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

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

[Benjamin R. Haskell]

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?

E.g., my latest guess:

name=foo
: ${${(P)name}::=something}
echo $foo
# should echo 'something'

Instead I get:
zsh: not an identifier:

Earlier attempt:
: ${${name}::=something}

Do I need to resort to 'eval'?

-- 
Best,
Ben

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

[Frank Terbeck]

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

Regards, Frank

-- 
In protocol design, perfection has been reached not when there is
nothing left to add, but when there is nothing left to take away.
                                                  -- RFC 1925

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

[Benjamin R. Haskell]

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'

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

[Frank Terbeck]

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

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

[Benjamin R. Haskell]

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

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

[Julius Plenz]

Hi, Benjamin!

* Benjamin R. Haskell <zsh@benizi.com> [2010-07-10 01:35]:
> but how do you *assign* to a parameter whose name is in a parameter?

> name=foo
> : ${${(P)name}::=something}
> echo $foo
> # should echo 'something'

How about:

    zsh> name=foo
    zsh> typeset $name=something
    zsh> echo $foo
    something

Julius

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

[Benjamin R. Haskell]

On Sat, 10 Jul 2010, Julius Plenz wrote:

> Hi, Benjamin!
> 
> * Benjamin R. Haskell [2010-07-10 01:35]:
> > but how do you *assign* to a parameter whose name is in a parameter?
> 
> > name=foo
> > : ${${(P)name}::=something}
> > echo $foo
> > # should echo 'something'
> 
> How about:
> 
>     zsh> name=foo
>     zsh> typeset $name=something
>     zsh> echo $foo
>     something

Thanks.  That'll do.  I often need the flexibility of the ${name=value} 
and ${name:=value} constructs, but not in this case.

Also, I tended to shy away from defining things via typeset, because it 
echoes their values if already set.  I didn't realize until just now 
rereading the man page that TYPESET_SILENT exists.  And I didn't realize 
until just testing it that an explicit assignment disables the echo.  
The usual case where this bites me is when I try to localize a variable 
that I used outside a function.

-- 
Thanks,
Ben

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

[Bart Schaefer]

On Fri, Jul 9, 2010 at 4:17 PM, Benjamin R. Haskell <zsh@benizi.com> 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?

Your first attempt wasn't complicated enough, but your second attempt
was too complicated. :-)

% name=foo
% : ${(P)name::=something}
% print $foo
something
% print $name
foo

The (P)name is interpreted before the ::= so you don't need nested expansion.