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Re: Why the l parameter flag in this case not work?
- X-seq: zsh-users 16913
- From: lilydjwg <lilydjwg@xxxxxxxxx>
- To: zsh-users@xxxxxxx
- Subject: Re: Why the l parameter flag in this case not work?
- Date: Tue, 20 Mar 2012 21:41:46 +0800
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- References: <20120320083214.GA3818@lilyforest> <CAEcrOvCsRGsfvCO_n6HfkjG=_REbGuu4nCTbb+zZK7w+_2o+xw@mail.gmail.com>
On Tue, Mar 20, 2012 at 09:57:08AM +0100, Damien Thébault wrote:
> On Tue, Mar 20, 2012 at 09:32, lilydjwg <lilydjwg@xxxxxxxxx> wrote:
> > I tried to print a line of `-' but not work:
> >
> > print ${(l.80..-.)}
> >
> > An empty line is printed. However, printing a line of `=' or other
> > strings do work:
> >
> > print ${(l.80..+.)}
> >
> > I tried to quote `-', but only 40 `-'s are printed:
> >
> > print ${(l.80..\-.)}
> >
>
> The "print" command is intepreting the expanded expression as an
> argument, you can use "echo" or use "--" to tell the "print" command
> that the end of arguments has been reached:
>
> echo ${(l.80..-.)}
> print -- ${(l.80..-.)}
Thanks!
--
Best regards,
lilydjwg
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