javascript - How to destructure option argument with all default values in ES6? - Stack Overflow

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I use ES6 features with babel compiler. I have a function which takes option object as an argument:

function myFunction({ option1 = true, option2 = 'whatever' }) {
    console.log(option1, option2);
    // do something...
}

When I call it, destructuring happens and everything works well. I want to call it with default options most of the time, so I do:

myFunction({}); // true 'whatever'

but it looks little bit strange. It would much more cleaner just call:

myFunction(); // TypeError: Cannot read property 'option1' of undefined

Is it possible?

I use ES6 features with babel compiler. I have a function which takes option object as an argument:

function myFunction({ option1 = true, option2 = 'whatever' }) {
    console.log(option1, option2);
    // do something...
}

When I call it, destructuring happens and everything works well. I want to call it with default options most of the time, so I do:

myFunction({}); // true 'whatever'

but it looks little bit strange. It would much more cleaner just call:

myFunction(); // TypeError: Cannot read property 'option1' of undefined

Is it possible?

• javascript
• ecmascript-6
• destructuring

Share Improve this question Follow asked Dec 14, 2015 at 20:26 madox2madox2 51.8k2020 gold badges104104 silver badges100100 bronze badges 1

• see also ES6 destructuring object assignment function parameter default value – Bergi Commented Mar 17, 2017 at 14:03

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1 Answer 1

Sorted by: Reset to default 29

Yes, you just have to provide a default value for the complete argument:

function myFunction({option1 = true, option2 = 'whatever'} = {}) {
//                                                         ^^^^
    console.log(option1, option2);
    // do something...
}

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