How to add two big numbers in javascript?

Question

I have two text boxes. Each will take input up to thousand digits.

Now i want to add these two numbers. My question is what data type should i use to store the result?

I have tried this:

<script>
   var x = 'Thousand digit of number'
    var y = 'Thousand digit of number'
    var z = x + y
</script>

but i am getting result in exponential form. How to store the result and display it?

i tried to put the number but it was not posting my question thats why i changed it. @LKTN.25 — Swapnil Patil, Jan 31 '17 at 08:20
Possible duplicate of [Javascript summing large integers](http://stackoverflow.com/questions/4557509/javascript-summing-large-integers) — Null, Jan 31 '17 at 08:21
Javascript will display anything larger than `10e21` in exponential form. My browser and Node.js both return `Infinity` for anything larger than (approximately) `1.7e308`. Javascript is not a good platform for dealing with very, very large numbers. — lonesomeday, Jan 31 '17 at 08:27
The biggest number you can work safely with in JavaScript is Number.MAX_SAFE_INTEGER. What are you calculating that takes 10 * googol sized numbers? Can it be simplified? There are libraries like http://mikemcl.github.io/decimal.js/ that deals with big numbers in JavaScript — Emil S. Jørgensen, Jan 31 '17 at 08:34

Qwerty · Answer 1 · 2020-11-08T15:24:09.370

13

Yet another solution, because it's faster and cleaner.

function add(A, B) {
  const AL = A.length
  const BL = B.length
  const ML = Math.max(AL, BL)

  let carry = 0, sum = ''

  for (let i = 1; i <= ML; i++) {
    let a = +A.charAt(AL - i)
    let b = +B.charAt(BL - i)

    let t = carry + a + b
    carry = t/10 |0
    t %= 10

    sum = (i === ML && carry)
      ? carry*10 + t + sum
      : t + sum
  }

  return sum
}

> add(
'9999999999999999999999999999999999999999999999999999999999999999999999999999',
'999999999999999999999999999999999999999'
)

> "10000000000000000000000000000000000000999999999999999999999999999999999999998"

edited Nov 08 '20 at 15:24

answered Nov 08 '20 at 15:17

Qwerty

29,062
22
108
136

return sum.startsWith("0") ? sum.substring(1):sum; To avoid starting with any ZERO – Devmaleeq Feb 11 '21 at 08:33
@Malik With what values did it return a leading `0` to you? – Qwerty Feb 12 '21 at 17:09

score 7 · Answer 2 · answered May 27 '21 at 20:58

7

Use BigInt as described here: https://stackoverflow.com/a/56370672/641913

const z = BigInt(x) + BigInt(y);

console.log(z.toString());

answered May 27 '21 at 20:58

Liam

143
2
7

score 2 · Answer 3 · answered Mar 16 '18 at 14:50

Here is another solution not so different from others you can find in the internet (consider that it doesn't work with negative numbers!):

function sums(arg1, arg2) {
 var sum = "";
 var r = 0;
 var a1, a2, i;

 // Pick the shortest string as first parameter and the longest as second parameter in my algorithm
 if (arg1.length < arg2.length) {
  a1 = arg1;
  a2 = arg2;
 }
 else {
  a1 = arg2;
  a2 = arg1;
 }
 a1 = a1.split("").reverse();
 a2 = a2.split("").reverse();

 // Sum a1 and a2 digits
 for (i = 0; i < a2.length; i++) {
  var t = ((i < a1.length) ? parseInt(a1[i]) : 0) + parseInt(a2[i]) + r;
  
  sum += t % 10;

  r = t < 10 ? 0 : Math.floor(t / 10);
 }
 // Append the last remain
 if (r > 0)
  sum += r;

 sum = sum.split("").reverse();
 
 // Trim the leading "0"
 while (sum[0] == "0")
  sum.shift();

 return sum.length > 0 ? sum.join("") : "0";
}

// Test
function testEquals(expected, actual) {
 if (expected == actual)
  console.log("OK: " + expected);
 else
  console.error("ERROR: " + expected + " != " + actual);
}

testEquals("100", sums("99", "1"));
testEquals("100", sums("00099", "0001"));
testEquals("10000000000", sums("9999999999", "1"));
testEquals("10000010101", sums("9999999999", "10102"));
testEquals("0", sums("0", "0"));
testEquals("1", sums("0", "1"));
testEquals("9", sums("8", "1"));
testEquals("9", sums("1", "8"));
testEquals("10000000000000000000000000000000000000000", sums("9999999999999999999999999999999999999999", "1"));

mudin · Accepted Answer · 2017-01-31T09:18:26.857

Input the numbers as string and add each characters each other as array something like this:

 function add() {
        document.getElementById("demo").innerHTML = "";
        var x = document.getElementById("txt1").value;
        var y = document.getElementById("txt2").value;
        var len;
        var lenx = x.length;
        var leny = y.length;
        var x1,y1,rem,div=0;
        if(lenx>leny) len = lenx; else len = leny;

        for(var i=0;i<len;i++){
            if(i>=lenx) x1  = 0;
            else x1 = parseInt(x[lenx-i-1]);
            if(i>=leny) y1 = 0;
            else y1 = parseInt(y[leny-i-1]);
            rem = (x1+y1+div)%10;
            div = Math.floor((x1 + y1+div)/10);
            document.getElementById("demo").innerHTML = rem + document.getElementById("demo").innerHTML;
        }
       if(div>0){
            document.getElementById("demo").innerHTML = div + document.getElementById("demo").innerHTML;
       }
    }

