JSON Definitions Using Internally Defined Variables

Question

I am trying to create a JSON object to store some parameters for a program. Some of the parameters need to be calculated from others as they are being defined. I would like to do this within the object definition but maybe this is not possible

var params = {
        a: 50,
        b: 70,
        c: this.a+this.b
    }

Result

What happens

>params.c
NaN

What I wished happened

>params.c
120

Edit

After doing some further reading, I think I am using Object Literal Notation instead of JSON.

Note this is not JSON, it is a object literal. [There's no such thing as a "JSON Object"](http://benalman.com/news/2010/03/theres-no-such-thing-as-a-json/) — Patrick Evans, Feb 13 '18 at 00:22

Ele · Accepted Answer · 2018-02-13T00:37:26.310

1

You can use this approach:

To avoid re-calculation, use the function Object.assign.

The get syntax binds an object property to a function that will be called when that property is looked up.

var params = {
  a: 50,
  b: 70,
  get c() {
    console.log('Called!');
    return this.a + this.b;
  }
};

console.log(params.c); // Prints 120 then Called!
console.log(params.c); // Prints 120 then Called!

var params = Object.assign({}, {
  a: 50,
  b: 70,
  get c() {
    console.log('Called from function Object.assign!');
    return this.a + this.b;
  }
});

params.a = 1000; // To illustrate.

console.log(params.c); // Prints 120
console.log(params.c); // Prints 120

.as-console-wrapper {
  max-height: 100% !important
}

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edited Feb 13 '18 at 00:37

answered Feb 13 '18 at 00:16

Ele

33,468
7
37
75

This is ok but I would like to avoid calculating each parameter every time it is called. – Cole Feb 13 '18 at 00:29
@Cole the attributes `a` and `b` will change in time? – Ele Feb 13 '18 at 00:31
@Cole see the updated answer. – Ele Feb 13 '18 at 00:37
No, this was intended for initialization purposes. Good to know that you can make it dynamic if needed though! – Cole Feb 13 '18 at 00:43

score 0 · Answer 2 · answered Feb 13 '18 at 00:15

0

Personally, I would create constants (since magic numbers are the devil), but this is an overly-simplistic example:

const FIFTY = 50;
const SEVENTY = 70;
var params = {
  a: FIFTY,
  b: SEVENTY,
  c: FIFTY + SEVENTY
};

answered Feb 13 '18 at 00:15

th3n3wguy

3,649
2
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30

score 0 · Answer 3 · edited Feb 13 '18 at 00:25

What I would recommend doing is starting with an object that does not contain c, and taking the calculation outside of the object. Once the calculation has occurred, simply add the sum back to the object as a new key/value pair:

var params = {
  a: 50,
  b: 70,
}

var sum = 0;

for (var el in params) {
  if (params.hasOwnProperty(el)) {
    sum += params[el];
  }
}

params['c'] = sum;
console.log(params);

JSON Definitions Using Internally Defined Variables

3 Answers3

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