Path: utzoo!attcan!uunet!husc6!uwvax!oddjob!gargoyle!att!chinet!mcdchg!usenet
From: abcscnge@csuna.UUCP (Scott Neugroschl)
Newsgroups: comp.unix
Subject: Re: Children's exit() status
Keywords: UNIX SVR3.1 fork() exec() exit() signal()
Message-ID: <7966@mcdchg.UUCP>
Date: 20 May 88 15:44:25 GMT
References: <4626@mcdchg.UUCP>
Sender: usenet@mcdchg.UUCP
Organization: California State University, Northridge
Lines: 33
Approved: usenet@mcdchg.UUCP

In article <4626@mcdchg.UUCP> lenb@houxs.UUCP writes:
>                The question is, is it possible that my program
>  be woken up only 20 times, for 30 children.  Ie. could I miss
>  child deaths because several occur "simultaneously".  (simultaneously
>  

As I understand it, wait() returns the PID of one child.  Therefore,
you should not get signals from the two children.  Another possibility:
nice() the children, so that the parent has a higher priority than the
child tasks. i.e.:

	for (i = 0  ; i < NUM_PROCS ; i++)
	{
	    if ((pid = fork()) == -1) 
	    {
		    /* BAD FORK() */
	    }
	    else if (pid == 0)   /* child process */
	    {
		nice(5);		/* lower my priority */
	    }
	    else 
	    {
		/* do parent process stuff */
	    }
	}

Just some thoughts...
-- 
Scott "The Pseudo-Hacker" Neugroschl
UUCP: {litvax,humboldt,sdcrdcf,rdlvax,ttidca,}\_ csun!abcscnge
      {psivax,csustan,nsc-sca,trwspf         }/
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