Question

Try this:

$test = array (2+2);
var_dump($test);

Then try the same but inside a class:

class test {
    public $test = array(2+2);
}

I just want to know why one gets a parser error and maybe how to solve this (in a class) as code-friendly and performant as possible.

Answer 1

You cannot use statements to initialize class fields. It must be a literal, a constant value. A workaround is to use a constructor:

class Test {
    public $test;

    public function __construct() {
        $this->test = array(2+2);
    }
}

From the manual:

Class member variables are called “properties”. You may also see them
referred to using other terms such as “attributes” or “fields”, but
for the purposes of this reference we will use “properties”. They are
defined by using one of the keywords public, protected, or private,
followed by a normal variable declaration. This declaration may
include an initialization, but this initialization must be a constant
value–that is, it must be able to be evaluated at compile time and
must not depend on run-time information in order to be evaluated.