Date: Thu, 27 Oct 2016 20:21:11 -0600 From: jd1008 <jd1008@gmail.com> To: freebsd-questions@freebsd.org Subject: Re: Interesting $0 Problem Message-ID: <5812B617.5070701@gmail.com> In-Reply-To: <a6931603-c2b2-1dc7-6997-ae896db90d88@tundraware.com> References: <b859f7a3-51d1-06f4-e793-332edd212068@tundraware.com> <20161028014923.GA11638@fedora24> <a6931603-c2b2-1dc7-6997-ae896db90d88@tundraware.com>
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That's because your PATH variable first searched /usr/local/bin so that is the name of the shell that is running. For example, from the command line: $ echo $0 -ksh and my PATH variable looks like this: $ echo $PATH /sbin:/usr/sbin:/bin:/usr/bin:/usr/libexec:/usr/local/bin:/opt/bin:/opt/schily/bin/: So, my entry in the password file says my shell is /bin/ksh $ grep jd /etc/passwd jd:x:108o:1080:jd:/home/jd:/bin/ksh HTH. On 10/27/2016 07:53 PM, Tim Daneliuk wrote: > On 10/27/2016 08:49 PM, Dutch Ingraham wrote: >> On Thu, Oct 27, 2016 at 08:30:40PM -0500, Tim Daneliuk wrote: >>> I was fidding with some shell code today and discovered it was breaking >>> because $0 was returning "-/usr/local/bin/bash". Why is there a leading >>> dash here? I've not seen that before. >> How are you invoking the expansion, i.e., from a file or the >> command-line? Is this a login shell? >> >> What do you get from the command-line with <echo "$0">? >> _______________________________________________ >> freebsd-questions@freebsd.org mailing list >> https://lists.freebsd.org/mailman/listinfo/freebsd-questions >> To unsubscribe, send any mail to "freebsd-questions-unsubscribe@freebsd.org" >> > > My .bashrc source as standard startup profile: > > . mystartup > > Inside mystartup the folloing statement exists: > > source foo.sh > > $0 as reported in foo.sh is coming back with "-/usr/local/bin/bash"... >
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