How to create a JS object within constructor and "parent object" as constructors argument?

Question

The question might sound a bit confusing so I'll let the code explain:

function Foo(arg) {
  const argument = arg;
  const fooPart = new FooPart(this);

  this.printArg = function() {
    console.log(argument);
  }
}

function FooPart(foo) {
  this.parent = foo;

  this.parent.printArg();

}
let foo = new Foo("this is the argument");

This is not working for me. How can I solve this problem or better - what would be the correct approach for this?

Thanks

It is this.parent.printArg() ... And you call printArg() before it is defined ... — Jonas Wilms
– Jonas Wilms, Commented Aug 14, 2019 at 13:57
You have not assigned a function to this.printArg by the time you're trying to call it. — deceze
– deceze ♦, Commented Aug 14, 2019 at 13:59
yeah it is like this in the original problem sorry I fixed it. @deceze yeah works when I create FooPart after defining printArg thanks! I'm used to Java so that feels strange because the class gets very untidy. is there an other way to do this? — BlakkM9
– BlakkM9, Commented Aug 14, 2019 at 14:00
@BlakkM9 It was already mentioned but just to clarify: On line 3, you call FooPart (which calls printArg). On line 5, you define printArg. You are calling printArg before defining it. — Tyler Roper
– Tyler Roper, Commented Aug 14, 2019 at 14:01

Alona · Accepted Answer · 2019-08-14 14:37:38Z

3

function Foo(arg) {
    this.argument = arg;
    this.fooPart = new FooPart(this);
}

Foo.prototype.printArg = function() {
    console.log(this.argument);
}

function FooPart(foo) {
    this.parent = foo;
    this.parent.printArg();
}

let foo = new Foo("this is the argument");

You should call FooPart after printArg definition
You should use this.parent to access parent

edited Aug 14, 2019 at 14:37

Alona

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answered Aug 14, 2019 at 14:00

George Wang

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3 Comments

BlakkM9 Over a year ago

I guess there is no way to make this more tidy, right?

George Wang Over a year ago

We can make it more tidy, yes. I will have updated one soon.

Alona Over a year ago

You should take a look for prototype and Class or defineProperty

deceze · Accepted Answer · 2019-08-14 14:25:22Z

The problem is that you define printArg after trying to call it.

The traditional way to define a "class" which doesn't have this problem would be:

function Foo(arg) {
  this.argument = arg;
  this.fooPart = new FooPart(this);
}

Foo.prototype.printArg = function() {
  console.log(this.argument);
}

function FooPart(foo) {
  this.parent = foo;
  this.parent.printArg();
}

let foo = new Foo("this is the argument");

The more modern version to define an "actual" class is:

class Foo {
  constructor(arg) {
    this.argument = arg;
    this.fooPart = new FooPart(this);
  }

  printArg() {
    console.log(this.argument);
  }
}

class FooPart {
  constructor(foo) {
    this.parent = foo;
    this.parent.printArg();
  }
}

let foo = new Foo("this is the argument");

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