Question

I’m really stuck with references in php. In my programm i’ve got something similiar to this:

1. Programm

   class Test
   {
       private $_property;
   
       function __construct($property)
       {
           $this->_property = $property;
       }
   
       public function setProperty($property)
       {
           $this->_property = $property;
       }
   
       public function getProperty()
       {
           return $this->_property;
       }
   }
   
   function doSmth(Test $var)
   {
       $newVar = new Test('test');
   
       //I need to do something here...
   }
   
   $var = new Test('original');
   
   doSmth($var);
   
   var_dump($var);
   

2. Question

What should i do to copy all contents of $newVar variable to my $var variable so that i will be able to see it after using var_dump() function that is outside of function doSmth(). And i can’t use getters and setters in my programm because i’ve got a lot of them and it will be a lot of code. Is it possible to solve this problem with my limitations?

UPDATE: I can’t return value in my function doSmth() and i also tried __clone but nothing works. Can someone show me how can i do it with __clone()?

Answer 1

$newVar only exists in the context of the function.

What you want to do is to create a return value like this:

function doSmth(Test $var)
{
    $newVar = new Test('test');

    //I need to do something here...

    return $newVar;
}

and call the function like this instead:

$myNewShinyVar=doSmth($var);

This way, your function returns the object of the class Test as a a return value.

var_dump($myNewShinyVar);

Edit: If you cannot return a value, you can pass by reference instead:

function doSmth(&$var)
{
    $var = new Test('test');
}

This will modify the variable itself that is passed as a parameter to it (rather than passing a copy of it)

doSmth($var);
var_dump($var);