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Request #11063 $var = function_returning_array(args)[0]; doesn't work
Submitted: 2001-05-23 16:35 UTC Modified: 2012-04-11 12:08 UTC
From: jeremy at deadbeef dot com Assigned: nikic (profile)
Status: Closed Package: *General Issues
PHP Version: 4.0.4pl1 OS: Windows NT 4.0 Build 1381
Private report: No CVE-ID: None
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 [2001-05-23 16:35 UTC] jeremy at deadbeef dot com 
$var = function_returning_array(args)[0];

gives an error but 

$var = function_returning_array(args);
$var=$var[0];

works fine.  I think that either should work.

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 [2001-05-23 17:33 UTC] derick@php.net 
This syntax does not make sense at all (to me). It is not present in any language AFAIK.
A function simply returns an array, and if you want to do something with that, do it later.

However, it is a valid feature request, but I don't think it wil be implemented in short time.
Changed type to Feature Request and status to Suspended.

Derick
 [2001-05-23 17:37 UTC] jeremy at deadbeef dot com 
I think that $var=(function_returning_array(args))[0]; should also work, but it does not either.

The real problem here is that the [] operator works only on variables, not on general expressions. 

Just trying to make php more c-like in it's expression syntax.
 [2012-04-11 12:08 UTC] nikic@php.net 
Closing as this was implemented in PHP 5.4.
 [2012-04-11 12:08 UTC] nikic@php.net
-Status: Suspended +Status: Closed -Package: Feature/Change Request +Package: *General Issues -Assigned To: +Assigned To: nikic
 
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