Using an even lesser known regex item... 
perl -e '
  $f = "\$foo \$(bar) \n";
  $f =~ s/\$ (?: (\w+) | \((\w+)\) )/<$^N>/xg;
  print $f;'

[download]

The nice variable $^N will return the value from the last matching group. When used with other less known constructs you can do some really neat things.

The following also works - but will allow for more matching groups after the $^N. 

perl -e '
  my $f = "\$foo \$(bar)";

  our $val;       # must use package global
                  # for temporization in regex

  $f =~ s{\$                        # the dollar
    (?:                             # outer altinator
      (\w+)   (?{ $val = $^N })     # match and then store
    | \((\w+) (?{ $val = $^N }) \)  # or match 2 and store
    )                               # close outer
  }{<$val>}xg; print "$f\n"'

[download]

my @a=qw(random brilliant braindead); print $a[rand(@a)];

In reply to Re: How do I avoid double substitution when replacing many patterns? by Rhandom
in thread How do I avoid double substitution when replacing many patterns? by kyle

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