A list of backreferences from a regex

ISAI student has asked for the wisdom of the Perl Monks concerning the following question:

Hello all. In Ruby & Python, one can have a regular expression, with backreferences, and then access the backrefernces as a list, and it's sub lists. For example, The Ruby code below returns the 1st backreference from a regexp

return line.match(/^integer.*?\"(\S+)\"/)[1]
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I am looking for something other than: @a =($1,$2,$3,$4) ; And rather:

@a=$foo[1..4];
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I could not find something similar to that in PERL, other than a reference to $+ $-, which is not what I wanted and has a warning about performance. Any ideas?

Comment on A list of backreferences from a regex Select or Download Code

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Re: A list of backreferences from a regex by Corion (Patriarch) on Mar 27, 2013 at 08:42 UTC
If you want variable access to the matches, you can only do that through `@-` and `@+`, as perlvar documents.	[reply] [d/l] [select]
Re: A list of backreferences from a regex by dave_the_m (Monsignor) on Mar 27, 2013 at 10:12 UTC
`my $string = "a b c"; my $middle = ($string =~ /(\w) (\w) (\w)/)[1]; print "middle=[$middle]\n"; # prints b` [download] Dave.	[reply] [d/l]
Re: A list of backreferences from a regex by hdb (Monsignor) on Mar 27, 2013 at 08:48 UTC
Something like `my $string = "put your own text here"; my @a = $string =~ /(\w+) (\w+) (\w+) (\w+) (\w+)/; print join ":", @a;` [download] ?	[reply] [d/l]