On Wed, 8 Oct 2025 11:41:23 +1100
David Gibson
On Wed, Oct 08, 2025 at 12:42:49AM +0200, Stefano Brivio wrote:
On Tue, 7 Oct 2025 10:34:01 +1100 David Gibson
wrote: On Tue, Oct 07, 2025 at 12:32:54AM +0200, Stefano Brivio wrote:
On Fri, 3 Oct 2025 13:43:32 +1000 David Gibson
wrote: [snip] @@ -1820,6 +1817,8 @@ static int tcp_data_from_tap(const struct ctx *c, struct tcp_tap_conn *conn, break; seq_from_tap += size; iov_i += count; + if (th->fin) + fin = 1;
if (keep == i) keep = -1;
We'd need to double check that the "accept data segment" path is safe with len == 0, of course.
For sure it's not before d2c33f45f7be ("tcp: Convert tcp_data_from_tap() to use iov_tail"), because we might add zero-length segments to the tcp_iov array, and that would make backporting an otherwise simple and critical fix to slightly older versions rather complicated.
Kinda. It's not that complicated to deal with that case, by wrapping the actual data processing in an `if (len) { ... }`
That's needed for sure, but do we risk looping forever on a particular batch of FIN segments without data with a given series of sequence numbers?
Right now the loop terminates because the sequence moves forward. I'm not sure what happens without data while moving 'keep' around. Maybe it takes a few minutes to figure out (I haven't tried) but I wouldn't call that trivial.
That should be fine, because we need to advance the sequence by one for the FIN anyway, so we will move forwards. I might have forgotten that in my quick example, which is a bug, but an easy one to fix.
But we don't, in that loop, and there's a specific reason for it. FIN segments are a special case in that, if you receive more than one, you don't advance the sequence by more than one, even if the segments themselves are in the expected sequence. If you want to move the sequence increase into that loop (which might make sense with some extra care), perhaps it's worth doing that together with the whole https://bugs.passt.top/show_bug.cgi?id=125 at this point. -- Stefano