On Sun, 30 Mar 2003, [ISO-8859-1] Mikko Työläjärvi wrote: > On Sun, 30 Mar 2003, Sean Hamilton wrote: > > > Dan Nelson wrote: > > | Just make sure your signal handler has the SA_RESTART flag unset > > | (either via siginterrupt() if the handler was installed with signal(), > > | or directly if the signal was installed with sigaction() ), and the > > | signal will interrupt the wait() call. > > > > Er, I think you've missed my problem. Or I'm not getting your solution. > > > > I'm concerned about this order of events: > > > > - alarm() > > - wait() returns successfully > > - if (alarmed...) [false] > > - SIGALRM is delivered, alarmed = true > > - loop > > - wait() waits indefinitely > > > > This is incredibly unlikely to ever happen, but it's irritating me somewhat > > that the code isn't airtight. Bad design. Surely there is some atomic means > > of setting a timeout on a system call. > > My stock solution to this kind of problem is to turn those pesky > signals into I/O and use an old fashioned select() loop to handle > them; create a pipe(2), let signal handlers write one-byte "messages" > (the signal number) into the pipe and then use select() to dequeue the > events (signals) from the pipe. > > Select() has a timeout parameter you can play with to your hearts > content, and provided you don't overflow the pipe, no events will > get lost. You'd have to install a hander for SIGCHLD, of course.
Or how about kqueue(2) with EVFILT_SIGNAL. That would seem to be a more elegant solution. No signal handlers or alarm() required. -Brian -- Brian Buchanan, CISSP [EMAIL PROTECTED] -------------------------------------------------------------------------- FreeBSD - The Power to Serve! http://www.freebsd.org _______________________________________________ [EMAIL PROTECTED] mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-hackers To unsubscribe, send any mail to "[EMAIL PROTECTED]"
