copying an array

copying an array

[Dominik Vogt]

I'm trying to write a function that

  * takes the name of a new array variable as its argument,
  * creates a global array variable with the given name
  * gets a copy of the values of another global array.

Sketch of the function:
--
typeset -g -a G
G=("a b c" d e)
copy_array () {
	typeset -g -a "$1"
	# how can this be done?
	$1="($G[@])"
}
--

The function needs to handle whitespace (or any special
characters) in the values of G gracefully.  The closest I've got so
far is

  eval $(echo "$1=($G[@])")

But this fails to handle whitespace properly.

Bonus points if this can be done without eval.  :-)

Any ideas?

Ciao

Dominik ^_^  ^_^

Re: copying an array

[Yuya Amemiya]

Hi,

Try this:

eval $1'=("${G[@]}")'

If the value of $1 is foo, this is substituted with eval foo=("${G[@]}") and
evaluate it.

regards

From: Dominik Vogt <vogt@linux.vnet.ibm.com>
Subject: copying an array
Date: Tue, 5 Aug 2014 06:44:34 +0100

> I'm trying to write a function that
> 
>   * takes the name of a new array variable as its argument,
>   * creates a global array variable with the given name
>   * gets a copy of the values of another global array.
> 
> Sketch of the function:
> --
> typeset -g -a G
> G=("a b c" d e)
> copy_array () {
> 	typeset -g -a "$1"
> 	# how can this be done?
> 	$1="($G[@])"
> }
> --
> 
> The function needs to handle whitespace (or any special
> characters) in the values of G gracefully.  The closest I've got so
> far is
> 
>   eval $(echo "$1=($G[@])")
> 
> But this fails to handle whitespace properly.
> 
> Bonus points if this can be done without eval.  :-)
> 
> Any ideas?
> 
> Ciao
> 
> Dominik ^_^  ^_^
> 
> 

Re: copying an array

[Bart Schaefer]

On Aug 5,  6:44am, Dominik Vogt wrote:
>
> typeset -g -a G
> G=("a b c" d e)
> copy_array () {
> 	typeset -g -a "$1"
> 	# how can this be done?
> 	$1="($G[@])"
> }

You probably don't need the function at all if this is the only thing it
is doing:

	set -A "$1" "$G[@]"

The only reason for the "typeset -g" would be if there is a local array
in scope whose name is the same as the value of $1.  Otherwise "set -A"
will create the global array for you.

Re: copying an array

[Dominik Vogt]

On Tue, Aug 05, 2014 at 08:55:27AM -0700, Bart Schaefer wrote:
> On Aug 5,  6:44am, Dominik Vogt wrote:
> >
> > typeset -g -a G
> > G=("a b c" d e)
> > copy_array () {
> > 	typeset -g -a "$1"
> > 	# how can this be done?
> > 	$1="($G[@])"
> > }
> 
> You probably don't need the function at all if this is the only thing it
> is doing:
> 
> 	set -A "$1" "$G[@]"

Ah, I did not have "set" on the radar.  Thanks!

> The only reason for the "typeset -g" would be if there is a local array
> in scope whose name is the same as the value of $1.  Otherwise "set -A"
> will create the global array for you.

Is there a specific reason why typeset cannot initialize arrays?

Ciao

Dominik ^_^  ^_^

-- 

Dominik Vogt
IBM Germany

Re: copying an array

[Bart Schaefer]

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On Thu, Aug 7, 2014 at 10:12 PM, Dominik Vogt <vogt@linux.vnet.ibm.com>
wrote:

>
> Is there a specific reason why typeset cannot initialize arrays?
>

In zsh, "typeset" is not a syntactic keyword, it's just an ordinary builtin
command.  So its arguments must follow the standard metacharacter,
word-break, and quoting rules that apply to the arguments of any command.
 Parenthesized array-assignment syntax can't be used, and unlike "set" the
arguments of "typeset" after the first are already given different
semantics so they can't be interpreted as values of the array.