How to avoid scientific notation for large numbers in JavaScript?

JavaScript converts integers with more than 21 digits to scientific notation when used in a string context. I'm printing an integer as part of a URL. How can I prevent the conversion from happening?

There's Number.toFixed, but it uses scientific notation if the number is >= 1e21 and has a maximum precision of 20. Other than that, you can roll your own, but it will be messy.

function toFixed(x) {
  if (Math.abs(x) < 1.0) {
    var e = parseInt(x.toString().split('e-')[1]);
    if (e) {
        x *= Math.pow(10,e-1);
        x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
    }
  } else {
    var e = parseInt(x.toString().split('+')[1]);
    if (e > 20) {
        e -= 20;
        x /= Math.pow(10,e);
        x += (new Array(e+1)).join('0');
    }
  }
  return x;
}

Above uses cheap-'n'-easy string repetition ((new Array(n+1)).join(str)). You could define String.prototype.repeat using Russian Peasant Multiplication and use that instead.

This answer should only be applied to the context of the question: displaying a large number without using scientific notation. For anything else, you should use a BigInt library, such as BigNumber, Leemon's BigInt, or BigInteger. Going forward, the new native BigInt (note: not Leemon's) should be available; Chromium and browsers based on it (Chrome, the new Edge [v79+], Brave) and Firefox all have support; Safari's support is underway.

Here's how you'd use BigInt for it: BigInt(n).toString()

Example:

const n = 13523563246234613317632;
console.log("toFixed (wrong): " + n.toFixed());
console.log("BigInt (right):  " + BigInt(n).toString());

Beware, though, that any integer you output as a JavaScript number (not a BigInt) that's more than 15-16 digits (specifically, greater than Number.MAX_SAFE_INTEGER + 1 [9,007,199,254,740,992]) may be be rounded, because JavaScript's number type (IEEE-754 double-precision floating point) can't precisely hold all integers beyond that point. As of Number.MAX_SAFE_INTEGER + 1 it's working in multiples of 2, so it can't hold odd numbers anymore (and similiarly, at 18,014,398,509,481,984 it starts working in multiples of 4, then 8, then 16, ...).

Consequently, if you can rely on BigInt support, output your number as a string you pass to the BigInt function:

const n = BigInt("YourNumberHere");

Example:

const n1 = BigInt(18014398509481985); // WRONG, will round to 18014398509481984
                                      // before `BigInt` sees it
console.log(n1.toString() + " <== WRONG");
const n2 = BigInt("18014398509481985"); // RIGHT, BigInt handles it
console.log(n2.toString() + " <== Right");

I know this is an older question, but shows recently active. MDN toLocaleString

const myNumb = 1000000000000000000000;
console.log( myNumb ); // 1e+21
console.log( myNumb.toLocaleString() ); // "1,000,000,000,000,000,000,000"
console.log( myNumb.toLocaleString('fullwide', {useGrouping:false}) ); // "1000000000000000000000"

you can use options to format the output.

Note:

Number.toLocaleString() rounds after 16 decimal places, so that...

const myNumb = 586084736227728377283728272309128120398;
console.log( myNumb.toLocaleString('fullwide', { useGrouping: false }) );

...returns...

586084736227728400000000000000000000000

This is perhaps undesirable if accuracy is important in the intended result.

For small number, and you know how many decimals you want, you can use toFixed and then use a regexp to remove the trailing zeros.

Number(1e-7).toFixed(8).replace(/\.?0+$/,"") //0.000

one more possible solution:

function toFix(i){
 var str='';
 do{
   let a = i%10;
   i=Math.trunc(i/10);
   str = a+str;
 }while(i>0)
 return str;
}

Here is my short variant of Number.prototype.toFixed method that works with any number:

Number.prototype.toFixedSpecial = function(n) {
  var str = this.toFixed(n);
  if (str.indexOf('e+') === -1)
    return str;

  // if number is in scientific notation, pick (b)ase and (p)ower
  str = str.replace('.', '').split('e+').reduce(function(b, p) {
    return b + Array(p - b.length + 2).join(0);
  });
  
  if (n > 0)
    str += '.' + Array(n + 1).join(0);
  
  return str;
};

console.log( 1e21.toFixedSpecial(2) );       // "1000000000000000000000.00"
console.log( 2.1e24.toFixedSpecial(0) );     // "2100000000000000000000000"
console.log( 1234567..toFixedSpecial(1) );   // "1234567.0"
console.log( 1234567.89.toFixedSpecial(3) ); // "1234567.890"

