On Thu, 2021-12-02 at 14:49 -0500, Khaled Elmously
wrote:> Hello,
>
> I have encountered situations where ipconfig takes longer than the
> user-specified timeout.
>
> The following is an easy way to reproduce the issue:
>
> # ip link add eth3 type dummy
> # date; /usr/lib/klibc/bin/ipconfig -t 2 eth3; date
> Thu Dec 2 14:26:15 EST 2021
> IP-Config: eth3 hardware address aa:8f:6f:bf:35:96 mtu 1500 DHCP RARP
> IP-Config: no response after 2 secs - giving up
> Thu Dec 2 14:26:25 EST 2021
>
>
>
> Notice that 'ipconfig' thinks it waited 2 seconds (which is the
> timeout value I specified), but in reality it waited 10 seconds as
> the
> 'date' values show.
>
> This seems to happen because of the "try again in 10 seconds"
that
> happens when ipconfig encounters an error on send.
Yes, I can see that.
> I believe that a fix such as the following is needed to avoid this
> issue:
>
> diff --git a/usr/kinit/ipconfig/main.c b/usr/kinit/ipconfig/main.c
> index 5472d0f0..0969e9cb 100644
> --- a/usr/kinit/ipconfig/main.c
> +++ b/usr/kinit/ipconfig/main.c
> @@ -477,6 +477,14 @@ static int loop(void)
> if (pending == 0 || (bringup_first && done))
> break;
>
> + /* Compensate for already-lost time */
> + /* Make sure to never exceed user-specified timeouts */
> + gettimeofday(&now, NULL);
> + if (now.tv_sec + timeout > start + loop_timeout) {
> + timeout = loop_timeout - (now.tv_sec - start);
> + printf("Lowered timeout to match user request
= (%d s) \n", timeout);
> + }
> +
> timeout_ms = timeout * 1000;
>
> for (x = 0; x < 2; x++) {
>
>
>
>
>
> Does that make sense? Any problems with this?
It does make some sense. (I think it would make more sense to get rid
of the inner loop as well; that would simplify this.)
However, I worry a bit that users might depend on the current
behaviour. In particular, it looks like a DHCP configuration will
currently work as long as the DHCPOFFER is received within the loop
timeout (and no packet timeout occurs). If we apply the loop timeout
before packet timeouts, the DHCPACK must also be received within the
loop timeout.
Another issue, if you really want to use short timeouts like 2 seconds,
is that we can currently time-out nearly 1 second too early due to
rounding down "start" to the last second. We would need to start
including the fractional seconds in timeout calculations.
Ben.
--
Ben Hutchings
It is easier to write an incorrect program
than to understand a correct one.
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