From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (qmail 22299 invoked by alias); 25 Mar 2011 21:51:12 -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: 15898 Received: (qmail 6558 invoked from network); 25 Mar 2011 21:51:11 -0000 X-Spam-Checker-Version: SpamAssassin 3.3.1 (2010-03-16) on f.primenet.com.au X-Spam-Level: X-Spam-Status: No, score=-2.0 required=5.0 tests=BAYES_00,DKIM_SIGNED, DKIM_VALID,DKIM_VALID_AU,FREEMAIL_FROM,RCVD_IN_DNSWL_NONE, T_TO_NO_BRKTS_FREEMAIL autolearn=ham version=3.3.1 Received-SPF: pass (ns1.primenet.com.au: SPF record at _spf.google.com designates 209.85.220.171 as permitted sender) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma; h=domainkey-signature:mime-version:in-reply-to:references:date :message-id:subject:from:to:cc:content-type; bh=GIJejhrANBl90b325jFaJRDHtfJco3u+VwNpVY4lMQw=; b=e8BSJqGXylZnV/6qNwXDGVzccZlfe7Zmh+NjERN8fFR8AZir2+3/J5Zy+iUhBaqWnd zhLS+/dIg0aIxWKfwqlfJsGiHomyeYwQ81cpwX4dxca7CRcHo6k7lo4L/LfDRHN5sz18 P9eCGhwPsjRs+5JCjlihYB+dhjy6ewzGBGd5Y= DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :cc:content-type; b=X1yp/H8jXwD5pLRF7MVNHI5jyoik1N/kpCDpTFPjgUaAL6jyDifLLb7w8Q5oyT1DKx UKtI/fIh3ju3ud6aZJv4rVHxXqOuIqZKKkxBGgDTE8s5h7RqmLBQiQ1a6Us01ZalTBKV O6a1rXDpviLzyYSjlRnAs4s6CsdnEJd/jjigI= MIME-Version: 1.0 In-Reply-To: <9861b13e6828d43041a5b489f016664a.squirrel@gameframe.net> References: <9861b13e6828d43041a5b489f016664a.squirrel@gameframe.net> Date: Fri, 25 Mar 2011 22:45:11 +0100 Message-ID: Subject: Re: How to add string to end of each array element without loops From: Mikael Magnusson To: nix@myproxylists.com Cc: zsh-users@zsh.org Content-Type: text/plain; charset=UTF-8 On 25 March 2011 22:32, wrote: > a=(1 2 3) > a=(foo$^a[@]) > > print -l "${(@n)a}" > > foo1 > foo2 > foo3 > > I would like to get the following output instead: > > 1foo > 2foo > 3foo a=(${^a}foo) ? -- Mikael Magnusson