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Subject: Re: how to refer to basename of $0
From: Tomasz Moskal <ramshackle.industries@gmail.com>
Reply-To: ramshackle.industries@gmail.com
To: zsh-users@zsh.org
Date: Fri, 29 Jul 2011 02:30:02 +0100
In-Reply-To: <20110729002402.GA41738@redoubt.spodhuis.org>
References: 
	<CADjGqHuKhx9A3HCUBjU4JgRsbj52s7DQ6HGewT=Y9uRMDGD-fg@mail.gmail.com>
	 <20110729002402.GA41738@redoubt.spodhuis.org>
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On Thu, 2011-07-28 at 20:24 -0400, Phil Pennock wrote:
> On 2011-07-28 at 18:55 -0400, TJ Luoma wrote:
> > the script "foo.sh" read .source like this:
> > 
> > 	. $HOME/.source
> > 
> > and then I did
> > 
> > 	echo "$NAME"
> > 
> > it would give me
> > 
> > 	foo.sh
> > 
> > but in zsh I get
> > 
> > 	zsh
> 
> Are you sure?
> 
> % cat -v foo
> . $HOME/bar
> % cat -v bar
> echo $0
> % zsh -f foo
> /home/me/bar
> % bash foo
> foo
> 
> The point is that in bash, sourcing a script does not change $0 while in
> zsh it does by default, because FUNCTION_ARGZERO is set.
> 
> % cat -v foo2
> unsetopt function_argzero
> . $HOME/bar
> % zsh -f foo2
> foo2
> 
> If you want to be portable to both bash and zsh, then:
> 
>   [[ -n $ZSH_VERSION ]] && unsetopt function_argzero
> 
> This does, unfortunately, have to be done in the script which does the
> including, so you can't have a common library used by both shells which
> assumes that $0 is the name of the original file and which can just be
> simply included.
> 
> -Phil

Damn, I was too late :-)