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Re: z flag in parameter expansion doesn't use $IFS ?

X-seq: zsh-users 17334
From: Han Pingtian <hanpt@xxxxxxxxxxxxxxxxxx>
To: zsh-users@xxxxxxx
Subject: Re: z flag in parameter expansion doesn't use $IFS ?
Date: Fri, 19 Oct 2012 16:53:27 +0800
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On Fri, Oct 19, 2012 at 09:26:22AM +0100, Peter Stephenson wrote:
> On Fri, 19 Oct 2012 06:18:26 +0800
> Han Pingtian <hanpt@xxxxxxxxxxxxxxxxxx> wrote:
> > It looks like the z flag of parameter expansion doesn't care what the
> > value of IFS is:
> 
> Yes, this is correct.  It's using the shell grammar for this.  If you
> type "echo foo", it doesn't matter what IFS is, it will always treat
> that as two words with space as separator.  The other splitting flags
> use IFS.
> 
> > I just found that it looks like the z flag won't cause "forced joining"
> > which stated in rules 10, like this:
> 
> Yes, (z) is not like the other splitting flags.  It's a utility for when you
> need something that obeys shell parsing rules.  It's not a simple
> word-splitting tool, which is what the other splitting flags are for.
> 
> pws
Thanks a lot.

References:
- z flag in parameter expansion doesn't use $IFS ?
  - From: Han Pingtian
- Re: z flag in parameter expansion doesn't use $IFS ?
  - From: Peter Stephenson

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