Bug #11063 Updated: $var = function_returning_array(args)[0]; doesn't work

ID: 11063 User Update by: jeremy@deadbeef.com Status: Suspended Bug Type: Feature/Change Request Operating system: Windows NT 4.0 Build 1381 PHP Version: 4.0.4pl1 Description: $var = function_returning_array(args)[0]; doesn't work 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. Previous Comments: --------------------------------------------------------------------------- [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@deadbeef.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. --------------------------------------------------------------------------- Full Bug description available at: http://bugs.php.net/?id=11063