Why does calling this function change my array?

Question

Perl seems to be killing my array whenever I read a file:

my @files = ("foo", "bar", "baz");
print "Files: " . join(" ", @files) . "\n";

foreach(@files) {
   print "The file is $_\n";
   func();
}

sub func {
   open(READ, "< test.txt");
   while(<READ>) {
   }
   close READ;
}

print "Files: " . join(" ", @files) . "\n";

produces:

Files: foo bar baz
The file is foo
The file is bar
The file is baz
Files:

but when I comment out func(), it gives what I would've expected:

Files: foo bar baz
The file is foo
The file is bar
The file is baz
Files: foo bar baz

Any ideas why this might be happening?

Leon Timmermans · Accepted Answer · 2008-12-03 13:53:08Z

12

You have to change foo to localize $_, or not to use $_ in your loop. Best yet, do both:

foreach my $filename (@files) {
    print "The file is $filename\n";
    func();
}

sub func {
    local $_;
    open my $read, '<', 'test.txt' or die "Couldn't open test.txt: $!";
    while(<$read>) {
    }
    close $read or die "Couldn't close file: $!";
}

The foreach loop aliases $_ to the current name of the file and the while(<READ>) assigns to $_. That's a bad combination of magic, so to say.

In general, it's a bad idea to rely on much on $_ for anything other than a one-liner.

edited Dec 3, 2008 at 13:53

answered Dec 3, 2008 at 3:46

Leon Timmermans

30.3k2 gold badges65 silver badges111 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

ysth Over a year ago

my $_; (5.10+ only) is a better way to do it

Leon Timmermans Over a year ago

True, but distributions are only beginning to move to 5.10. I assume people are still on 5.8 for the most part.

Leon Timmermans Over a year ago

Brad: that is correct, but personally I'm only using it since last week, so I'm not assuming other people are also using it. My Debian machines don't have it yet, for example. For that matter, plenty of people are still stuck on Perl 5.6 :-|.

Collectives™ on Stack Overflow

Why does calling this function change my array?

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related