Date: Fri, 30 Nov 2001 15:54:17 -0600 From: Mike Meyer <mwm@mired.org> To: David Miller <dmiller@sparks.net> Cc: freebsd-hackers@freebsd.org Subject: Re: [OT] alarm() question Message-ID: <15368.9.298714.294055@guru.mired.org> In-Reply-To: <Pine.BSF.4.21.0111301452350.27497-100000@search.sparks.net> References: <Pine.BSF.4.21.0111301452350.27497-100000@search.sparks.net>
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David Miller <dmiller@sparks.net> types:
> Apologies for this being more C than freebsd, but I did say OT in
> the subject...
>
> In the most basic use of an alarm, like this:
>
> #include <stdio.h>
> #include <unistd.h>
> #include <signal.h>
>
> sig_t
> signal(int sig, sig_t func);
>
> static void bzzt() {
> printf("In routine bzzt now, timer expired after 3 seconds\n");
> }
>
> main() {
>
> signal(SIGALRM, bzzt);
> alarm(3);
> system("/usr/bin/host -t soa 111.0.12.in-addr.arpa");
> printf("Done\n");
> }
>
> Why does the alarm go off but not interrupt the system call? bzzt() is
> executed, but the program doesn't print Done and exit for a minute plus.
>
> Pointers to FM to RT welcome.
Try the system() man page. system() does a fork, then exec's a shell
with the string. So in the child process, the ALARM handling will be
done by the shell, and I'm pretty sure it ignores them. As you
noticed, the parent process gets the alarm.
Checking the wait system page says that wait system calls - like the
one done by the system() library routine - may either be interrupted,
or restarted after the signal handler runs. Guess which one is
happening here.
<mike
--
Mike Meyer <mwm@mired.org> http://www.mired.org/home/mwm/
Q: How do you make the gods laugh? A: Tell them your plans.
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