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From: bing@galbp.UUCP
Newsgroups: comp.os.minix
Subject: bug in alarm() system call
Message-ID: <1767@galbp.LBP.HARRIS.COM>
Date: Wed, 30-Sep-87 10:18:03 EDT
Article-I.D.: galbp.1767
Posted: Wed Sep 30 10:18:03 1987
Date-Received: Fri, 2-Oct-87 05:45:37 EDT
Reply-To: bing@galbp.UUCP (Bing Bang)
Organization: Harris/Lanier, Atlanta, GA
Lines: 41

there seems to be a bug in the way alarms are handled in minix. i noticed the
problem only by accident. on my minix i have gotten rid of the ram disk and use
my hard disk as root. this means every 30 seconds when update does a sync(), 
my hard disk light blinks. i was playing around with a small program that made
an alarm() call when i noticed update stopped running.

upon further investigation, i found out that if two processes are waiting
on an alarm, under an unknown set of circumstances, the second process (the one
with the longer alarm time) never receives a signal. i am trying to find the
bug now, below is the test program i use. it should run forever if alarm()
works ok, but now it stops after 1 or 2 itterations.

------------------------------------------------------------------------
#include <signal.h>

int pid, catcher();

main()
{
	pid = fork();

	for( ; ; )
		{
		signal(SIGALRM, catcher);
		alarm(pid?5:7);
		pause();
		}
}

int
catcher()
{
	if(pid)
		write(1, "\nparent alarm.\n", 15);
	else
		write(1, "\nchild alarm.\n", 14);
}
-- 
Bing H. Bang                         +-------------------+
Harris/Lanier                        |MSDOS: just say no.|
Atlanta GA                           +-------------------+