pass 0 argument (executable filename by default) to called programs

Question

By default bash passes executable filename as first (0 to be more precise) argument while invoking programs

Is there any special form for calling programs that can be used to pass 0 argument?

It is usefull for bunch of programs that behave in different ways depending on location where they were called from

While the endeavor in the linked duplicate is different, the semantics are pretty much the same. — Tim Post, Mar 03 '14 at 06:53

score 3 · Accepted Answer · answered Mar 19 '13 at 16:54

I think the only way to set argument 0 is to change the name of the executable. For example:

$ echo 'echo $0' > foo.sh
$ ln foo.sh bar.sh
$ sh foo.sh
foo.sh
$ sh bar.sh
bar.sh

Some shells have a non-POSIX extension to the exec command that allow you to specify an alternate value:

$ exec -a specialshell bash
$ echo $0
specialshell

I'm not aware of a similar technique for changing the name of a child process like this, other than to run in a subshell

$ ( exec -a subshell-bash bash )

Update: three seconds later, I find the argv0 command at http://cr.yp.to/ucspi-tcp/argv0.html.

the first methos is not suitable since file with same name already exists (it is the bash script itself :). argv0 command is not found on my system. But there is bash and -a working for it. There is only need in method of brancing exec in different subprocess that I can find easily (I hope). Thanks — ayvango, Mar 19 '13 at 17:15

1 Answers1