CODEWITHSUNDEEP
php
Richard Knop
Richard Knop

Reputation: 83697

PHP use function return value as array

Why is it that this works:

        $cacheMatchesNotPlayed = $cache->load('externalData');
        $cacheMatchesNotPlayed = $cacheMatchesNotPlayed['matchesNotPlayed'];

But this doesn't work:

        $cacheMatchesNotPlayed = $cache->load('externalData')['matchesNotPlayed'];

Is there some reason for it? The second bit is easier to write.

Upvotes: 2

Views: 544

Answers (3)

salathe
salathe

Reputation: 51950

PHP does not support the function array dereferencing, which you want to do.

See an RFC on the subject http://wiki.php.net/rfc/functionarraydereferencing, and the associated mailing list conversations, which was declined roughly this time last year. While there was support for this, and it's still a common request (usually "why doesn't this work?"), there are no plans in motion to introduce this syntax.

As for why, quite simply the code implementing it has not been submitted for approval (if it has been written at all).

Update

This feature has been implemented in the trunk (main development) branch of PHP and will likely be included in the next non-bugfix version (5.4.0). For anyone wanting to play with FAD, feel free to download a trunk snapshot.

Upvotes: 6

user7675
user7675

Reputation:

PHP just does not support array dereferencing in the way you have described. Also covered here:

PHP syntax for dereferencing function result

Upvotes: 5

Tyler Carter
Tyler Carter

Reputation: 61557

Because PHP does not support that syntax.

Upvotes: 7

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