How can I combine backreferences and regex interpolation?

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

I'd like to interpolate a RHS that contains backrefs. Like this:

$lhs = '(.)oo';
$rhs = '\1ar';
s/$lhs/$rhs/;
# I want to turn foo into far
[download]

(except that doesn't work.)

Comment on How can I combine backreferences and regex interpolation? Download Code

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Re: How can I combine backreferences and regex interpolation? by mdillon (Priest) on Sep 20, 2000 at 19:16 UTC
this should work: `$_ = 'foo'; $lhs = '(.)oo'; $rhs = '$1."ar"'; s/$lhs/$rhs/ee;` [download] or this: `$_ = 'foo'; $lhs = '(.)oo'; $rhs = '${1}ar'; s/$lhs/qq{qq{$rhs}}/ee;` [download] p.s. '\1' is the proper way to do a backreference (which is part of the left-hand side), but it is not the right way to refer to a capture variable in the right-hand side. for that (and anywhere else outside the regular expression part of an s/// or m// expression), you're supposed to use '$1'.	[reply] [d/l] [select]
RE: How can I combine backreferences and regex interpolation? by knight (Friar) on Sep 20, 2000 at 19:25 UTC
You can eval it: `% cat x.pl $_ = 'foo'; $lhs = '(.)oo'; $rhs = '\1ar'; eval "s/$lhs/$rhs/"; print "$_\n"; % perl x.pl far %` [download]	[reply] [d/l]
Re: How can I combine backreferences and regex interpolation? by rpc (Monk) on Sep 21, 2000 at 01:01 UTC
if you don't mind the original value being changed: my $foo = "foo"; $foo =~ s/(.)oo/$1ar/;	[reply]