CODEWITHSUNDEEP
javascriptecmascript-6ecmascript-next
TheRealFakeNews
TheRealFakeNews

Reputation: 8153

How to use the rest and spread operators together in a function definition?

I have the following:

const a = (...args) =>{ return {...args}}
const abc = a('lol', 'rofl', 'lmao');
console.log('abc', abc);

However, this prints out

Object {0: "lol", 1: "rofl", 2: "lmao"}

But I expected

Object {lol: "lol", rofl: "rofl", lmao: "lmao"}

since

{lol, rofl, lmao}

produces the line above.

Is there a way to spread the arguments so that I can get this result?

Upvotes: 1

Views: 100

Answers (3)

Redu
Redu

Reputation: 26181

You may do as follows;

function F(...args){
  args.forEach(a => this[a] = a, this);
}
var abc = new F('lol', 'rofl', 'lmao');
console.log('abc', abc);

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386650

You could use Object.assign and computed properties.

const a = (...args) => args.reduce((obj, v) => Object.assign(obj, { [v]: v }), {});
const abc = a('lol', 'rofl', 'lmao');

console.log('abc', abc);

Upvotes: 0

Pranav C Balan
Pranav C Balan

Reputation: 115232

I don't think that there is any build in method to do that, the spread syntax will not work as you expected. Anyway, you can use Array#reduce method.

const a = (...args) => args.reduce((obj, v) => (obj[v] = v, obj), {})
const abc = a('lol', 'rofl', 'lmao');
console.log('abc', abc);

Upvotes: 1

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