[Oorexx-bugs] [ oorexx-Bugs-2988155 ] queue use crashes rexx after sysfork child exits

[SourceForge.net <no...@so...>]

Bugs item #2988155, was opened at 2010-04-16 02:41
Message generated for change (Tracker Item Submitted) made by dangerdan
You can respond by visiting: 
https://sourceforge.net/tracker/?func=detail&atid=684730&aid=2988155&group_id=119701

Please note that this message will contain a full copy of the comment thread,
including the initial issue submission, for this request,
not just the latest update.
Category: None
Group: v4.0
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: DangerDan (dangerdan)
Assigned to: Nobody/Anonymous (nobody)
Summary: queue use crashes rexx after sysfork child exits

Initial Comment:
In ooRexx 4.0.0 compiled and run on Slackware Linux 13 (only 64 bit libraries installed), using the SysFork() RexxUtil function sets up the interpreter to crash when the parent process uses a queue after the child (spawned) process exits.

#!/opt/ooRexx/bin/rexx
PID=sysfork()
if PID=0 then do    --CHILD
  say 'bye bye'
end; else do        --PARENT
  say 'Queued()='queued();
  say 'Queued()='queued();
  say 'Hurrah!'
  queue 'one'
  parse pull T; say T
end
exit 0

On Slackware 12 (32 bit Linux) using ooRexx 3.2.0, the exit code is 0 and the result displayed is
bye bye
Queued()=0
Queued()=0
Hurrah!
one

With ooRexx 4.0.0, I get an exit code from rexx of 141 and the following display:
bye bye
Queued()=0

The only work-around I currently have is to create and request a mutex semaphore before the fork and not allow the child process to exit until parent releases it, which curiously doesn't work for ooRexx 3.2.0, because SysRequestMutexSem() could return 0 before the semaphore was released (and possibly before the timeout expired).

----------------------------------------------------------------------

You can respond by visiting: 
https://sourceforge.net/tracker/?func=detail&atid=684730&aid=2988155&group_id=119701