Messages in this thread Patch in this message | | | Date | Fri, 12 Apr 2013 17:42:20 +0800 | From | Chen Gang <> | Subject | Re: [PATCH] kernel: auditfilter: looping issue, memory leak if has 2 or more AUDIT_FILTERKEYs |
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On 2013年04月11日 12:10, Chen Gang wrote: > On 2013年04月11日 05:19, Eric Paris wrote: >> ----- Original Message ----- >> >>>> b. has an new issue for AUDIT_DIR: >>>> after AUDIT_DIR succeed, it will set rule->tree. >>>> next, the other case fail, then will call audit_free_rule. >>>> but audit_free_rule will not free rule->tree. >> Definitely a couple of leaks here... >> >> I'm seeing leaks on size 8, 64, and 128. >> >> Al, what do you think? Should I be calling audit_put_tree() in the error case if entry->tree != NULL? The audit trees are some of the most complex code in the kernel I think. >> >>
it seems, your way is the only executable way (if not change code much). what my original idea is incorrect.
we need add related code at failure process area in audit_data_to_entry. and another functions need not add these code (should not add). 'watch' also need be processed, since audit_to_watch let ref count = 2. (it just like the function audit_del_rule has done)
please help check thanks.
:-)
diff --git a/kernel/auditfilter.c b/kernel/auditfilter.c index 81f63f9..f5327ce 100644 --- a/kernel/auditfilter.c +++ b/kernel/auditfilter.c @@ -594,6 +594,10 @@ exit_nofree: return entry; exit_free: + if (entry->rule.watch) + audit_put_watch(entry->rule.watch); /* matches initial get */ + if (entry->rule.tree) + audit_put_tree(entry->rule.tree); /* that's the temporary one */ audit_free_rule(entry); return ERR_PTR(err); }
> > can we add it in audit_free_rule ? > > maybe like this: > > @@ -75,6 +75,8 @@ static inline void audit_free_rule(struct audit_entry *e) > /* some rules don't have associated watches */ > if (erule->watch) > audit_put_watch(erule->watch); > + if (erule->tree) > + audit_put_tree(erule->tree); > if (erule->fields) > for (i = 0; i < erule->field_count; i++) { > struct audit_field *f = &erule->fields[i]; > > > thanks. > > :-) >
-- Chen Gang
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