[Bug 262622] [pf][patch] fix showing rules in (some) nested anchors

From: <bugzilla-noreply_at_freebsd.org>
Date: Thu, 17 Mar 2022 17:39:43 UTC
https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=262622

--- Comment #7 from Matteo Riondato <matteo@FreeBSD.org> ---
(In reply to Kristof Provost from comment #6)

This fixes the issue I outline here (but my patch already did that, and it is
simpler: I don't think copying anchor_name is needed, as least for this
specific issue).

The issue in bug #262590 is a little subtler: 

1) we don't recursively print rules in anchors included in pf.conf with a
wildcard, for example, using "anchor blacklistd/*". Here is the output from
"pfctl -a "*" -sr" with a pfctl patched with your patch:

anchor "*" in on lagg0 all {
pfctl: DIOCGETRULES: Invalid argument
}

Note that that "*" is wrong, it should be "blacklistd", but the assignment of p
around line 1250 of pfctl.c is too naive, as it should take into account the
current values of path and of anchor_call before deciding what path/anchorname
to use next (indeed the error is due to pfctl_get_rules_info being called with
a path of "*")

2) we don't recursively print rules in other cases, but it is a little weird
when it happens, as it only happens in some cases. For example (with pfctl
patched with you patch):

On one hand:

#  /usr/obj/usr/src/amd64.amd64/sbin/pfctl/pfctl -a "blacklistd/*" -sr
#
(no output)

but there is a nested anchor "2200" under "blacklistd", and its rules should be
printed by the above command. Indeed, if we try to print its rules directly it
works:

#  /usr/obj/usr/src/amd64.amd64/sbin/pfctl/pfctl -a "blacklistd/2200" -sr
block drop in quick proto tcp from <port2200> to any port = 2200
#

Although, it is really weird, because the following works:

# /usr/obj/usr/src/amd64.amd64/sbin/pfctl/pfctl -a "test/*" -sr
anchor "test2" all {
  block drop in proto udp from any to any port = 3999
}
#

In any case, the situation there is more complex than what is addressed with
this patch, which is still needed independently.

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