Pass variable by reference with func_get_args() in PHP5?

Question

I currently have a class method/function in this form:

function set_option(&$content,$opt ,$key, $val){

   //...Some checking to ensure the necessary keys exist before the assignment goes here.

   $content['options'][$key][$opt] = $val;

}

Now, I am looking into modification the function a bit to make the first argument optional, allowing me to pass just 3 parameters. In which case, a class property content is used in place of the one I omit.

The first thing that comes to mind is using func_num_args() & func_get_args() in conjunction with this, something like:

function set_option(){

    $args = func_get_args();

    if(func_num_args() == 3){

        $this->set_option($this->content,$args[0],$args[1],$args[2]);

    }else{

       //...Some checking to ensure the necessary keys exist before the assignment goes here.

       $args[0]['options'][$args[1]][$args[2]] = $args[3];

   }

}

How can I specify that I am passing the first argument for this as a reference? (I am using PHP5 so specifying that the variable is passed by reference on function call isn't really one of my better options.)

(I know I can just modify the parameter list so that the last parameter would be optional, doing it like function set_option($opt,$key,$val,&$cont = false), but I'm curious if passing by reference is possible in conjunction with function definitions like above. If it is I'd rather use it.)

Answer 1

Without a parameter list in the function declaration, there's no way to have an argument used as a reference. What you'd need to do is something like

function set_option(&$p1, $p2, $p3, $p4=null){

    if(func_num_args() == 3){
        $this->set_option($this->content,$p1, $p2, $p3);
    }else{
        $p1['options'][$p2][$p3] = $p4;
    }
}

So, depending on the result of func_num_args(), interpret what each parameter really is.

Pretty ugly, and makes for code you wouldn't want to maintain later :)