How to get the same behavior in an object when the value is copied to another variable

Question

In this console:

When B is set to A but A was destroyed afterwards, B.b() throws an error because A is truly undefined.

How can one avoid this?

PS: I am aware that I can simply return this (or change the function in some way) but that doesn't fulfill my purposes.

EDIT: How do I 'localise' and somehow tell javascript that by A, I mean B even if A is undefined, WITHOUT ALTERING THE FUNCTION ITSELF?

Could you elaborate your purposes in the question, seeing as `this` keyword is used for precisely that? — Anurag Srivastava, Mar 26 '20 at 18:21
Images are fine, but text is better. Imagine if you couldn't see the image; your text wouldn't make sense. Please include either just text, or both. — Heretic Monkey, Mar 26 '20 at 18:22
You could also just change `B.b` to look at `B` instead of `A`. I.e., `B.b = function () { return B.a; };`. But @Anurag's question is a good one; why not avail yourself of the obvious and purpose-built mechanism for this functionality? — Heretic Monkey, Mar 26 '20 at 18:30
@HereticMonkey and everyone else, Can you check [this](https://stackoverflow.com/questions/60876579/how-to-clone-a-function-but-handle-the-variables-too) out? Thanks. — Angshu31, Mar 26 '20 at 21:46

score 0 · Answer 1 · answered Mar 26 '20 at 18:34

0

You should change your A to look like the following:

let A = {
  a: "a",
  b: function() {
    // change A.a to this.a
    return this.a;
  }
};

Right now since it references A when you destroy A Even though B is already equal to A the b method still needs the original A to exist. Swapping it for this will fix the issue.

answered Mar 26 '20 at 18:34

mikerojas

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Thanks, for your contribution, can you check [this](https://stackoverflow.com/questions/60876579/how-to-clone-a-function-but-handle-the-variables-too) out as well? Thanks. – Angshu31 Mar 26 '20 at 21:45

Jason Lydon · Answer 2 · 2020-03-31T15:59:07.393

0

The way the question and example is written that "error" cannot be avoided. return A.a; will and SHOULD be undefined based on your example because you are assigning A = undefined;

Also, if you are trying to clone an object, there are a couple of other options you should look into like:

let B = JSON.parse(JSON.stringify(A));
let B = Object.assign({}, A); MDN
Javascript classes

And maybe check out this answer: How do I correctly clone a JavaScript object?

edited Mar 31 '20 at 15:59

answered Mar 26 '20 at 19:01

Jason Lydon

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Which 'this' are you talking about? :) – Anurag Srivastava Mar 26 '20 at 19:04
#1 in your list would remove the `b` method altogether... #2 suffers from the same problem as the OP's code, since it copies a reference to the `b` method to `B`. – Heretic Monkey Mar 26 '20 at 20:42
1

Even if I do any of that, as soon as I set `A` to undefined, the same problem occurs, doesn't it? – Angshu31 Mar 26 '20 at 20:53
Anyway, please see [this question](https://stackoverflow.com/questions/60876579/how-to-clone-a-function-but-handle-the-variables-too) that I asked as well. – Angshu31 Mar 26 '20 at 21:43

score 0 · Accepted Answer · answered Jun 01 '20 at 09:12

To the new viewers, if you want to do this without altering the function:

var A = (function() {
  var A = {a: 90, b: function() { return A.a }}; // same object as in the Q
  return A;
})()

console.log(A.b()) // 90

var B = A;
A = undefined;

console.log(B.b()) // Also 90

B.a = 89;
console.log(B.b()) // 89

How to get the same behavior in an object when the value is copied to another variable

3 Answers3