Busting out the regular expressions. This has no precision issues and is not a lot of code.

function toPlainString(num) {
  return (''+ +num).replace(/(-?)(\d*)\.?(\d*)e([+-]\d+)/,
    function(a,b,c,d,e) {
      return e < 0
        ? b + '0.' + Array(1-e-c.length).join(0) + c + d
        : b + c + d + Array(e-d.length+1).join(0);
    });
}

console.log(toPlainString(12345e+12));
console.log(toPlainString(12345e+24));
console.log(toPlainString(-12345e+24));
console.log(toPlainString(12345e-12));
console.log(toPlainString(123e-12));
console.log(toPlainString(-123e-12));
console.log(toPlainString(-123.45e-56));
console.log(toPlainString('1e-8'));
console.log(toPlainString('1.0e-8'));

The question of the post was avoiding e notation numbers and having the number as a plain number.

Therefore, if all is needed is to convert e (scientific) notation numbers to plain numbers (including in the case of fractional numbers) without loss of accuracy, then it is essential to avoid the use of the Math object and other javascript number methods so that rounding does not occur when large numbers and large fractions are handled (which always happens due to the internal storage in binary format).

The following function converts e (scientific) notation numbers to plain numbers (including fractions) handling both large numbers and large fractions without loss of accuracy as it does not use the built-in math and number functions to handle or manipulate the number.

The function also handles normal numbers, so that a number that is suspected to become in an 'e' notation can be passed to the function for fixing.

The function should work with different locale decimal points.

94 test cases are provided.

For large e-notation numbers pass the number as a string.

Examples:

eToNumber("123456789123456789.111122223333444455556666777788889999e+50");
// output:
"12345678912345678911112222333344445555666677778888999900000000000000"

eToNumber("123.456123456789123456895e-80");
// output:
"0.00000000000000000000000000000000000000000000000000000000000000000000000000000123456123456789123456895"

eToNumber("123456789123456789.111122223333444455556666777788889999e-50");
// output:
"0.00000000000000000000000000000000123456789123456789111122223333444455556666777788889999"

Valid e-notation numbers in Javascript include the following:

123e1   ==> 1230
123E1   ==> 1230
123e+1  ==> 1230
123.e+1 ==> 1230
123e-1  ==> 12.3
0.1e-1  ==> 0.01
.1e-1   ==> 0.01
-123e1  ==> -1230

/******************************************************************
 * Converts e-Notation Numbers to Plain Numbers
 ******************************************************************
 * @function eToNumber(number)
 * @version  1.00
 * @param   {e nottation Number} valid Number in exponent format.
 *          pass number as a string for very large 'e' numbers or with large fractions
 *          (none 'e' number returned as is).
 * @return  {string}  a decimal number string.
 * @author  Mohsen Alyafei
 * @date    17 Jan 2020
 * Note: No check is made for NaN or undefined input numbers.
 *
 *****************************************************************/
function eToNumber(num) {
  let sign = "";
  (num += "").charAt(0) == "-" && (num = num.substring(1), sign = "-");
  let arr = num.split(/[e]/ig);
  if (arr.length < 2) return sign + num;
  let dot = (.1).toLocaleString().substr(1, 1), n = arr[0], exp = +arr[1],
      w = (n = n.replace(/^0+/, '')).replace(dot, ''),
    pos = n.split(dot)[1] ? n.indexOf(dot) + exp : w.length + exp,
    L   = pos - w.length, s = "" + BigInt(w);
    w   = exp >= 0 ? (L >= 0 ? s + "0".repeat(L) : r()) : (pos <= 0 ? "0" + dot + "0".repeat(Math.abs(pos)) + s : r());
  L= w.split(dot); if (L[0]==0 && L[1]==0 || (+w==0 && +s==0) ) w = 0; //** added 9/10/2021
  return sign + w;
  function r() {return w.replace(new RegExp(`^(.{${pos}})(.)`), `$1${dot}$2`)}
}
//*****************************************************************

