Array reference

$VAR = [ [ '04', '01' ], 1 ];

how do i print "04" without using any temp variables?

Re: Array reference

On Wed, 26 Dec 2007 03:04:56 -0800 (PST), "vijay@iavian.com"
wrote:

>$VAR = [ [ '04', '01' ], 1 ];
>
>how do i print "04" without using any temp variables?

print $VAR->[0][0]; # perldoc perlref


Michele
--
{$_=pack'B8'x25,unpack'A8'x32,$a^=sub{pop^pop}->(map substr
(($a||=join'',map--$|x$_,(unpack'w',unpack'u','G^ ..'KYU;*EVH[.FHF2W+#"\Z*5TI/ER 256),7,249);s/[^\w,]/ /g;$ \=/^J/?$/:"\r";print,redo}#JAPH,

Re: Array reference

vijay@iavian.com wrote:
> $VAR = [ [ '04', '01' ], 1 ];
>
> how do i print "04" without using any temp variables?

One way:

print $VAR->[0][0];

--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl

Re: Array reference

another way is.

${$VAR}[0][0]

or slightly shorter

$$VAR[0][0]


A bit of explanation to all the suggested solutions:

Remember how you subscript a regular array: $array[0].

If you have a reference to an array instead of a simple array, you can
just replace 'array' with your reference, which gives you $
$RefToArray[0].

To be sure that you don't get into precedence problems, you can also
write ${$RefToArray}[0]. That's because "anywhere you'd put an
alphanumeric identifier as part of a variable or subroutine name, you
can replace the identifier with a block returning a reference of the
correct type." [quoted from "Programming Perl"]

An alternative syntax to $$RefToArray[0] (i.e. syntactic sugar) is to
use the -> operator as in $RefToArray->[0]. The left operand of ->
must be a reference to an array/hash. The right is an array/hash
subscript (never slice). Thus you could even also write ($RefToArray)-
>[0].

Putting this together for your case, where the you have a reference to
an array of references: To get the '04', you want to subscript the the
reference to the inner array, which is given by, as learned above,
$VAR->[0] (or $$VAR[0] or ${$VAR}[0] ). Subscripting is always of the
form $...[...], thus, in a first version, you get ${$VAR->[0]}[0]. If
you use the syntax sugar notation, you get ($VAR->[0])->[0], or
without the superfluous parentheses $VAR->[0]->[0]. (or $$VAR[0]->[0]
or ${$VAR}[0]->[0]). To be honest, I don't know why exactly the
parentheses are superfluous, I strongly assume because of operator
precedence/associativity rules.

The second -> is optional too, so you get the shorter $VAR->[0][0] (or
$$VAR[0][0] or ${$VAR}[0][0]) as suggested by Gunnar and Michele. To
quote Programming Perl again: "The arrow is optional between brackets
or braces, or between a closing bracket or brace and a parenthesis for
and indirect function call"

Flo