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Re: subprocess.Popen zombie

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Op Wednesday 20 May 2015 15:16 CEST schreef Robin Becker:

> As part of a long running PyQT process running as a window app in
> Arch linux I needed an alert sound, I decided to use the beep
> command and the app code then looked like
>
> pid = Popen(['/home/robin/bin/mybeep', '-r3', '-f750', '-l100',
> '-d75']).pid
>
> the mybeep script handles module loading if required etc etc.
>
> Anyhow, this works with one slight oddity. When this code is
> executed it works fine, but leaves behind a single zombie process,
> when next executed the zombie disappears and a new zombie replaces
> it.
>
> Is this because I'm not waiting? Does the process module reap

Yes, you should do a:
     pid.wait()

Should not be a problem, because the program you are calling takes
very little time.

-- 
Cecil Westerhof
Senior Software Engineer
LinkedIn: http://www.linkedin.com/in/cecilwesterhof

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subprocess.Popen zombie Robin Becker <robin@reportlab.com> - 2015-05-20 14:16 +0100
  Re: subprocess.Popen zombie Cecil Westerhof <Cecil@decebal.nl> - 2015-05-20 16:48 +0200
  Re: subprocess.Popen zombie Alain Ketterlin <alain@universite-de-strasbourg.fr.invalid> - 2015-05-20 17:42 +0200
    Re: subprocess.Popen zombie Robin Becker <robin@reportlab.com> - 2015-05-20 17:44 +0100
      Re: subprocess.Popen zombie Thomas Rachel <nutznetz-0c1b6768-bfa9-48d5-a470-7603bd3aa915@spamschutz.glglgl.de> - 2015-05-21 08:35 +0200
        Re: subprocess.Popen zombie Ian Kelly <ian.g.kelly@gmail.com> - 2015-05-21 08:13 -0600

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