Populating a Hash: Can someone help me to understand?

[Limo]

What I'd like to do is create:

my %hash;

where:

@keys = hash keys
@values = key values

I've read about references from several sources; I think referencing the arrays are my answer, but I'm just not getting it.

[ar0n]

(ugly stuff removed, replaced with less ugly)


map { $hash{$keys[$_]} = $vals[$_]  } (0..$#keys);


update
but there are nicer, prettier ways to do this; as you'll probably see below...

[merlyn]

You said:

while (my $key = shift @keys and my $value = shift @values) {
  $hash{$key} = $value;
}

And your code stops too quickly if either the key or the value is false.

---

You updated:

while (defined (my $key = shift @keys) and defined(my $value = shift @
+values)) {
  $hash{$key} = $value;
}

But even adding the defined test as you did doesn't help. Consider @keys = (1, 2, undef, 3, 4);. You'll never get to 3 or 4.

What would work is:

while (@keys and @values) {
  $hash{shift @keys} = shift @values;
}

-- Randal L. Schwartz, Perl hacker

[Limo]

Hmmm. Something strange here.

my @array = subroutine($file1, $list1);
    #foreach my $item(@array) {
        #print "$item\n";
   # }

Now, when I uncomment the foreach/print block, it prints as intended. However, when I add:

   my %hash;
while (@keys and @values) {
          $hash{shift @keys} = shift @values;
        } 
   
 foreach my $k (sort keys %hash) {
print "$k => $hash{$k}\n";     

 }

the program returns only one value per key, while @values contains several whitespace delimited elements.

[japhy]

@hash{@keys} = @values;

$_="goto+F.print+chop;\n=yhpaj";F1:eval

[Adam]

Japhy's got it in Re: Populating a Hash: Can someone help me to understand?, but I figured you might want the reverse too:

my @values = values %hash;
my @keys = keys %hash;

Not that I would do that (since I like to keep the data in one place, and values and keys are such nice hooks. But I tell you this in case you didn't know about them.

[Guildenstern]

Why can't you just use a hash slice? Like this:@hash{@keys} = @values;I imagine this would require that both arrays have equal numbers of values, but then any other kind of hash population would likely assume that as well.
Update: Okay, so I missed the post by japhy. At least merlyn had the common courtesy to explain why he --'d it.

Guildenstern
Negaterd character class uber alles!