Issues with .link()

[Twinny]

Hey guys,

I'm trying to use .link() in a script i'm working on. I would assume it would work like this,

PHP Code:

function onCreated() {
  this.number = 5;
  temp.clone = this.number;
  temp.clone = 10;
  printf("this.number = %i", this.number); //Outputs 5
  printf("temp.clone = %i", temp.clone); //Outputs 10
  echo("--------");
  temp.link = this.number.link();
  printf("temp.link = %i", temp.link); //Outputs 5
  temp.link = 20;
  printf("this.number = %i", this.number); //Outputs 5 -- Expected 20?
  printf("temp.clone = %i", temp.clone);  //Ouputs 10
  printf("temp.link = %i", temp.link);  //Outputs 20
} 


I also tried a different example,

PHP Code:

function onCreated() {
  this.num1 = 1;
  this.num2 = 2;
  this.array = {this.num1.link(), this.num2.link()};
  printf("Pre-Change: %s", this.array);
  
  for (i=0; i<2; i++)
    this.array[i] = 20;
  
  printf("Post-Change: %s", this.array);
  
  echo("this.num1 = " @ this.num1);
  echo("this.num2 = " @ this.num2);
}

Pre-Change: 1,2
Post-Change: 20,20
this.num1 = 1
this.num2 = 2 


This was tested on Testbed.

I would have thought temp.link = this.number.link(); would make temp.link reference this.number rather than copying the object but the output shows opposite.

Is .link() supposed to be returning a reference or is it providing another use?