const list = [1, 2, 3, 4, 5];

console.log(list.reduce((number, nextNumber) => number + nextNumber));

// --> 15

var array = [36, 25, 6, 15];

​

array.reduce(function(accumulator, currentValue) {

  return accumulator + currentValue;

}, 0); // 36 + 25 + 6 + 15 = 82

const arr = [1, 2, 3]

const result = arr.reduce((acc, curr) => acc + curr, 0)

// 1 + 2 + 3 = 6

console.log(result)

const sum = array.reduce((accumulator, element) => {

  return accumulator + element;

}, 0);

// define a reusable function

const calculateSum = (arr) => {

    return arr.reduce((total, current) => {

        return total + current;

    }, 0);

}

​

// try it

console.log(calculateSum([1, 2, 3, 4, 5]));

/*

This method takes a callback that takes two parameters,

one that represents the element from the last iteration and the other is the current

element of the iteration

*/

​

let nums = [2,4,6,8,3,5]

​

let result = nums.reduce((prev,curr)=>prev+curr)

console.log(result); // 28

[3, 2.1, 5, 8].reduce((total, number) => total + number, 0)

​

// loop 1: 0 + 3

// loop 2: 3 + 2.1

// loop 3: 5.1 + 5

// loop 4: 10.1 + 8

// returns 18.1

​

​

You can return whatever you want from reduce method, reduce array method good for computational problems

const arr = [36, 25, 6, 15];

array.reduce(function(accumulator, currentValue) {

  return accumulator + currentValue;

}, 0); // this will return Number

array.reduce(function(accumulator, currentValue) {

  return accumulator + currentValue;

}, []); // this will return array

let {total} = array.reduce(function(accumulator, currentValue) {

  let {currentVal} = currentValue

  return accumulator + currentVal;

}, {

total: 0

}); // this will return object

And one additional rule is always always retrun the first param, which do arithmetic operations.

let numbers = [1, 2, 3];

let sum = numbers.reduce(function (previousValue, currentValue) {

    return previousValue + currentValue;

});

reduce function sample solution