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'.

In reply to Re: How can I combine backreferences and regex interpolation? by mdillon
in thread How can I combine backreferences and regex interpolation? by Anonymous Monk

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