I'm trying to take each line of input, run a substitution according to user-specified "before" and "after" patterns, and output the result - it's for use in a typical "Perl rename"-type script. Because the patterns are user-specified, I end up doing basically $input =~ s/$old/$new/ - however, this seems to break subexpression references on the replacement side.

Demonstration code:

#!/usr/bin/perl -w

use strict;

my ($oldpat, $newpat);
my ($oldname, $newname);

$oldname = 'foo-64-bar';   # would normally be read from stdin

# basic substitution
$newname = $oldname;
$newname =~ s/-([0-9]+)-/_$1_/;
print "static patterns: $oldname becomes $newname\n";

# the same, but with the patterns in variables
$oldpat = '-([0-9]+)-';
$newpat = '_$1_';

$newname = $oldname;
$newname =~ s/$oldpat/$newpat/e;
print "variable patterns: $oldname becomes $newname\n";

[download]

As far as I can see, this should perform the same substitution in both cases, but it produces this:

static patterns: foo-64-bar becomes foo_64_bar
variable patterns: foo-64-bar becomes foo_$1_bar

[download]

I've checked the perlop and perlre manpages, as well as any likely-looking threads here, but didn't see anything suggesting that the contents of a scalar used as a replacement pattern are somehow special, or that $var references in such a string won't be interpolated (unless you use single quotes as the pattern delimiter, which I'm not). I've tried wrapping the subsitution in an eval{}, as well as leaving off the /e modifier, but neither made any difference.

I'm sure I'm missing something obvious. Anyone care to tell me what it might be?

In reply to s/// ignoring subexpressions with variable replacement patterns by Anonymous Monk

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