Javascript execute optional function of object

Question

If I have an object like this

 const obj = {
     func: props.func, // we don't know if func is a function or undefined...
 }

and I have this in my code:

 obj.func(params);

What is the good way to execute the optional function of the object?

I have thought to do something like this:

if(typeof obj.func === "function")
   obj.func(params);

or

try {
    obj.func(params);
} catch (err) {}

But I'm not sure if both forms are correct to do it (or if there is a better way) as I don't have much experience with javascript.

Someone who knows? Thank you.

Either could be good, depends on the type of check you want to accomplish. Also, consider reading about EAFP vs LBYL and check what's preferred approach is used in your application. — Eugene Naydenov, Oct 27 '20 at 09:38
https://stackoverflow.com/questions/5999998/check-if-a-variable-is-of-function-type check this — Artem Arkhipov, Oct 27 '20 at 09:39
ES2020 [optional chaining](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Optional_chaining). `obj.func?.()` — Yury Tarabanko, Oct 27 '20 at 09:40

Victor Molina · Answer 1 · 2020-10-27T09:42:29.163

-1

Read about optional chaining in JS here

Personally, I would use this way from ES2020:

const props = {
  func: Math.random() >= 0.5 ? undefined : (params) => console.log(params),
};

const obj = {
   func: props.func,
};

And then

obj.func?.("hello world!");

This way is recommended here

edited Oct 27 '20 at 09:42

answered Oct 27 '20 at 09:34

Victor Molina

2,353
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19
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score -1 · Answer 2 · answered Oct 27 '20 at 09:36

-1

I personally would go with

if (typeof obj.func === "function") {
    obj.func();
}

because if you use a try catch, and an error occurred within the function itself, it might appear as though the function was never called in the first place.

answered Oct 27 '20 at 09:36

Simon

369
1
9

Javascript execute optional function of object

2 Answers2