//================================================
//             Test Cases
//================================================
let r = 0; // test tracker
r |= test(1, "123456789123456789.111122223333444455556666777788889999e+50", "12345678912345678911112222333344445555666677778888999900000000000000");
r |= test(2, "123456789123456789.111122223333444455556666777788889999e-50", "0.00000000000000000000000000000000123456789123456789111122223333444455556666777788889999");
r |= test(3, "123456789e3", "123456789000");
r |= test(4, "123456789e1", "1234567890");
r |= test(5, "1.123e3", "1123");
r |= test(6, "12.123e3", "12123");
r |= test(7, "1.1234e1", "11.234");
r |= test(8, "1.1234e4", "11234");
r |= test(9, "1.1234e5", "112340");
r |= test(10, "123e+0", "123");
r |= test(11, "123E0", "123");
// //============================
r |= test(12, "123e-1", "12.3");
r |= test(13, "123e-2", "1.23");
r |= test(14, "123e-3", "0.123");
r |= test(15, "123e-4", "0.0123");
r |= test(16, "123e-2", "1.23");
r |= test(17, "12345.678e-1", "1234.5678");
r |= test(18, "12345.678e-5", "0.12345678");
r |= test(19, "12345.678e-6", "0.012345678");
r |= test(20, "123.4e-2", "1.234");
r |= test(21, "123.4e-3", "0.1234");
r |= test(22, "123.4e-4", "0.01234");
r |= test(23, "-123e+0", "-123");
r |= test(24, "123e1", "1230");
r |= test(25, "123e3", "123000");
r |= test(26, -1e33, "-1000000000000000000000000000000000");
r |= test(27, "123e+3", "123000");
r |= test(28, "123E+7", "1230000000");
r |= test(29, "-123.456e+1", "-1234.56");
r |= test(30, "-1.0e+1", "-10");
r |= test(31, "-1.e+1", "-10");
r |= test(32, "-1e+1", "-10");
r |= test(34, "-0", "-0");
r |= test(37, "0e0", "0");
r |= test(38, "123.456e+4", "1234560");
r |= test(39, "123E-0", "123");
r |= test(40, "123.456e+50", "12345600000000000000000000000000000000000000000000000");

r |= test(41, "123e-0", "123");
r |= test(42, "123e-1", "12.3");
r |= test(43, "123e-3", "0.123");
r |= test(44, "123.456E-1", "12.3456");
r |= test(45, "123.456123456789123456895e-80", "0.00000000000000000000000000000000000000000000000000000000000000000000000000000123456123456789123456895");
r |= test(46, "-123.456e-50", "-0.00000000000000000000000000000000000000000000000123456");

r |= test(47, "-0e+1", "-0");
r |= test(48, "0e+1", "0");
r |= test(49, "0.1e+1", "1");
r |= test(50, "-0.01e+1", "-0.1");
r |= test(51, "0.01e+1", "0.1");
r |= test(52, "-123e-7", "-0.0000123");
r |= test(53, "123.456e-4", "0.0123456");

r |= test(54, "1.e-5", "0.00001"); // handle missing base fractional part
r |= test(55, ".123e3", "123"); // handle missing base whole part

// The Electron's Mass:
r |= test(56, "9.10938356e-31", "0.000000000000000000000000000000910938356");
// The Earth's Mass:
r |= test(57, "5.9724e+24", "5972400000000000000000000");
// Planck constant:
r |= test(58, "6.62607015e-34", "0.000000000000000000000000000000000662607015");

r |= test(59, "0.000e3", "0");
r |= test(60, "0.000000000000000e3", "0");
r |= test(61, "-0.0001e+9", "-100000");
r |= test(62, "-0.0e1", "-0");
r |= test(63, "-0.0000e1", "-0");


r |= test(64, "1.2000e0", "1.2000");
r |= test(65, "1.2000e-0", "1.2000");
r |= test(66, "1.2000e+0", "1.2000");
r |= test(67, "1.2000e+10", "12000000000");
r |= test(68, "1.12356789445566771234e2", "112.356789445566771234");

