Interesting $0 Problem

[Tim Daneliuk]

On 10/27/2016 09:21 PM, jd1008 wrote:
> 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.

I guess I am dense.  I get that the full qualified path to
the shell should be there.  I don't get why the "-" is prepended.
This is not behavior I have seen on non-FreeBSD systems.

> 
> 
> 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">?
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>>
>> 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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Tim Daneliuk     tundra at tundraware.com
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