JavaScript (ES6): Named parameters and default values

Question

I am coming from Python and am really liking the way to set named parameters and default values—now it seems that ES6 allows me to do similar. But I can't see why this last call breaks:

fun = ({first=1, last=1}) => (1*first+2*last)

console.log("-----------")

console.log( fun({first:1, last:2}) )

console.log("-----------")

console.log( fun({last:1, first:2}) )

console.log("-----------")

console.log( fun() ) // Breaks

Javascript doesn't have named arguments. Object deconstruction is not the same, even if it can be used to implement something close to named arguments. You shouldn't approach them the same way. — deceze
– deceze ♦, Commented May 24, 2018 at 14:07
it's because you are thinking of them as kwargs. It's not the same. Good answers below. — Adam Hopkins
– Adam Hopkins, Commented May 24, 2018 at 14:10

Nina Scholz · Accepted Answer · 2018-05-24 14:06:54Z

36

You need a default object.

var fun = ({ first = 1, last = 1 } = {}) => 1 * first + 2 * last;
//                                 ^^^^

console.log(fun({ first: 1, last: 2 }));
console.log(fun({ last: 1, first: 2 }));
console.log(fun());

answered May 24, 2018 at 14:06

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Jonas Wilms · Accepted Answer · 2018-05-24 14:07:42Z

3

Cause you need an object which can be deconstructed:

fun({})

answered May 24, 2018 at 14:07

Jonas Wilms

139k20 gold badges164 silver badges164 bronze badges

1 Comment

Norfeldt Over a year ago

Like this answer because it forces me to remember that I always must pass an object

Collectives™ on Stack Overflow

JavaScript (ES6): Named parameters and default values

2 Answers 2

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related