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copied!<p>Perl always passes by reference. It's just that sometimes the caller passes temporary scalars.</p> <p>The first thing you have to realise is that the arguments of subs can be one and only one thing: a list of scalars.* One cannot pass arrays or hashes to them. Arrays and hashes are evaluated, returning a list of their content. That means that</p> <pre><code>f(@a) </code></pre> <p>is the same** as</p> <pre><code>f($a[0], $a[1], $a[2]) </code></pre> <p>Perl passes by reference. Specifically, Perl aliases each of the arguments to the elements of <code>@_</code>. Modifying the elements <code>@_</code> will change the scalars returned by <code>$a[0]</code>, etc. and thus will modify the elements of <code>@a</code>.</p> <p>The second thing of importance is that the key of an array or hash element determines where the element is stored in the structure. Otherwise, <code>$a[4]</code> and <code>$h{k}</code> would require looking at each element of the array or hash to find the desired value. This means that the keys aren't modifiable. Moving a value requires creating a new element with the new key and deleting the element at the old key.</p> <p>As such, whenever you get the keys of an array or hash, you get a <em>copy</em> of the keys. Fresh scalars, so to speak.</p> <p>Back to the question,</p> <pre><code>f(%h) </code></pre> <p>is the same** as</p> <pre><code>f( my $k1 = "a", $h{a}, my $k2 = "b", $h{b}, my $k2 = "c", $h{c}, ) </code></pre> <p><code>@_</code> is still aliased to the values returned by <code>%h</code>, but some of those are just temporary scalars used to hold a key. Changing those will have no lasting effect.</p> <p>* — Some built-ins (e.g. <code>grep</code>) are more like flow control statements (e.g. <code>while</code>). They have their own parsing rules, and thus aren't limited to the conventional model of a sub.</p> <p>** — Prototypes can affect how the argument list is evaluated, but it will still result in a list of scalars.</p>

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