// ------------- testing for Non e-Notation Numbers -------------
r |= test(69, "12345.7898", "12345.7898") // no exponent
r |= test(70, 12345.7898, "12345.7898") // no exponent
r |= test(71, 0.00000000000001, "0.00000000000001") // from 1e-14
r |= test(72, 0.0000000000001, "0.0000000000001") // from 1e-13
r |= test(73, 0.000000000001, "0.000000000001") // from 1e-12
r |= test(74, 0.00000000001, "0.00000000001") // from 1e-11
r |= test(75, 0.0000000001, "0.0000000001") // from 1e-10
r |= test(76, 0.000000001, "0.000000001") // from 1e-9
r |= test(77, 0.00000001, "0.00000001") // from 1e-8
r |= test(78, 0.0000001, "0.0000001") // from 1e-7
r |= test(79, 1e-7, "0.0000001") // from 1e-7
r |= test(80, -0.0000001, "-0.0000001") // from 1e-7
r |= test(81, 0.0000005, "0.0000005") // from 1e-7
r |= test(82, 0.1000005, "0.1000005") // from 1e-7
r |= test(83, 1e-6, "0.000001") // from 1e-6

r |= test(84, 0.000001, "0.000001"); // from 1e-6
r |= test(85, 0.00001, "0.00001"); // from 1e-5
r |= test(86, 0.0001, "0.0001"); // from 1e-4
r |= test(87, 0.001, "0.001"); // from 1e-3
r |= test(88, 0.01, "0.01"); // from 1e-2
r |= test(89, 0.1, "0.1") // from 1e-1
r |= test(90, -0.0000000000000345, "-0.0000000000000345"); // from -3.45e-14
r |= test(91, -0, "0");
r |= test(92, "-0", "-0");
r |= test(93,2e64,"20000000000000000000000000000000000000000000000000000000000000000");
r |= test(94,"2830869077153280552556547081187254342445169156730","2830869077153280552556547081187254342445169156730");

if (r == 0) console.log("All 94 tests passed.");

//================================================
//             Test function
//================================================
function test(testNumber, n1, should) {
  let result = eToNumber(n1);
  if (result !== should) {
    console.log(`Test ${testNumber} Failed. Output: ${result}\n             Should be: ${should}`);
    return 1;
  }
}

The following solution bypasses the automatic exponentional formatting for very big and very small numbers. This is outis's solution with a bugfix: It was not working for very small negative numbers.

function numberToString(num)
{
    let numStr = String(num);

    if (Math.abs(num) < 1.0)
    {
        let e = parseInt(num.toString().split('e-')[1]);
        if (e)
        {
            let negative = num < 0;
            if (negative) num *= -1
            num *= Math.pow(10, e - 1);
            numStr = '0.' + (new Array(e)).join('0') + num.toString().substring(2);
            if (negative) numStr = "-" + numStr;
        }
    }
    else
    {
        let e = parseInt(num.toString().split('+')[1]);
        if (e > 20)
        {
            e -= 20;
            num /= Math.pow(10, e);
            numStr = num.toString() + (new Array(e + 1)).join('0');
        }
    }

    return numStr;
}

// testing ...
console.log(numberToString(+0.0000000000000000001));
console.log(numberToString(-0.0000000000000000001));
console.log(numberToString(+314564649798762418795));
console.log(numberToString(-314564649798762418795));

The answers of others do not give you the exact number!
This function calculates the desired number accurately and returns it in the string to prevent it from being changed by javascript!
If you need a numerical result, just multiply the result of the function in number one!

function toNonExponential(value) {
    // if value is not a number try to convert it to number
    if (typeof value !== "number") {
        value = parseFloat(value);

        // after convert, if value is not a number return empty string
        if (isNaN(value)) {
            return "";
        }
    }

    var sign;
    var e;

    // if value is negative, save "-" in sign variable and calculate the absolute value
    if (value < 0) {
        sign = "-";
        value = Math.abs(value);
    }
    else {
        sign = "";
    }

    // if value is between 0 and 1
    if (value < 1.0) {
        // get e value
        e = parseInt(value.toString().split('e-')[1]);

        // if value is exponential convert it to non exponential
        if (e) {
            value *= Math.pow(10, e - 1);
            value = '0.' + (new Array(e)).join('0') + value.toString().substring(2);
        }
    }
    else {
        // get e value
        e = parseInt(value.toString().split('e+')[1]);

        // if value is exponential convert it to non exponential
        if (e) {
            value /= Math.pow(10, e);
            value += (new Array(e + 1)).join('0');
        }
    }

    // if value has negative sign, add to it
    return sign + value;
}

You can use from-exponential module. It is lightweight and fully tested.

import fromExponential from 'from-exponential';

fromExponential(1.123e-10); // => '0.0000000001123'

Use .toPrecision, .toFixed, etc. You can count the number of digits in your number by converting it to a string with .toString then looking at its .length.

