How can I use a variable in the replacement side of the Perl substitution operator?

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On the replacement side, you must use $1, not \1.

And you can only do what you want by making replace an evalable expression that gives the result you want and telling s/// to eval it with the /ee modifier like so:

$find="start (.*) end";
$replace=""foo $1 bar"";

$var = "start middle end";
$var =~ s/$find/$replace/ee;

print "var: $var\n";

To see why the “” and double /e are needed, see the effect of the double eval here:

$ perl
$foo = "middle";
$replace=""foo $foo bar"";
print eval('$replace'), "\n";
print eval(eval('$replace')), "\n";
__END__
"foo $foo bar"
foo middle bar

(Though as ikegami notes, a single /e or the first /e of a double e isn’t really an eval(); rather, it tells the compiler that the substitution is code to compile, not a string. Nonetheless, eval(eval(...)) still demonstrates why you need to do what you need to do to get /ee to work as desired.)

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