Thanks Anonyrnous Monk! Does this mean that, irrespective of the number of regexes I write within the IF statement, $3 would point to the 3rd group of the last regex? Out of curiosity,what do I have to do if I need to print the 5 digit number in the first regex as well? Please bear with my questions if at all they sound silly. I'm new to perl and I really want to learn! Thanks :-)

http://search.cpan.org/~jesse/perl-5.12.2/pod/perlre.pod#Capture_buffers

The numbered match variables ($1, $2, $3, etc.) and the related punctuation set ($+, $&, $`, $', and $^N) are all dynamically scoped until the end of the enclosing block or until the next successful match

Out of curiosity,what do I have to do if I need to print the 5 digit number in the first regex as well?

store it, or print it before performing another match operation

Numbering of captures counts per match/regex, of which you have two here - the one with the 8-digit capture being evaluated last.

If you wanted to extract the 5-digit number from the first regex, you could simply reverse the order of the &&-combined tests:

if (($arr =~ / (\w*)(def)(\d{8})/) && ($arr =~ /(\w*)(abc)(\d{5})/)) {
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Now, $3 is the 5-digit number.

If you wanted to keep all captures, you could assign them to variables, e.g.:

if (( my ($c1, $c2, $c3) = $arr =~ /(\w*)(abc)(\d{5})/)
 && ( my ($c4, $c5, $c6) = $arr =~ / (\w*)(def)(\d{8})/)) {
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Great! Just what I was looking for! Can I use the below mentioned approach to arrive at the same result?

if ($arr =~ /(\w*)(abc)(\d{5})/){
$a = $3
 if (($arr =~ / (\w*)(def)(\d{8})/){
 $b = $3
  }
}
print $a;
prnit $b;
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