Req #11063 [Sus->Csd]: $var = function_returning_array(args)[0]; doesn't work

Edit report at https://bugs.php.net/bug.php?id=11063&edit=1 ID: 11063 Updated by: nikic@php.net Reported by: jeremy at deadbeef dot com Summary: $var = function_returning_array(args)[0]; doesn't work -Status: Suspended +Status: Closed Type: Feature/Change Request -Package: Feature/Change Request +Package: *General Issues Operating System: Windows NT 4.0 Build 1381 PHP Version: 4.0.4pl1 -Assigned To: +Assigned To: nikic Block user comment: N Private report: N New Comment: Closing as this was implemented in PHP 5.4. Previous Comments: ------------------------------------------------------------------------ [2001-05-23 17:37:21] 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. ------------------------------------------------------------------------ [2001-05-23 17:33:22] 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 16:35:48] 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. ------------------------------------------------------------------------ -- Edit this bug report at https://bugs.php.net/bug.php?id=11063&edit=1