Re: Assign to parameter in parameter -- opposite of ${(P)name}?

["Benjamin R. Haskell" <zsh@xxxxxxxxxx>]

On Sat, 10 Jul 2010, Frank Terbeck wrote:

> 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. :)

Ah, gotcha.  I hadn't seen Julius's reply yet, and didn't take the 
logical step on my own.

-- 
Thanks,
Ben