From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (qmail 8678 invoked by alias); 19 Oct 2012 23:45:30 -0000 Mailing-List: contact zsh-users-help@zsh.org; run by ezmlm Precedence: bulk X-No-Archive: yes List-Id: Zsh Users List List-Post: List-Help: X-Seq: 17336 Received: (qmail 11300 invoked from network); 19 Oct 2012 23:45:16 -0000 X-Spam-Checker-Version: SpamAssassin 3.3.2 (2011-06-06) on f.primenet.com.au X-Spam-Level: X-Spam-Status: No, score=-6.9 required=5.0 tests=BAYES_00,RCVD_IN_DNSWL_HI autolearn=ham version=3.3.2 Received-SPF: none (ns1.primenet.com.au: domain at linux.vnet.ibm.com does not designate permitted sender hosts) Date: Sat, 20 Oct 2012 07:34:51 +0800 From: Han Pingtian To: zsh-users@zsh.org Subject: Re: flag z won't cause forced joining? Message-ID: <20121019233451.GD6056@localhost.localdomain> References: <20121019080315.GB6056@localhost.localdomain> <121019083818.ZM3914@torch.brasslantern.com> MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: <121019083818.ZM3914@torch.brasslantern.com> User-Agent: Mutt/1.5.21 (2010-09-15) X-Content-Scanned: Fidelis XPS MAILER x-cbid: 12101923-7606-0000-0000-000004A691D8 On Fri, Oct 19, 2012 at 08:38:18AM -0700, Bart Schaefer wrote: > On Oct 19, 4:03pm, Han Pingtian wrote: > } > } I just found that it looks like the z flag won't cause "forced joining" > } which stated in rules 10 > > Yes, this is a case of the documentation being wrong; it was recently > corrected. Rule 10 now says: > > 10. _Forced joining_ > If the `(j)' flag is present, or no `(j)' flag is present but the > string is to be split as given by rule 11., and joining did not > take place at step 5., any words in the value are joined together > using the given string or the first character of $IFS if none. > Note that the `(F)' flag implicitly supplies a string for joining > in this manner. > > 11. _Simple word splitting_ > If one of the `(s)' or `(f)' flags are present, or the `=' > specifier was present (e.g. ${=VAR}), the word is split on > occurrences of the specified string, or (for = with neither of the > two flags present) any of the characters in $IFS. > > If no `(s)', `(f)' or `=' was given, but the word is not quoted > and the option SH_WORD_SPLIT is set, the word is split on > occurrences of any of the characters in $IFS. Note this step, too, > takes place at all levels of a nested substitution. > > Note that the reference to rule 17 [the (z) flag] has been removed from > rule 10 (and rule 11 is new). Thanks a lot.