Perl: array reference versus anonymous array

Question

This may be a silly question... The following code outputs the contents of @arrayref and @arraycont respectively. Note the difference between them

Answer 1

This creates a shallow copy of the array:

$arraycont[0] = [@array];

Whereas this just creates a reference to it:

$arrayref[0] = \@array;

Since you later modify the array:

@array = qw(5 6 7 8);

arrayref still points to the same array location in memory, and so when dereferenced in the print statements it prints the current array values 5 6 7 8.

Answer 2

You made stored references to a single array in both $arrayref[0] and $arrayref[1]. You should have used different arrays.

my @refs;

my @array1 = qw(1 2 3 4);
push @refs, \@array1;

my @array2 = qw(5 6 7 8);
push @refs, \@array2;

In practice, my is executed in each pass of a loop, creating a new array each time.

my @refs;
while ( my @row = get() ) {
   push @refs, \@row;
}

In rare occasions where you have to clone an array, you can use:

use Storable qw( dclone );

push @refs, [ @row ];       # Shallow clone
push @refs, dclone(\@row);  # Deep clone

Answer 3

The first block stores the address of @array. REferences are like 'live streaming', you get current status. So if you create a reference to @array, like \@array, when you de-reference it you will always get what @array points at the moment of de-reference. When you de-refer @array was having (5 6 7 8)

When you do [@array] its like recording the live streaming into your disk. So when you (re)play the recorded content you get what was streamed at the time of recording. So when you refer $arraycont[0] you get what @array was having at the time of copying that is
(1 2 3 4)