You can loop over the number and achieve the rounding

> // functionality to replace char at given index

String.prototype.replaceAt=function(index, character) {
    return this.substr(0, index) + character + this.substr(index+character.length);
}

> // looping over the number starts

var str = "123456789123456799.55";
var arr = str.split('.');
str = arr[0];
i = (str.length-1);
if(arr[1].length && Math.round(arr[1]/100)){
  while(i>0){
    var intVal = parseInt(str.charAt(i));
    
   if(intVal == 9){
      str = str.replaceAt(i,'0');
      console.log(1,str)
   }else{
      str = str.replaceAt(i,(intVal+1).toString()); 
      console.log(2,i,(intVal+1).toString(),str)
      break;
   }
   i--;
 }
}

This is what I ended up using to take the value from an input, expanding numbers less than 17digits and converting Exponential numbers to x10^y

// e.g.
//  niceNumber("1.24e+4")   becomes 
// 1.24x10 to the power of 4 [displayed in Superscript]

function niceNumber(num) {
  try{
        var sOut = num.toString();
      if ( sOut.length >=17 || sOut.indexOf("e") > 0){
      sOut=parseFloat(num).toPrecision(5)+"";
      sOut = sOut.replace("e","x10<sup>")+"</sup>";
      }
      return sOut;

  }
  catch ( e) {
      return num;
  }
}

Your question:

number :0x68656c6c6f206f72656f
display:4.9299704811152646e+23

You can use this: https://github.com/MikeMcl/bignumber.js

> A JavaScript library for arbitrary-precision decimal and non-decimal arithmetic.

like this:

let ten =new BigNumber('0x68656c6c6f206f72656f',16);
console.log(ten.toString(10));
display:492997048111526447310191

I think there may be several similar answers, but here's a thing I came up with

// If you're gonna tell me not to use 'with' I understand, just,
// it has no other purpose, ;( andthe code actually looks neater
// 'with' it but I will edit the answer if anyone insists
var commas = false;

function digit(number1, index1, base1) {
    with (Math) {
        return floor(number1/pow(base1, index1))%base1;
    }
}

function digits(number1, base1) {
    with (Math) {
        o = "";
        l = floor(log10(number1)/log10(base1));
        for (var index1 = 0; index1 < l+1; index1++) {
            o = digit(number1, index1, base1) + o;
            if (commas && i%3==2 && i<l) {
                o = "," + o;
            }
        }
        return o;
    }
}

// Test - this is the limit of accurate digits I think
console.log(1234567890123450);

Note: this is only as accurate as the javascript math functions and has problems when using log instead of log10 on the line before the for loop; it will write 1000 in base-10 as 000 so I changed it to log10 because people will mostly be using base-10 anyways.

This may not be a very accurate solution but I'm proud to say it can successfully translate numbers across bases and comes with an option for commas!

I know it's many years later, but I had been working on a similar issue recently and I wanted to post my solution. The currently accepted answer pads out the exponent part with 0's, and mine attempts to find the exact answer, although in general it isn't perfectly accurate for very large numbers because of JS's limit in floating point precision.

