Question

I want to pass a lexical file handle to a subroutine using a named argument, but the following does not compile:

#!/usr/bin/perl -w
use strict;

my $log_fh;
my $logname = "my.log";

sub primitive {
   my ($fh, $m) = @_;
   print $fh $m;
}

sub sophisticated {
   my ($args) = @_;
   print $args->{m};
   print $args->{fh} $args->{m} ;
}

open $log_fh, ">", $logname;

print $log_fh "Today I learned ...\n";

primitive($log_fh,"... the old way works ...\n");

sophisticated({
   fh=>$log_fh, 
   m=>"... and the new way requires an intervention by SO.",
   });
close $log_fh;

The complaint is:

Scalar found where operator expected at ./lexical.file.handle.pl line 15, near
} $args"
(Missing operator before  $args?)

$ perl --version

This is perl, v5.10.1

It works O.K. when I use the primitive technique of passing arguments, and the named-argument hash technique works for the message portion, just not for the file handle portion. Do I need a new version of print ?

Answer 1

When you’ve got a complex expression that returns a filehandle (like $args->{fh}) you’ll need to disambiguate the syntax a bit by adding some extra curlies:

print { $args->{fh} } $args->{m};

This is due to the weird way the print operator is designed, with no comma between the filehandle and the list of stuff to print.

Alternatively, you could grab the filehandle out of your arguments hashref first, e.g.

my $fh = $args->{fh};
print $fh $args->{m};