Pass by null reference in PHP - php

Question

In PHP I need to pass some arguments to a function by reference.
I don't want to write 2 different methods for similar behaviour.
So i need to select behaviour by argument.
But I can't pass null by reference.
So I created a dummy array.
So i run it either by
$temp[0]=-1;
$this->doSomething($bigIds, $temp);
or
$temp[0]=-1;
$this->doSomething($temp, $smallIds);
public function doSomething(&$bigIds, &$smallIds) {
if ($bigIds[0] != -1) {
// make some thing
}
if ($smallIds[0] != -1) {
// make some thing
}
}
Is there a better/ elegant way to do this?

Answer 1

I would suggest an enum but this is PHP. So this should do it for you:
class YourClass
{
const DoSomethingSmall = 0;
const DoSomethingBig = 1;
public function doSomething(&$ids, $actionType) {
// can also use a switch here
if($actionType == self::DoSomethingSmall) {
// do small
}
else if($actionType == self::DoSomethingBig) {
// do big
}
}
}
Then you can do:
$this->doSomething($bigIds, self::DoSomethingBig);
$this->doSomething($smallIds, self::DoSomethingSmall);
From outside the class you can use YourClass::DoSomethingBig and YourClass::DoSomethingSmall

Answer 2

There could be loads of things you might rather do, for instance what #ad7six says in a comment, and you could also just give it some sort of setting and just one array..
public function doSomething(&$bIds, $mode) {
switch($mode){
case 1: do smallthing;break;
case 2: do bigthing;break;
case 3: do both;break;
default: do nothing;break;
}
It all depends on what you need really