From: Frank Terbeck <ft@bewatermyfriend.org>
To: "Benjamin R. Haskell" <zsh@benizi.com>
Cc: Zsh Users <zsh-users@zsh.org>
Subject: Re: Assign to parameter in parameter -- opposite of ${(P)name}?
Date: Sat, 10 Jul 2010 17:44:34 +0200 [thread overview]
Message-ID: <874og7tk7x.fsf@ft.bewatermyfriend.org> (raw)
In-Reply-To: <alpine.LNX.2.01.1007101133060.4808@hp.internal> (Benjamin R. Haskell's message of "Sat, 10 Jul 2010 11:36:05 -0400 (EDT)")
Benjamin R. Haskell wrote:
> On Sat, 10 Jul 2010, Frank Terbeck wrote:
>> Benjamin R. Haskell wrote:
>> > It's the end of the week, and I'm tired, so I'm sure I'm completely
>> > overlooking something obvious, but how do you *assign* to a
>> > parameter whose name is in a parameter?
>> [...]
>> > Do I need to resort to 'eval'?
>>
>> % typeset foobar=baz
>> % print ${foobar}
>> baz
>>
>
> I was tired... but not thaaat tired... :-)
>
> Using different variable names, I was looking for:
>
> name=xyzzy
> value=asdf
>
> # <-- something that doesn't involve the string xyzzy
>
> echo $xyzzy # echoes 'asdf'
Yes. But the "foobar=baz" part is just a parameter to a builtin
command. Hence you can do this:
% name=foobar
% typeset ${name}=baz
% print $foobar
baz
I thought that was what you were after. If not, I must have
misunderstood the problem entirely. :)
Regards, Frank
next prev parent reply other threads:[~2010-07-10 15:52 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2010-07-09 23:17 Benjamin R. Haskell
2010-07-10 1:13 ` Frank Terbeck
2010-07-10 15:36 ` Benjamin R. Haskell
2010-07-10 15:44 ` Frank Terbeck [this message]
2010-07-10 15:54 ` Benjamin R. Haskell
2010-07-10 14:57 ` Julius Plenz
2010-07-10 15:52 ` Benjamin R. Haskell
2010-07-10 17:39 ` Bart Schaefer
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