Perl array in scalar context

Question

I am a newbie in Perl. I'm trying to understand Perl context. I've the following Perl code.

use strict;
use warnings;
use diagnostics;

my @even = [ 0, 2, 4, 6, 8 ];
my @odd = [ 1, 3, 5, 7, 9 ];
my $even1 = @even;
print "$even1\n";

When I execute the code, I get the following output ...

But, as I've read, the following scalar context should places the number of elements in the array in the scalar variable.

my $even1 = @even;

So, this is bizarre to me. And, what's going inside the code?

friedo · Accepted Answer · 2014-08-01 00:09:31Z

8

The correct syntax for defining your arrays is

my @even = ( 0, 2, 4, 6, 8 );
my @odd  = ( 1, 3, 5, 7, 9 );

When you use square brackets, you're actually creating a reference (pointer) to an anonymous array, and storing the reference in @even and @odd. References are scalars, so the length of @even and @odd is one.

See the Perl references tutorial for more on references.

edited Aug 1, 2014 at 0:09

answered Jul 31, 2014 at 7:20

friedo

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Jaap Joris Vens · Accepted Answer · 2014-07-31 07:34:02Z

4

By using square brackets in Perl, you are creating an array reference rather than an actual array. You can read up on how references work in the manual: perldoc perlreftut. Replace the square brackets with round parentheses and the code will do what you expect:

my @even = ( 0, 2, 4, 6, 8 );
my @odd = ( 1, 3, 5, 7, 9 );
my $scalar = @even;
print "$scalar\n";

will print

edited Jul 31, 2014 at 7:34

answered Jul 31, 2014 at 7:22

Jaap Joris Vens

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