I was writing code, fast and dirty, and I needed a test for values in a hash that would tell me if they existed and had content that wasn't white space. So I rapidly threw together a test that looked a bit like this:

if ( defined($hash{$key}) && hash{$key} =~ m/\w+/ ) {
    $hash{$key} = "Numbers: " . $hash{$key};
}

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It tested fine using perl -cw but of course fails when it is run. The second 'hash' statement in the "if" should be $hash. Taking this kind of construct out and putting it into a small test program:

#!/usr/bin/perl
use warnings;
use strict;

my %hash = ();

$hash{a} = "123";
$hash{b} = "456";
$hash{c} = "789";

my $key = "c";

if ( defined($hash{$key}) && hash{$key} =~ m/\w+/ ) {
    $hash{$key} = "Numbers: " . $hash{$key};
}

print $hash{$key}, "\n";

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And running this code using Active State Perl (my servers all have older Perls than the Active State version), I again get the syntax check success and the run time failure.

C:\>perl -cw bareword.pl
bareword.pl syntax OK

C:\>perl bareword.pl
Can't locate object method "hash" via package "c" (perhaps you forgot 
+to load "c"?) at bareword.pl line 13.

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And knowing nothing about all permutations of object methods, shouldn't the method bar() of an object $foo be accessed by $foo->bar()? What am I missing here?

In reply to Why does this code pass a syntax check? by dwm042

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