Can PHP static class variables be defined using functions?

Question

Creating a class with variables like this works fine:

class Example {
    public static $example = array('simple', 'example');
    // ... 
}

But, if I use a function, when defining the variable, I get an unexpected '(', expecting ')' error:

class Example {
    public static $example = explode(' ', 'simple example');
    // ... 
}

I tried it without the static keyword and still got the same error. Is it possible to use functions, when defining class variables like that? What is the alternative?

Just make a namespace and put inside of it a global variable .. it will be exactly the same. And you wouldn't be pretending to "do OOP". — tereško
– tereško, Commented Apr 20, 2012 at 22:37

mellamokb · Accepted Answer · 2012-04-20 16:31:14Z

4

According to the documentation (emphasis mine):

Like any other PHP static variable, static properties may only be initialized using a literal or constant; expressions are not allowed. So while you may initialize a static property to an integer or array (for instance), you may not initialize it to another variable, to a function return value, or to an object.

answered Apr 20, 2012 at 16:31

mellamokb

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Kato · Accepted Answer · 2012-04-20 16:32:30Z

2

Array is not really a function, but an ~~operator~~ literal, which is why it works. To use a function, just do it with a setter or external to the class:

class Example {
    public static $example = null;
    // ... 

    public static function setE($val) {
       self::$example = $val;
    }
}

Example::$example = explode(' ', 'simple example');

// or

Example::setE(explode(' ', 'nudder example'));

answered Apr 20, 2012 at 16:32

Kato

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2 Comments

ryanve Over a year ago

Choosing this answer b/c it came with an alternative way to do it. Thx :]

Kato Over a year ago

Sadly, you can't make it final this way, but if you're calling a function on it, it's unlikely you really want it to be anything you compute on the fly anyways.

Beenish Khan · Accepted Answer · 2012-04-20 16:31:21Z

1

You should be able to do the following,

public static $example = new array('simple','example');

answered Apr 20, 2012 at 16:31

Beenish Khan

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