lddzjwwy:

You can answer most simple questions like this by:

• Reading the documentation--in this case perldoc perlvar, and perldoc perlre
• Trying it out to see what happens

Generally when you read the documentation, it will mention various things that you might not understand, so when you write your experiment, try to cover as many different types of cases as possible. This will help you understand the feature better as well as giving you some practice on testing boundary conditions in your programs.

Update: Fixed links/formatting. Thanks to Happy-the-monk and ww for the catches.

...roboticus

When your only tool is a hammer, all problems look like your thumb.

Your use of escaped brackets in the second regex leaves me uncertain about your original data... because my first thought was that you misunderstood the use of a character class. But on second thoughts, and taking a wild guess what at what the data might be (a number inside square brackets, in the original string), this example may clarify:

#!/usr/bin/perl
use 5.016;
use strict;
use warnings;

# 1039350

$_ = "  this is a number [12345]   ";

if(/^\s*(.+)\s*/){
    say "in outter 'if', \$1: $1";
    $_ = $1;
    if($1 =~ /\[(\d+)\]/){
           say "\t But in inner if, \$1 : $1";
       }
 }

=head output:

in outter 'if', $1: this is a number [12345]
         But in inner if, $1 : 12345

=cut
[download]

A couple points, beyond those made above:

• In your title, "Perl $" is not particularly good shorthand for "Perl vars" as the $ sigil specifies only one of several kinds of variables
• Including a data sample is usually useful, and often clears up ambiguities
• perldoc perlre and company will be helpful.

Is the (\d+) inside the second if $1 or $2?.

its a different regular expression, its a different m//atch operation

capture groups are numbered per match operation

so all these 4 match operations overwrite $1 /(.)/; $foo=~ /(.)/; $bar=~ /(.)/; $baz=~ /(.)/;

As do these $one = qr/(.)/ ; m{$one}; $foo=~ m{$one}; $bar =~ m{$one}; $baz =~ m{$one};

perlvar#$<_digits_> ($1, $2, ...) , perlrequick, perlretut

Thanks dude, very clear explanation.