Re^2: Data structure / hash reference question

[stevieb]

You were close :)

my $class = 'regex1';

my $color = ( first { $class =~ /$_->{'re'}/ } @rules )->{ color };   
+             

print "$color\n";

[roboticus]

stevieb:

You'd think that as long as I've been toying around with perl, I'd've figured it out or stumbled across it by now.

I tried several "obvious" things, none of which worked:

$color = ${ ... }->{color};
$color = $${ ... }->{color};
$color = ${%{ ... }}{color};

The parenthesis are a surprise to me, but I'm glad it's possible to avoid the two-step!

...roboticus

When your only tool is a hammer, all problems look like your thumb.

[stevieb]

I love learning new things :)

If I recall correctly, I found the basis for this trick years ago after figuring there must be a simpler way to do this:

# want fourth word
my $str   = "one two three four five";
my @array = split / /, $str;
my $word  = $array[3];

I monkeyed about and accidentally came up with this:

my $str  = "one two three four five";
my $word = ( split / /, $str )[3];

Combined with my current knowledge of refs and anonymous data and where they are handy, I revisited the parens and came to the conclusion that the value after the expression is evaluated within the parens can be operated on directly.

[tobyink]

You were close with the first one - you just didn't need the dereferencing arrow ->. This should work:

$color = ${ ... }{color};

Why? Because if ... is a hashref, then %{...} is a plain old hash.

If our hash was called %foo then we'd access a value in it as $foo{color} - that is, change the sigil from % to $, and add the key in curlies. Our hash is called %{...} but we can still use the same technique.

[stevieb]

I was curious as to how this could work, so I got playing around.

I forgot that you could substitute the -> deref operator for ${}

perl -E '$h={a=>1,b=>2}; say $h->{a}'

...is the same as:

perl -E '$h={a=>1,b=>2}; say ${$h}{a}'

In roboticus' attempt that you pointed out, you'd still need to use the parens inside of the deref block, or else perl will try to use the bareword 'first' as the name of the hash instead of evaluating the expression to its final value (I only found this after testing it):

my $color = ${( first { $class =~ /$_->{'re'}/ } @rules )}{ color };

[hperange]

Thanks, this works :)