Question

I would like to do the following:

$find = "start (.*) end"; $replace = "foo \1 bar";  $var = "start middle end"; $var =~ s/$find/$replace/; 

I would expect $var to contain "foo middle bar", but it does not work. Neither does:

$replace = 'foo \1 bar'; 

Somehow I am missing something regarding the escaping.

Answer 1

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.)