After further examination, the examples with "fg $$" and "fg $!" clearly do not bring the subshell into the foreground, as they are evaluated prior to the subshells background execution. I'm trying to bring the subshell to the foreground to perform an exit, after a delay. Ultimately, it will be used as part of a terminal emulator inactivity timeout. I suspected there are advantages to exiting the emulator vs. killing the process. Clearly, I misunderstood. Thanks again.
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- Re: possible bash bug bringing job. to fo... Greg Wooledge
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