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Javascript iife functions

I was following a beginners course of JavaScript and I've come across this code. But I don't understand why sum needs to be an IIFE function. Could you please help me understand how this code works exactly?

const sum = function() {
  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))

about 4 years ago · Juan Pablo Isaza
3 Respuestas
Responde la pregunta

0

In this case the variable args takes the arguments

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

The reduce() method executes a reducer function for each value of an array. In this case you are taking first two elements of arrays, finding the sum of these two and the reduce method goes one by one to other elements.

Other example to explain you better

const numbers = [175, 50, 25];

function myFunc(total, num) {
   return total - num;
} // returns 100 (175 - 50 - 25)
about 4 years ago · Juan Pablo Isaza Denunciar

0

In that function looks like is just to show you how to work the IIFE(the parentheses at the end of the function) that parenthesis is to auto-execute the function, the use is if you want to execute a function when this load you can add the "()" at the end or "+" at the beginning of the clause "function" like this "+function" to see more in details check the following link:

Immediately-Invoked Function Expression (IIFE)

about 4 years ago · Juan Pablo Isaza Denunciar

0

There is no current reason the function is inside another executed function.

These are equivalent except that one gets returned immediately by a function. If I had to guess, the next instruction may talk about scoping or currying?, who knows.

const sum = function() {
  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))

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

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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