Use the Rest Operator with Function Parameters!

JavaScript
1

Tell us what’s happening:
Why can’t I use spread operator in const args ? “n” is also an array of arguments right?

Your code so far


const sum = (function() {
  "use strict";
  return function sum(...n) {
    const args = [...n];
    return args.reduce((a, b) => a + b, 0);
  };
})();
console.log(sum(1, 2, 3)); // 6

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.80 Safari/537.36.

Link to the challenge:
https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/es6/use-the-rest-operator-with-function-parameters

2

Well, args should be the function parameter . And you really don’t need const args = [...n]. You already have an array on which you can use reduce() on

3

You don’t need to declare another args array, just spread the function parameter “args” and then reduce your values in args to accumulate. In other words, delete const args = […n]; and replace your …n parameter with …args

1 Like
4

and what if i did the code this way:

const sum = (function() {
“use strict”;
return function sum(…s) {
return s.reduce((a, b) => a + b, 0);
};
})();
console.log(sum(1, 2, 3));

why not giving same result as i just replaced args with s ? ( im really new to coding :upside_down_face: )

5

the result is the same, and it works exactly the same, the tests ask specifically for the args parameter tho

1 Like
6

Perfect! thanks Prof.