
+ for (*res = 0; isdigit((unsigned char)v[p]); p++)
+ *res = 10 * *res + v[p] - '0';
+ if (p == *pos + n)
+ return 0;
+
+ if (n)
+ *res = -*res;
+
+ *pos = p;
+ return 1;
+}
+
+/*
+ * Evaluate a string comparison condition.
+ * The first character is the delimiter.
+ * Succeed if the string up to its second occurrence
+ * matches the string up to its third occurence.
+ * Advance the cursor after the third occurrence
+ * or lacking that, to the end of the line.
+ */
+static int
+roff_evalstrcond(const char *v, int *pos)
+{
+ const char *s1, *s2, *s3;
+ int match;
+
+ match = 0;
+ s1 = v + *pos; /* initial delimiter */
+ s2 = s1 + 1; /* for scanning the first string */
+ s3 = strchr(s2, *s1); /* for scanning the second string */
+
+ if (NULL == s3) /* found no middle delimiter */
+ goto out;
+
+ while ('\0' != *++s3) {
+ if (*s2 != *s3) { /* mismatch */
+ s3 = strchr(s3, *s1);
+ break;
+ }
+ if (*s3 == *s1) { /* found the final delimiter */
+ match = 1;
+ break;
+ }
+ s2++;