Re: not able to get last index of array

[ww]

This is a FAQ: Answer: How do I find the index of the last element in an array? (There are multiple answers here)

This may not be the answer you're seeking: if not, please clarify your question.

Update: And this may clarify the answers (here and above):

#!/usr/bin/perl
use 5.018;
use strict;
use warnings;
# 1164677 

my @numbers = (15,5,7,3,9,1,20,13,9,8,15,16,2,6,12,90);
(my @newstring) = sort {$a <=> $b} @numbers;
for (@newstring) {
     print "$_ ";
}

say "\nNumber of elements: " . @newstring;               # number of e
+lements
my $lastindex = $#newstring;
say "\$lastindex: $lastindex";
say "minus 2: " . $newstring[-2];    # "@a" provokes an "unintended in
+terpolation" warning
say "minus 1: " . $newstring[-1];

=head output

1 2 3 5 6 7 8 9 9 12 13 15 15 16 20 90
Number of elements: 16
$lastindex: 15
minus 2: 20
minus 1: 90

=cut

[Lotus1]

Hi ww,

When I run the OP's script with use warnings; added I don't see a warning for printing "@a" but I do get one for print $a[-2] since there is only one element in the OP's array. Here is the warning I get:

Use of uninitialized value in print at C:\usr\pm\sort\1164677.pl line 7.

When you print an array inside of double quotes it adds a space by default between the elements or whatever the output field list separator, "$," '$"', is set to. How do I print an array with commas separating each element?

Update: I forgot to mention also at the line (my @newstring) = sort {$a <=> $b} @numbers; you don't need the parentheses since defining an array is already in array context. These are nitpicky comments since your program works and answered the OPs questions, but it's my 2 cents worth.

Update: Fixed the separator as pointed out by AnomalousMonk

[AnomalousMonk]

When you [interpolate] an array inside of double quotes it adds ... the output field separator, "$,", ...

Array interpolation adds the list separator  $" between elements. See perlvar.

[Lotus1]

Thanks AnomalousMonk

If I had read more closely the node I linked to it explains the difference between $, and $" . I just knew there was a variable involved and grabbed the first one I saw.