Breaking out of loop causes different results [duplicate] - regex

Question

This question already has answers here:
Why doesn't Perl's each() iterate through the entire hash the second time?
(2 answers)
Closed 7 years ago.
I have this code:
#!/usr/bin/perl
use strict;
use warnings;
my $judge_exes = {
"^A" => "foo",
"^B" => "bar",
"^C" => "baz",
};
sub get_judge {
my ($test_id) = #_;
my $exe = undef;
while (my ($regex, $judge) = each %$judge_exes) {
if ($test_id =~ /$regex/) {
$exe = $judge;
last;
}
}
if ($exe) {
return $exe;
} else {
return "Undefined!";
}
}
print get_judge("A1");
print get_judge("B2");
print get_judge("C3");
(ideone: http://ideone.com/slxebG)
I expect to get the output foobarbaz, but I end up getting fooUndefined!baz. However, when I comment out the last statement, I get the correct behavior:
#!/usr/bin/perl
use strict;
use warnings;
my $judge_exes = {
"^A" => "foo",
"^B" => "bar",
"^C" => "baz",
};
sub get_judge {
my ($test_id) = #_;
my $exe = undef;
while (my ($regex, $judge) = each %$judge_exes) {
if ($test_id =~ /$regex/) {
$exe = $judge;
# last;
}
}
if ($exe) {
return $exe;
} else {
return "Undefined!";
}
}
print get_judge("A1");
print get_judge("B2");
print get_judge("C3");
(ideone: http://ideone.com/QJpxbK)
Why is this happening? (I'm on Perl 5.16.2, but the issue is also present on 5.10.1 and whatever ideone is using.)
As far as I understand, last just breaks out of the while loop, which is what I want.
$exe doesn't seem to be an alternative falsey value causing me to hit the wrong if-branch. (I could return early, which would be better, but I still wouldn't understand the cause of this.)
I think I'm dereferencing the hash reference correctly.
I didn't think that regex matching would have any side effects relating to loop termination.
The regexes seem to be matching correctly, since I can actually get them to match appropriately under some circumstances, so I don't think it's an interpolation issue.
I'm not modifying the container that I'm iterating over.
Am I just making some silly non-Perl related mistake?

Answer 1

You have become confused. I am not sure where the confusion lies, but you must be aware that the each operator maintains its state between calls. That means your last inside the while loop won't terminate the iteration. Instead it will continue where it last left off when it is next encountered.
You should also be sparing with double-quotes, as they interpolate any scalar or array variables, or backslashed control characters.
Here is how I suggest that you write your algorithm
#!/usr/bin/perl
use strict;
use warnings;
use 5.010;
my $judge_exes = {
'^A' => 'foo',
'^B' => 'bar',
'^C' => 'baz',
};
say get_judge('A1', $judge_exes);
say get_judge('B2', $judge_exes);
say get_judge('C3', $judge_exes);
say get_judge('D4', $judge_exes);
say get_judge('E5', $judge_exes);
sub get_judge {
my ($test_id, $judges) = #_;
for my $re ( keys %$judges ) {
return 1 if $test_id =~ /$re/;
}
'Undefined!';
}
output
1
1
1
Undefined!
Undefined!

Answer 2

Oh, it looks like the behavior is explained here:
Using keys %hash in scalar context returns the number of keys in the hash and resets the iterator associated with the hash. You may need to do this if you use last to exit a loop early so that when you re-enter it, the hash iterator has been reset.
Wasn't expecting that — I had assumed that each would reset the iterator.