Use the Rest Operator with Function Parameters (Please Help Me)

Use the Rest Operator with Function Parameters (Please Help Me)
0

#1

Tell us what’s happening:

Your code so far


const sum = (function() {
  "use strict";
  return function sum(...args) {
    const args = [ x, y, z ];
    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/68.0.3440.106 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

You are already using spread operator on the function parameter which is correct.

Just delete the line under it where you initialize args again because it’s overlapping it.


#3

I’m not undertand it. Where is delete line ?


#4

My Code is Working. Thanks