JavaScript optional destructuring argument in function

Question

I have this function signature

const foo = (arg, { opt1, opt2, opt3 }) => {
   ...
};

but I'd like to have this second argument optional, such as calling the function like

foo("Hello");

However, I get

TypeError: Cannot destructure property opt1 of 'undefined' or 'null'.

So, I'm tempted to fix this with changing the function such as :

const foo = (arg, options = {}) => {
   const { opt1, opt2, opt3 } = options;

   ...
};

But was wondering if there was a more inline alternative?

Nina Scholz · Accepted Answer · 2020-06-20 16:32:01Z

13

You could assign a default object and take a destructuring at the same time.

The result is undefined for all three destructured properties, if no second parameter or undefined.

const foo = (arg, { opt1, opt2, opt3 } = {}) => {
   ...
};

answered Jun 20, 2020 at 16:32

Nina Scholz

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Comments

Yash Goyal · Accepted Answer · 2020-06-20 16:32:44Z

6

You can do:-

const foo = (arg, { opt1, opt2, opt3 } = {}) => {

   ...
};

answered Jun 20, 2020 at 16:32

Yash Goyal

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Saurabh Singh · Accepted Answer · 2020-06-20 16:33:02Z

6

You can do this { opt1, opt2, opt3 } = {} when declaring the function.

answered Jun 20, 2020 at 16:33

Saurabh Singh

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