bugzilla-daemon@bugzilla.netfilter.org
2004-Sep-09 01:14 UTC
[Bug 119] iptc_flush_entries causes segmentation fault
https://bugzilla.netfilter.org/bugzilla/show_bug.cgi?id=119 netfilter@linuxace.com changed: What |Removed |Added ---------------------------------------------------------------------------- CC| |netfilter@linuxace.com ------- Additional Comments From netfilter@linuxace.com 2004-09-09 01:14 CEST ------- Wesley -- this report is over 1 year old. Have you since succeeded in your libiptc programming efforts? -- Configure bugmail: https://bugzilla.netfilter.org/bugzilla/userprefs.cgi?tab=email ------- You are receiving this mail because: ------- You are on the CC list for the bug, or are watching someone who is.
bugzilla-daemon@bugzilla.netfilter.org
2004-Sep-14 00:59 UTC
[Bug 119] iptc_flush_entries causes segmentation fault
https://bugzilla.netfilter.org/bugzilla/show_bug.cgi?id=119 netfilter@linuxace.com changed: What |Removed |Added ---------------------------------------------------------------------------- Status|ASSIGNED |RESOLVED Resolution| |INVALID ------- Additional Comments From netfilter@linuxace.com 2004-09-14 00:59 CEST ------- Suggest you follow this example from current libiptc: /* Flushes the entries in the given chain (ie. empties chain). */ int TC_FLUSH_ENTRIES(const IPT_CHAINLABEL chain, TC_HANDLE_T *handle) { unsigned int startindex, endindex; STRUCT_ENTRY *startentry, *endentry; struct chain_cache *c; int ret; iptc_fn = TC_FLUSH_ENTRIES; if (!(c = find_label(chain, *handle))) { errno = ENOENT; return 0; } startindex = offset2index(*handle, c->start_off); endindex = offset2index(*handle, c->end_off); startentry = offset2entry(*handle, c->start_off); endentry = offset2entry(*handle, c->end_off); ret = delete_rules(endindex - startindex, (char *)endentry - (char *)startentry, c->start_off, startindex, handle); return ret; } -- Configure bugmail: https://bugzilla.netfilter.org/bugzilla/userprefs.cgi?tab=email ------- You are receiving this mail because: ------- You are on the CC list for the bug, or are watching someone who is.