This does work for Math.pow(2, 100), returning the correct value of 1267650600228229401496703205376.

function toFixed(x) {
  var result = '';
  var xStr = x.toString(10);
  var digitCount = xStr.indexOf('e') === -1 ? xStr.length : (parseInt(xStr.substr(xStr.indexOf('e') + 1)) + 1);
  
  for (var i = 1; i <= digitCount; i++) {
    var mod = (x % Math.pow(10, i)).toString(10);
    var exponent = (mod.indexOf('e') === -1) ? 0 : parseInt(mod.substr(mod.indexOf('e')+1));
    if ((exponent === 0 && mod.length !== i) || (exponent > 0 && exponent !== i-1)) {
      result = '0' + result;
    }
    else {
      result = mod.charAt(0) + result;
    }
  }
  return result;
}

console.log(toFixed(Math.pow(2,100))); // 1267650600228229401496703205376

If you are just doing it for display, you can build an array from the digits before they're rounded.

var num = Math.pow(2, 100);
var reconstruct = [];
while(num > 0) {
    reconstruct.unshift(num % 10);
    num = Math.floor(num / 10);
}
console.log(reconstruct.join(''));

I tried working with the string form rather than the number and this seemed to work. I have only tested this on Chrome but it should be universal:

function removeExponent(s) {
	var ie = s.indexOf('e');
	if (ie != -1) {
		if (s.charAt(ie + 1) == '-') {
			// negative exponent, prepend with .0s
			var n = s.substr(ie + 2).match(/[0-9]+/);
			s = s.substr(2, ie - 2); // remove the leading '0.' and exponent chars
			for (var i = 0; i < n; i++) {
				s = '0' + s;
			}
			s = '.' + s;
		} else {
			// positive exponent, postpend with 0s
			var n = s.substr(ie + 1).match(/[0-9]+/);
			s = s.substr(0, ie); // strip off exponent chars			
			for (var i = 0; i < n; i++) {
				s += '0';
			}		
		}
	}
	return s;
}

Currently there is no native function to dissolve scientific notation. However, for this purpose you must write your own functionality.

Here is my:

function dissolveExponentialNotation(number)
{
    if(!Number.isFinite(number)) { return undefined; }
    
    let text = number.toString();
    let items = text.split('e');
    
    if(items.length == 1) { return text; }
    
    let significandText = items[0];
    let exponent = parseInt(items[1]);
    
    let characters = Array.from(significandText);
    let minus = characters[0] == '-';
    if(minus) { characters.splice(0, 1); }
    let indexDot = characters.reduce((accumulator, character, index) =>
    {
        if(!accumulator.found) { if(character == '.') { accumulator.found = true; } else { accumulator.index++; } }
        return accumulator;
    }, { index: 0, found: false }).index;
    
    characters.splice(indexDot, 1);
    
    indexDot += exponent;
    
    if(indexDot >= 0 && indexDot < characters.length - 1)
    {
        characters.splice(indexDot, 0, '.');
    }
    else if(indexDot < 0)
    {
        characters.unshift("0.", "0".repeat(-indexDot));
    }
    else
    {
        characters.push("0".repeat(indexDot - characters.length));
    }
    
    return (minus ? "-" : "") + characters.join("");
}

If you don't mind using Lodash, it has toSafeInteger()

_.toSafeInteger(3.2);
// => 3
 
_.toSafeInteger(Number.MIN_VALUE);
// => 0
 
_.toSafeInteger(Infinity);
// => 9007199254740991
 
_.toSafeInteger('3.2');
// => 3

You can also use YourJS.fullNumber. For instance YourJS.fullNumber(Number.MAX_VALUE) results in the following: 179769313486231570000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

It also works for really small numbers. YourJS.fullNumber(Number.MIN_VALUE) returns this: 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005

It is important to note that this function will always return finite numbers as strings but will return non-finite numbers (eg. NaN or Infinity) as undefined.

You can test it out in the YourJS Console here.

If you want to convert scientific notation to integer :

parseInt("5.645656456454545e+23", 10)

result: 5

function printInt(n) { return n.toPrecision(100).replace(/\..*/,""); }

with some issues:

• 0.9 is displayed as "0"
• -0.9 is displayed as "-0"
• 1e100 is displayed as "1"
• works only for numbers up to ~1e99 => use other constant for greater numbers; or smaller for optimization.

You can use number.toString(10.1):

console.log(Number.MAX_VALUE.toString(10.1));

Note: This currently works in Chrome, but not in Firefox. The specification says the radix has to be an integer, so this results in unreliable behavior.

I had the same issue with oracle returning scientic notation, but I needed the actual number for a url. I just used a PHP trick by subtracting zero, and I get the correct number.

for example 5.4987E7 is the val.

newval = val - 0;

newval now equals 54987000