Why has to be ....args, not diferrent parameter

JavaScript
#1

Tell us what’s happening:
I have one question, In the parameters, I was using for example (…z), why this parameter can’t be returned.
obligatory the example has to use (…args)?

Your code so far


const sum = (function() {
  "use strict";
  return function sum(...args) {
    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 6.3; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.121 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

no, you can use z as long as you are also using z.reduce() and it will be the same effect as using args, the issue is that the tester expects you to use args specifically for the problem to pass, but it is not a requirement and args is not a reserved word. (However arguments is a reserved word)

1 Like
#3