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

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} = {};
}

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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.

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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} || {}};

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    A user level that continues to overstate my experience :-))

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[d/l]

      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.

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[d/l]
[select]
        Thanx for that ikegami, - as usual you've spotted the deliberate mistake ;-)

        A user level that continues to overstate my experience :-))
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