From: Han Pingtian <hanpt@linux.vnet.ibm.com>
To: zsh-users@zsh.org
Subject: Re: z flag in parameter expansion doesn't use $IFS ?
Date: Fri, 19 Oct 2012 16:53:27 +0800 [thread overview]
Message-ID: <20121019085327.GC6056@localhost.localdomain> (raw)
In-Reply-To: <CAECNH1TXtAQgefN0HwF6Ybr_gbpw=qX+oYhWBVP2GyCJ7CH6pA@mail.gmail.com>
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@linux.vnet.ibm.com> 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.
prev parent reply other threads:[~2012-10-19 9:04 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
2012-10-18 22:18 Han Pingtian
2012-10-19 8:26 ` Peter Stephenson
2012-10-19 8:53 ` Han Pingtian [this message]
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