Question

Perl docs recommend this:

$foo = $bar =~ s/this/that/r;

However, I get this error:

Bareword found where operator expected near
    "s/this/that/r" (#1)

This is specific to the r modifier, without it the code works.
However, I do not want to modify $bar.
I can, of course, replace

my $foo = $bar =~ s/this/that/r;

with

my $foo = $bar;
$foo =~ s/this/that/;

Is there a better solution?

Answer 1

As ruakh wrote, /r is new in perl 5.14. However you can do this in previous versions of perl:

(my $foo = $bar) =~ s/this/that/;