Using the Rest Operator with Function Parameter example

JavaScript
1

Hello, strangers.
I’m new to programming so I’m quite confused and clueless most of the time. Can someone explain why there’s an opening parenthesis before the function keyword? Also, why create a function called sum and then return that exact function? I’ve never seen anything like this before. Thanks in advance.


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

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

2

Hello,

the parenthesis make it an Immediately Invoked Function Expression (IIFE). “A common pattern in JavaScript is to execute a function as soon as it is declared.” They are explained later in the curriculum ( JavaScript Algorithms And Data Structures Certification > Object Oriented Programming > exercises:
Understand the Immediately Invoked Function Expression (IIFE) and Use an IIFE to Create a Module) There is also a video by Beau: Immediately Invoked Function Expression - Beau teaches JavaScript

I am not sure about the return.

2 Likes
3

Hello
Good question : why create a function called sum and then return that exact function?
const sum define a constant reference to function sum(...args){...}.
if we try this :

function sum(...args) {
  return -1;
}

we get this error :

SyntaxError: redeclaration of const sum

this protect sum from being accidentally changed when code grow.

1 Like