audit: fix a memory leak bug
In audit_rule_change(), audit_data_to_entry() is firstly invoked to translate the payload data to the kernel's rule representation. In audit_data_to_entry(), depending on the audit field type, an audit tree may be created in audit_make_tree(), which eventually invokes kmalloc() to allocate the tree. Since this tree is a temporary tree, it will be then freed in the following execution, e.g., audit_add_rule() if the message type is AUDIT_ADD_RULE or audit_del_rule() if the message type is AUDIT_DEL_RULE. However, if the message type is neither AUDIT_ADD_RULE nor AUDIT_DEL_RULE, i.e., the default case of the switch statement, this temporary tree is not freed. To fix this issue, only allocate the tree when the type is AUDIT_ADD_RULE or AUDIT_DEL_RULE. Signed-off-by: Wenwen Wang <wang6495@umn.edu> Reviewed-by: Richard Guy Briggs <rgb@redhat.com> Signed-off-by: Paul Moore <paul@paul-moore.com>
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@ -1114,22 +1114,24 @@ int audit_rule_change(int type, int seq, void *data, size_t datasz)
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int err = 0;
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struct audit_entry *entry;
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entry = audit_data_to_entry(data, datasz);
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if (IS_ERR(entry))
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return PTR_ERR(entry);
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switch (type) {
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case AUDIT_ADD_RULE:
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entry = audit_data_to_entry(data, datasz);
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if (IS_ERR(entry))
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return PTR_ERR(entry);
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err = audit_add_rule(entry);
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audit_log_rule_change("add_rule", &entry->rule, !err);
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break;
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case AUDIT_DEL_RULE:
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entry = audit_data_to_entry(data, datasz);
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if (IS_ERR(entry))
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return PTR_ERR(entry);
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err = audit_del_rule(entry);
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audit_log_rule_change("remove_rule", &entry->rule, !err);
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break;
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default:
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err = -EINVAL;
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WARN_ON(1);
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return -EINVAL;
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}
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if (err || type == AUDIT_DEL_RULE) {
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