How to find the sum of an array of numbers

This'd be exactly the job for reduce.

If you're using ECMAScript 2015 (aka ECMAScript 6):

const sum = [1, 2, 3].reduce((partialSum, a) => partialSum + a, 0); console.log(sum); // 6

For older JS:

const sum = [1, 2, 3].reduce(add, 0); // with initial value to avoid when the array is empty function add(accumulator, a) { return accumulator + a; } console.log(sum); // 6

Isn't that pretty? :-)

---

Recommended (reduce with default value)

Array.prototype.reduce can be used to iterate through the array, adding the current element value to the sum of the previous element values.

console.log( [1, 2, 3, 4].reduce((a, b) => a + b, 0) ) console.log( [].reduce((a, b) => a + b, 0) )

Without default value

You get a TypeError

console.log( [].reduce((a, b) => a + b) )

Prior to ES6's arrow functions

console.log( [1,2,3].reduce(function(acc, val) { return acc + val; }, 0) ) console.log( [].reduce(function(acc, val) { return acc + val; }, 0) )

Non-number inputs

If non-numbers are possible inputs, you may want to handle that?

console.log( ["hi", 1, 2, "frog"].reduce((a, b) => a + b) ) let numOr0 = n => isNaN(n) ? 0 : n console.log( ["hi", 1, 2, "frog"].reduce((a, b) => numOr0(a) + numOr0(b)) )

Non-recommended dangerous eval use

We can use eval to execute a string representation of JavaScript code. Using the Array.prototype.join function to convert the array to a string, we change [1,2,3] into "1+2+3", which evaluates to 6.

console.log( eval([1,2,3].join('+')) ) //This way is dangerous if the array is built // from user input as it may be exploited eg: eval([1,"2;alert('Malicious code!')"].join('+'))

Of course displaying an alert isn't the worst thing that could happen. The only reason I have included this is as an answer Ortund's question as I do not think it was clarified.

---

Why not reduce? It's usually a bit counter intuitive, but using it to find a sum is pretty straightforward:

var a = [1,2,3];
var sum = a.reduce(function(a, b) { return a + b; }, 0);

---

var arr = [1, 2, 3, 4];
var total = 0;
for (var i in arr) {
  total += arr[i];
}

---

var total = 0;
$.each(arr,function() {
    total += this;
});

---

Anyone looking for a functional oneliner like me?

Assuming:

const arr = [1, 2, 3, 4];

Here's the oneliner for modern JS:

sum = arr.reduce((a, b) => a + b, 0);

(If you happen to have to support ye olde IE without arrow functions:)

sum = arr.reduce(function (a, b) {return a + b;}, 0);

Note that 0 is the initial value here, so you can use that as offset if needed. Also note that this initial value is needed, otherwise calling the function with an empty array will error.

---

If you happen to be using Lodash you can use the sum function

array = [1, 2, 3, 4];
sum = _.sum(array); // sum == 10

---

This is possible by looping over all items, and adding them on each iteration to a sum-variable.

var array = [1, 2, 3];

for (var i = 0, sum = 0; i < array.length; sum += array[i++]);

JavaScript doesn't know block scoping, so sum will be accesible:

console.log(sum); // => 6

The same as above, however annotated and prepared as a simple function:

function sumArray(array) {
  for (
    var
      index = 0,              // The iterator
      length = array.length,  // Cache the array length
      sum = 0;                // The total amount
      index < length;         // The "for"-loop condition
      sum += array[index++]   // Add number on each iteration
  );
  return sum;
}

---

arr.reduce(function (a, b) {
    return a + b;
});

Reference: Array.prototype.reduce()

---

OK, imagine you have this array below:

const arr = [1, 2, 3, 4];

Let's start looking into many different ways to do it as I couldn't find any comprehensive answer here:

1) Using built-in reduce()

function total(arr) {
  if(!Array.isArray(arr)) return;
  return arr.reduce((a, v)=>a + v);
}

2) Using for loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0;
  for (let i=0,l=arr.length; i<l; i++) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

3) Using while loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0, i=-1;
  while (++i < arr.length) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

4) Using array forEach

function total(arr) {
  if(!Array.isArray(arr)) return;
  let sum=0;
  arr.forEach(each => {
    sum+=each;
  });
  return sum;
};

and call it like this:

total(arr); //return 10

It's not recommended to prototype something like this to Array...

---

You can also use reduceRight.

[1,2,3,4,5,6].reduceRight(function(a,b){return a+b;})

which results output as 21.

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/ReduceRight

---

Funny approach:

eval([1,2,3].join("+"))

---

A standard JavaScript solution:

var addition = [];
addition.push(2);
addition.push(3);

var total = 0;
for (var i = 0; i < addition.length; i++)
{
    total += addition[i];
}
alert(total);          // Just to output an example
/* console.log(total); // Just to output an example with Firebug */

This works for me (the result should be 5). I hope there is no hidden disadvantage in this kind of solution.

---

I am a beginner with JavaScript and coding in general, but I found that a simple and easy way to sum the numbers in an array is like this:

    var myNumbers = [1,2,3,4,5]
    var total = 0;
    for(var i = 0; i < myNumbers.length; i++){
        total += myNumbers[i];
    }

Basically, I wanted to contribute this because I didn't see many solutions that don't use built-in functions, and this method is easy to write and understand.

