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+++ This bug was initially created as a clone of Bug #589266 +++

Description of problem:

If a defined qdisk heuristic "program" does not have a defined timeout
the "program" will run till it completes. The drawback to this is that
the time to complete could be longer than (interval*tko) time.

An example of this would:
<heuristic interval="2" program="sleep 100" score="1"/>

What is needed is a timer around the calling function to terminate the
"program" as failed since it exceeded (interval*tko) time and declare
that node failed heuristic.

static int check_heuristic(struct h_data *h, int block) {
   ...
   ret = waitpid(h->childpid, &status, block?0:WNOHANG);
   ...
} 

Version-Release number of selected component (if applicable):
cman-2.0.115-34.el5

How reproducible:
Everytime

Steps to Reproduce:
1. Setup cluster with qdisk
2. Define a heuristic "program" that will exceed (interval*tko) time.
3. Start qiskd
  
Actual results:
The qdisk daemon does not noticed that heuristic "program" has ran for
too long and exceeded (interval*tko) time.

Expected results:
The qdisk daemon should noticed that heuristic "program" has ran for
too long and exceed (interval*tko) time.

Additional info:

Created attachment 448781 [details]
Fix

This has only been unit-tested.

Any heuristic which takes longer than (interval*(tko-1)) seconds to return is considered failed.

http://git.fedorahosted.org/git?p=cluster.git;a=commit;h=c51005e99964c0020511cb953b7eeacb8bc1bbdf

STABLE31 commit

An advisory has been issued which should help the problem
described in this bug report. This report is therefore being
closed with a resolution of ERRATA. For more information
on therefore solution and/or where to find the updated files,
please follow the link below. You may reopen this bug report
if the solution does not work for you.

http://rhn.redhat.com/errata/RHBA-2011-0537.html