What the heck does "tests->{$stype} ||= {};" do?

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks,

I am having the great joy of figuring out someone else code. And I haven't seen this before ...

$tests->{$stype} ||= {};
my %tests = %{$tests->{$stype}};
[download]

... what the heck would that code be trying to do?

The $test is a big ugly global hash ref at the top of the code that is about 100 lines long so I will only post a small version that might have typos in it here:

my $tests = {
    all => {
        cost      => \&cost_checker
    },
    YEAM => {
        lipmons => sub { check_lipmons(@_, 'SD'  => 3,
                                           'ron' => 2,
                                           'SP' => 3); },
        dd          => \&check_dd,
        stype     => sub { check_stypes(@_, "EM"); },
        heat_monr => sub { check_heatmonr(@_, 1); },
        heat_pgm     => sub { check_hpgm(@_, 'Solar'); },
    },
    ...
[download]

Thanks

Comment on What the heck does "tests->{$stype} \|\|= {};" do? Select or Download Code

Replies are listed 'Best First'.
Re: What the heck does "tests->{$stype} \|\|= {};" do? by kyle (Abbot) on Sep 19, 2008 at 19:40 UTC
You actually have two different variables, both named "tests". One is the hash, `%tests`, and the other is a scalar, `$tests`, which holds a reference to a hash. If you're not familiar with references, swing on over to perlreftut and perlref. After that, References quick reference is correctly named. The first line (`$tests->{$stype} \|\|= {};`) checks to see if `$tests->{$stype}` is a true value. If not, it sets it to "`{}`", which is a new, empty, anonymous hash reference. The intent of this is (probably) to ensure that `$tests->{$stype}` is a hash ref (in case it's empty), but it will actually do nothing if there's some other (true) value there already. It's equivalent to: `if ( ! $tests->{$stype} ) { $tests->{$stype} = {}; }` [download] If you're interested in the "`\|\|=`" operator, have a look at perlop. The second line (`my %tests = %{$tests->{$stype}};`) dereferences the hash reference in `$tests->{$stype}` and copies it to `%tests`. This is probably the reason for the first line. If there's no hash reference in there, the second line will die.	[reply] [d/l] [select]
Re^2: What the heck does "tests->{$stype} \|\|= {};" do? by Bloodnok (Vicar) on Sep 20, 2008 at 10:38 UTC
... the 2 lines in question could be restated, IMCO without losing (maybe even introducing) clarity, to: `my %tests = %{$tests->{$stype} \|\| {}};` [download] A user level that continues to overstate my experience :-))	[reply] [d/l]
Re^3: What the heck does "tests->{$stype} \|\|= {};" do? by ikegami (Patriarch) on Sep 20, 2008 at 11:23 UTC
That's not equivalent. In your code, `$tests->{$stype}` is never assigned anything. But it's probably better. Assigning to `$tests->{$stype}` was probably a side-effect.	[reply] [d/l] [select]
Re^4: What the heck does "tests->{$stype} \|\|= {};" do? by Bloodnok (Vicar) on Sep 20, 2008 at 13:34 UTC
Re: What the heck does "tests->{$stype} \|\|= {};" do? by jwkrahn (Abbot) on Sep 19, 2008 at 19:39 UTC
`$tests->{$stype} \|\|= {};` is short for `$tests->{$stype} = $tests->{$stype} \|\| {};` which says that if `$tests->{$stype}` is true keep the same value but if it is false then assign the value `{}` to it.	[reply] [d/l] [select]


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