---

You can try the following code:

[1, 2, 3, 4].reduce((pre,curr)=>pre+curr,0)

---

Use a for loop:

const array = [1, 2, 3, 4];
let result = 0;

for (let i = 0; i < array.length - 1; i++) {
  result += array[i];
}

console.log(result); // Should give 10

Or even a forEach loop:

const array = [1, 2, 3, 4];
let result = 0;

array.forEach(number => {
  result += number;
})

console.log(result); // Should give 10

For simplicity, use reduce:

const array = [10, 20, 30, 40];
const add = (a, b) => a + b
const result = array.reduce(add);

console.log(result); // Should give 100

---

var totally = eval(arr.join('+'))

That way you can put all kinds of exotic things in the array.

var arr = ['(1/3)','Date.now()','foo','bar()',1,2,3,4]

I'm only half joking.

---

A few people have suggested adding a .sum() method to the Array.prototype. This is generally considered bad practice so I'm not suggesting that you do it.

If you still insist on doing it then this is a succinct way of writing it:

Array.prototype.sum = function() {return [].reduce.call(this, (a,i) => a+i, 0);}

then: [1,2].sum(); // 3

Note that the function added to the prototype is using a mixture of ES5 and ES6 function and arrow syntax. The function is declared to allow the method to get the this context from the Array that you're operating on. I used the => for brevity inside the reduce call.

---

ES6 for..of

let total = 0;

for (let value of [1, 2, 3, 4]) {
    total += value; 
}

---

A short piece of JavaScript code would do this job:

var numbers = [1,2,3,4];
var totalAmount = 0;

for (var x = 0; x < numbers.length; x++) {

    totalAmount += numbers[x];
}

console.log(totalAmount); //10 (1+2+3+4)

---

Use reduce

let arr = [1, 2, 3, 4]; let sum = arr.reduce((v, i) => (v + i)); console.log(sum);

---

No need to initial value! Because if no initial value is passed, the callback function is not invoked on the first element of the list, and the first element is instead passed as the initial value. Very cOOl feature :)

[1, 2, 3, 4].reduce((a, x) => a + x) // 10
[1, 2, 3, 4].reduce((a, x) => a * x) // 24
[1, 2, 3, 4].reduce((a, x) => Math.max(a, x)) // 4
[1, 2, 3, 4].reduce((a, x) => Math.min(a, x)) // 1

---

Here's an elegant one-liner solution that uses stack algorithm, though one may take some time to understand the beauty of this implementation.

const getSum = arr => (arr.length === 1) ? arr[0] : arr.pop() + getSum(arr);

getSum([1, 2, 3, 4, 5]) //15

Basically, the function accepts an array and checks whether the array contains exactly one item. If false, it pop the last item out of the stack and return the updated array.

The beauty of this snippet is that the function includes arr[0] checking to prevent infinite looping. Once it reaches the last item, it returns the entire sum.

---

Accuracy

Sort array and start sum form smallest numbers (snippet shows difference with nonsort)

[...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0)

arr=[.6,9,.1,.1,.1,.1] sum = arr.reduce((a,c)=>a+c,0) sortSum = [...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0) console.log('sum: ',sum); console.log('sortSum:',sortSum); console.log('sum==sortSum :', sum==sortSum); // we use .sort((a,b)=>a-b) instead .sort() because // that second one treat elements like strings (so in wrong way) // e.g [1,10,9,20,93].sort() --> [1, 10, 20, 9, 93]

For multidimensional array of numbers use arr.flat(Infinity)

arr= [ [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ], [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ] ]; sum = arr.flat(Infinity).reduce((a,c)=> a+c,0); console.log(sum); // 60

---

Those are really great answers, but just in case if the numbers are in sequence like in the question ( 1,2,3,4) you can easily do that by applying the formula (n*(n+1))/2 where n is the last number

---

You can combine reduce() method with lambda expression:

[1, 2, 3, 4].reduce((accumulator, currentValue) => accumulator + currentValue);

---

With reduce()

[1, 2, 3, 4].reduce((a, b) => a + b, 0); // 10

With forEach()

let sum = 0;
[1, 2, 3, 4].forEach(n => sum += n);
sum; // 10

With Parameter

function arrSum(arr) { 
  sum = 0;  
  arr.forEach(n => sum += n); 
  return sum; 
}

arrSum([1, 2, 3, 4]) // 10

---

i saw all answers going for 'reduce' solution

var array = [1,2,3,4]
var total = 0
for (var i = 0; i < array.length; i++) {
    total += array[i]
}
console.log(total)

---

very simple

step 1 we should have an array like :

const arrayNumber = [500,152,154,1555,12445];

step 2 (you can ignore this step if) step is to be sur that all values in table are number for that

let newArray = [];
for (let i = 0; i < arrayNumber.length; i++) {
        newArray.push(parseInt(arrayNumber[i], 10));
      }

step 3

const sumInArray = dataData.reduce( (a, b) => a + b);

finally

console.log(sumInArray);

---

Simplest answer to understand underlying process:

let array = [10, 20, 30, 40, 50]
let total = 0

for(let i in array)
{
    total += array[i]
}

console.log(total)

& if you're already familiar with underlying process then built-in method can save you time:

let array = [10, 20, 30, 40, 50]
let total = array.reduce((x, y) => x + y)
console.log(total)

---