From the php documentation,we access global classes like
<?php
namespace A\B\C;
class Exception extends \Exception {}
$a = new Exception('hi'); // $a is an object of class A\B\C\Exception
$b = new \Exception('hi'); // $b is an object of class Exception
$c = new ArrayObject; // fatal error, class A\B\C\ArrayObject not found
?>
However,i am a lost when the doc says that
<?php
$a = new \stdClass;
?>
is functionally equivalent to:
<?php
$a = new stdClass;
?>
here http://php.net/manual/en/language.namespaces.faq.php#language.namespaces.faq.shouldicare
Can someone kindly explain what the doc is saying here.
Thanks.
Right above those two code examples in the docs is the important heading:
If I don't use namespaces, should I care about any of this?
No. Namespaces do not affect any existing code in any way, or any as-yet-to-be-written code that does not contain namespaces. You can write this code if you wish:
So the implication here is not that you can write new stdClass; inside a namespace and have it be equivalent, but rather that if you are not using namespaces at all, then you needn't worry about the backslash new \stdClass;
When working in a namespace, the rule does apply and you will need to use new \stdClass;.
And this doesn't just apply to the generic stdClass, but to any class.
// Without namespaces:
$mysqli = new MySQLi(...);
// With namespaces
$mysqli = new \MySQLi(...);
The documentation says that \stdClass is functionally equivalent to stdClass only in:
as-yet-to-be-written code that does not contain namespaces
I.e. if you are using namespaces - as your code indicates - you will have you define the class within the current namespace (to use new stdClass); or you will just have to use the native class which does not require one (new \stdClass).