I'm very new to perl. Need some assistance in the below scenario.
I have declared an array and trying to initialize like
my @array1;
$array1[0] = 1;
$array1[1] = 2;
$array1[3] = 4;
Now I want to insert another array lets say my @data = [10,20,30]; at index 2 of array1.
So after insert array1 should look like [1, 2, [10,20,30],4]
How can I do that?
You will need to use array references.
In short you can do
$array1[2] = \@data;
This will make the second member of $array1 a reference to @data. Because this is a reference you can't access it in quite the same way as a normal array. To access @data via array1 you would use @{$array1[2]}. To access a particular member of the reference to @data you can use -> notation. See for example the simple program below:
use strict;
use warnings;
my @array1;
$array1[0] = 1;
$array1[1] = 2;
$array1[3] = 4;
my @data = (10,20,30);
$array1[2] = \@data;
print "A @array1\n"; #Note that this does not print contents of data @data
print "B @{$array1[2]}\n"; #Prints @data via array1
print "C $array1[2]->[0]\n"; #Prints 0th element of @data
You can write
$array1[2] = [10,20,30];
Be aware that what you inserting into the host array is actually an array reference. Hence the syntax: [10,20,30] is a reference while (10,20,30) is a list proper.
Perl doesn't have nested arrays.
Perl doesn't have nested array? it's absolutely wrong, Perl supports multi-dimensional arrays well.
You can use splice for doing so.
use Data::Dumper;
splice @array1, 2, 0, @data;
print Dumper \@array1;
the splice prototype is:
splice [origin array], [index], [length of elements to replace], [replacement]
splice doesn't return the modified array. splice modifies the array in place and returns the removed elements (or the last removed element in scalar context).
splice @array1, 2, 0, @data? Will it update?
[] characters, since those have meaning in Perl. Or use some other characters. I'm also not sure if this actually answers OP's question, since they want a nested array, and I believe this won't nest the replacement.