Here the code: https://jsfiddle.net/mtsL1k2x/5/

Note: this is only for natural numbers. You can modify depending on your inputs

score 0 · Answer 5 · answered Jan 31 '17 at 08:57

0

Either use a big number library like https://mathjs.org/docs/datatypes/bignumbers.html , or you can use something lighter weight (but easy to understand) like http://www.discoversdk.com/knowledge-base/arbitrary-length-integer-addition-in-javascript

answered Jan 31 '17 at 08:57

ephraim

21
2

score 0 · Answer 6 · answered Sep 09 '20 at 05:33

Well, if you want to do this without using BigInt or any third-party Library, then I don't think you need to convert to an array, you can use the charAt() function to add the individual characters at each point in the string. You would have to use the for loop starting from its maximum value and reducing till its lowest. The code snippet is below;

function add(a, b) {
let sum='';
 let z,x;
 let r=0;
 if (a.length>=b.length){
   z=a;
   x=b;
 }
 else{
   z=b;
   x=a;
 };
 let p=x.length;
 for (let i=z.length;i>0;i--){
   let t=((p>0)?parseInt(x.charAt(p-1)):0)+parseInt(z.charAt(i-1))+r;
   sum=(t%10)+sum;
   r=t<10?0:Math.floor(t/10);
   p=p-1;
 };
 if (r>0){sum=r+sum};
 return sum;

};

Riaz Bappy · Answer 7 · 2022-06-27T12:08:12.567

function add(a, b) {
    a = a.split("").reverse();
    b = b.split("").reverse();
    let maxLen=Math.max(a.length, b.length);
    let sum = [];
    let remainder = 0;
    for (let i = 0; i < maxLen; i++) {
      let x = parseInt(a[i]) ? parseInt(a[i]) : 0;
      let y = parseInt(b[i]) ? parseInt(b[i]) : 0;
      let digit = (x + y + remainder) % 10;
      remainder = Math.floor((x + y + remainder) / 10);
      sum.unshift(digit);
    }
    if (remainder) {sum.unshift(remainder)}
    
    return sum.join("");
  }

score -2 · Answer 8 · answered Oct 16 '20 at 11:55

function add(x, y) {
    //this function adds two extremely large numbers, negative and/or positive
    var temp, borrow=false, bothNeg=false, oneNeg=false, neg=false;
    if (x < 0 && y < 0) { bothNeg = true; x = -x; y = -y; } 
    else if (x < 0 || y < 0) {
        oneNeg = true;
        if (Math.abs(x) == Math.abs(y)) { x = 0; y = 0; }
        else if (x < 0 && Math.abs(x) > Math.abs(y)) { neg = true; x = -x; y = -y; }
        else if (x < 0 && Math.abs(x) < Math.abs(y)) { temp = y; y = x; x = temp; }
        else if (y < 0 && Math.abs(x) < Math.abs(y)) { neg = true; temp = y; y = -x; x = -temp; }
    }
    x = parseInt(x*1000000000/10).toString();
    y = parseInt(y*1000000000/10).toString();
    var lenx=x.length, leny=y.length, len=(lenx>leny)?lenx:leny, sum="", div=0, x1, y1, rem;
    for (var i = 0; i < len; i++) {
        x1 = (i >= lenx) ? 0 : parseInt(x[lenx-i-1]);
        y1 = (i >= leny) ? 0 : parseInt(y[leny-i-1]);
        y1 = (isNaN(y1)) ? 0 : y1;
        if (oneNeg) y1 = -y1;
        if (borrow) x1 = x1 - 1;
        if (y < 0 && x1 > 0 && Math.abs(x1) >= Math.abs(y1)) { borrow=false; div=0; }
        if (y < 0 && y1 <= 0 && (x1 < 0 || Math.abs(x1) < Math.abs(y1))) { borrow=true; rem=(x1+y1+div+10)%10; div=10; }
        else { rem=(x1+y1+div)%10; div=Math.floor((x1+y1+div)/10); }
        sum = Math.abs(rem).toString() + sum;
    }
    if (div > 0) sum = div.toString() + sum;
    sum = parseFloat(sum*10/1000000000);
    if (bothNeg || neg) sum = -sum;
    return sum;
}

How does it work? This will not give an expected result `add(9999999999999999999999999999999999999999999999999999999999999999999999999999,999999999999999999999999999999999999999)` — Qwerty, Nov 04 '20 at 16:43

score -3 · Answer 9 · answered Jan 31 '17 at 08:32

-3

 <body>
<p>Click the button to calculate x.</p>
<button onclick="myFunction()">Try it</button>
<br/>
<br/>Enter first number:
<input type="text" id="txt1" name="text1">Enter second number:
<input type="text" id="txt2" name="text2">
<p id="demo"></p>
<script>
  function myFunction() {
    var y = document.getElementById("txt1").value;
    var z = document.getElementById("txt2").value;
    var x = +y + +z;
    document.getElementById("demo").innerHTML = x;
  }
</script>

https://jsfiddle.net/Sanjeevgaut/mtsL1k2x/

answered Jan 31 '17 at 08:32

Sanjeev Gautam

353
3
15

Can you explain why would `+y` and `+z` make a difference ? – Null Jan 31 '17 at 08:39
`input up to thousand digits` – mudin Jan 31 '17 at 08:39
If you will not put + then it will treat as string, If we add +var then only it will treat as number. – Sanjeev Gautam Jan 31 '17 at 08:41
hey i hope my solution is working fine? I have share the working link also. – Sanjeev Gautam Jan 31 '17 at 09:40
1

he doesnt want to get result in exponential form. try to input large numbers – mudin Jan 31 '17 at 10:31

How to add two big numbers in javascript?

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