Setting default values for properties in Javascript

Question

How can I set default values for the properties p1,p2,p3 and p4` of a class:

class O extends AnotherClass {

    constructor(p1,p2,p3,p4) {
    super(); \\ for the this-reference
    if (p1) this.p1 = p1;
    if (p2) this.p2 = p2;
    if (p3) this.p3 = p3;
    if (p4) this.p3 = p4;

    }

do I have to write one by one

O.prototype.p1 = "default1"
O.prototype.p2 = "default2"
O.prototype.p3 = "default3"
O.prototype.p4 = "default4"

or is there a more elegant way, like

O.prototype = {p1: "default1", p2 : "default1", p3 : "default3", p4 : "default4"}

but the latter does not seem to work...

Note that if (p1) will evaluate to false for values other than undefined, such as null, 0, false, etc. You might be better off with a typeof test. — RobG
– RobG, Commented Mar 14, 2017 at 22:39

Simon Mo · Accepted Answer · 2017-03-14 22:21:48Z

5

You can set default properties in es6 when you declare params in your constructor like this constructor(p1 = 'Default Variable',p2 = 'Default Variable',p3 = 'Default Variable',p4 = 'Default Variable')

answered Mar 14, 2017 at 22:21

Simon Mo

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Matías Fidemraizer · Accepted Answer · 2017-03-14 23:41:02Z

1

In addition to plain ES2015+ default parameter values as @Fried_Chicken has already answered, there's still an alternate approach.

Why don't you accept arguments as an object and then, you use ES2015+ destructuring capabilities? This is a great choice as you can provide the whole arguments in any order, or even provide just one or some of them.

Also, you won't need to provide null/undefined to some given parameters:

doStuff(1, null, null, 2);

See the following runnable code snippet and play with it. The same solution can be applied in your scenario as destructuring can be used on class constructors.

function doStuff({ arg0 = null, arg1 = null } = {}) {
  if (arg0 !== null) {
      console.log(arg0);
  } else {
      console.log("not provided");
  }
}

doStuff();
doStuff({ arg0: "hello world" });

answered Mar 14, 2017 at 23:41

Matías Fidemraizer

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2 Comments

Michael Hewson Over a year ago

Whoa, since when does StackOverflow have runnable code snippets?

Matías Fidemraizer Over a year ago

@MichaelHewson Hey, I'm not sure about that, but I believe it was more than a year ago!

Davoud Ramezani · Accepted Answer · 2022-02-15 23:53:39Z

0

  constructor(p1, p2, p3, p4) {
    super();
    if (p1) this.p1 = p1 === undefined ? "p1Default" : p1;
    if (p2) this.p2 = p2 === undefined ? "p2Default" : p2;
    if (p3) this.p3 = p3 === undefined ? "p3Default" : p3;
    if (p4) this.p3 = p4 === undefined ? "p4Default" : p4;
  }

answered Feb 15, 2022 at 23:53

Davoud Ramezani

1

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