CODEWITHSUNDEEP
perlbareword
sds
sds

Reputation: 60084

perl s/this/that/r ==> "Bareword found where operator expected"

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?

Upvotes: 8

Views: 4642

Answers (3)

ruakh
ruakh

Reputation: 183612

There's no better solution, no (though I usually write it on one line, since the s/// is essentially serving as part of the initialization process:

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

By the way, the reason for your error-message is almost certainly that you're running a version of Perl that doesn't support the /r flag. That flag was added quite recently, in Perl 5.14. You might find it easier to develop using the documentation for your own version; for example, http://perldoc.perl.org/5.12.4/perlop.html if you're on Perl 5.12.4.

Upvotes: 2

wiredcontrol
wiredcontrol

Reputation: 168

For completeness. If you are stuck with an older version of perl. And really want to use the s/// command without resorting to using a temporary variable. Here is one way:

perl -E 'say map { s/_iter\d+\s*$//; $_ } $ENV{PWD}'

Basically use map to transform a copy of the string and return the final output. Instead of what s/// does - of returning the count of substitutions.

Upvotes: -1

bvr
bvr

Reputation: 9697

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/;

Upvotes: 20

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