Return by reference

[doebaladoo]

I was thought that (when returning by reference) you should always add a & when assigning the return value. For instance:

 

function & a () { }
$a = & a();

 

and not:

 

$a = a();

 

The follow works hower (without the added & ):

 

class a {
	public $a = 1;
}

class b {
	static $var;

	public static function & gvar($i) {
		if(!(isset(b::$var[$i]))) {
			b::$var[$i] = new a;
		}

		return b::$var[$i];
	}
}

$b = new b;
$a = b::gvar(1);
$a->a = 14;

$c = b::gvar(1);
echo $c->a; // 14

 

Notice the

 

$a = b::gvar(1);

 

without the &.

 

Am I missing something???

[J.Daniels]

In the function definition, you are specifying that it return by reference:

 

      public static function & gvar($i) {

 

This is from Passing By References in the PHP manual

[genericnumber1]

Daniels, I think he knows that.

 

doebaladoo, as of php5 all objects are passed by reference (they're instantiated on the heap if that means anything to you), so even specifying that the method returns a reference is not needed.

 

Try removing the & from the function prototype and you should still get the same result.

[doebaladoo]

Thanks genericnumber1!

 

That solves it. I didn't even know that. I was struggling with the &'s all the time.