Initialising a static variable in PHP

Question

Taken from the PHP manual:

Like any other PHP static variable, static properties may only be initialized using a literal or constant; expressions are no

Answer 1

Because an array is not a function.

While array(1,2) looks like you are calling a function that is called array, you are doing none of the sorts. You are just making an array, which is not a function call. it is closer to saying $a = 1.

Answer 2

A good question! The simple ansvar is that array() only looks like a function, but in reality isn't one.

From the PHP documentation:

Note: array() is a language construct used to represent literal arrays, and not a regular function.

Answer 3

array() is not a function, it's an initializer. Unlike ordinary functions, it's interpreted at compile time, so it can be used to initialize a static.

For the reference, this is what is allowed after the static keyword:

static_scalar_value:
    common_scalar  (e.g. 42)
    static_class_name_scalar (Foo::class)
    namespace_name      (Foo)
    T_NAMESPACE T_NS_SEPARATOR namespace_name (namespace \Foo)
    T_NS_SEPARATOR namespace_name (\Foo)
    T_ARRAY '(' static_array_pair_list ')' e.g. array(1,2,3)
    '[' static_array_pair_list ']' e.g. [1,2,3]
    static_class_constant e.g. Foo::bar
    T_CLASS_C (__CLASS__)

http://lxr.php.net/xref/PHP_5_5/Zend/zend_language_parser.y#945

Php 5.6 adds "static operations" to this list, which makes it possible to use expressions for statics, as long as these ultimately resolve to static scalars.

class X {
    static $foo = 11 + (22/11); // syntax error in 5.5, valid in 5.6
}