Develop Reference

Javascript equivalent of Java's UUID class

In java, you can do something like
UUID id = UUID.fromString("eb66c416-4739-465b-9af3-9dc33ed8eef9");
long msb = id.getMostSignificantBits();
long lsb = id.getLeastSignificantBits();
System.out.println(msb + ", " + lsb);
// -1484283427208739237, -7281302710629372167
System.out.println(new UUID(msb, lsb));
// eb66c416-4739-465b-9af3-9dc33ed8eef9
This same example is referenced in another question which is pretty similar, so this would be a follow up. While in the related question problem of lsb, msb -> string was solved, I cannot find solution for reverse problem, string -> msb, lsb
The original solution was
function toUuidString(lsb, msb) {
return `${digits(msb >> 32n, 8n)}-${digits(msb >> 16n, 4n)}-${digits(
msb,
4n
)}-${digits(lsb >> 48n, 4n)}-${digits(lsb, 12n)}`
}
function digits(value, ds) {
const hi = 1n << (ds * 4n)
return (hi | (value & (hi - 1n))).toString(16).slice(1)
}
Now I'd like to have a function that takes in string and returns msb and lsb.
Following original questions' paths, I've discovered java source code and tried to do the equivalent, which would be:
function fromString(name) {
let components = name.split('-')
if (components.length !== 5) {
throw new Error(`Invalid UUID string: ${name}`)
}
for (let index = 0; index < 5; index++) {
components[index] = `0x${components[index]}`
}
let mostSigBits = Number.parseInt(components[0], 16)
mostSigBits <<= 16
mostSigBits |= Number.parseInt(components[1], 16)
mostSigBits <<= 16
mostSigBits |= Number.parseInt(components[2], 16)
let leastSigBits = Number.parseInt(components[3], 16)
leastSigBits <<= 48
leastSigBits |= Number.parseInt(components[4], 16)
return {
leastSigBits,
mostSigBits,
}
}
However, when I try to test this with something like:
const originalUuid = 'eb66c416-4739-465b-9af3-9dc33ed8eef9'
const parts = fromString(originalUuid)
const newUUid = toUuidString(
BigInt(parts.leastSigBits),
BigInt(parts.mostSigBits)
)
console.log('Original', originalUuid)
console.log('New', newUUid)
I do not get equivalent uuids. They have equivalent parts but some parts are missing
Original eb66c416-4739-465b-9af3-9dc33ed8eef9
New 00000000-4739-465b-ffff-ffffbefbeef9
Any ideas what went wrong?

Finally I found the problem - two codes were not strictly equivalent, java source code declared mostSigBits and leastSigBits as long, which cannot be represented in javascript, so we need to use BigInt.
To sum up my question and previous question, javascript equivalent for java's UUID operations would be:
string -> msb, lsb
function fromString(name) {
let components = name.split('-')
if (components.length !== 5) {
throw new Error(`Invalid UUID string: ${name}`)
}
for (let index = 0; index < 5; index++) {
components[index] = `0x${components[index]}`
}
let mostSigBits = BigInt(Number.parseInt(components[0], 16))
mostSigBits <<= 16n
mostSigBits |= BigInt(Number.parseInt(components[1], 16))
mostSigBits <<= 16n
mostSigBits |= BigInt(Number.parseInt(components[2], 16))
let leastSigBits = BigInt(Number.parseInt(components[3], 16))
leastSigBits <<= 48n
leastSigBits |= BigInt(Number.parseInt(components[4], 16))
return {
leastSigBits,
mostSigBits,
}
}
msb, lsb -> string (from the referenced question)
function toUuidString(lsb, msb) {
return `${digits(msb >> 32n, 8n)}-${digits(msb >> 16n, 4n)}-${digits(
msb,
4n
)}-${digits(lsb >> 48n, 4n)}-${digits(lsb, 12n)}`
}
function digits(value, ds) {
const hi = 1n << (ds * 4n)
return (hi | (value & (hi - 1n))).toString(16).slice(1)
}

convert decimal number to fraction in javascript or closest fraction [duplicate]

This question already has answers here:
How to simplify a decimal into the smallest possible fraction?
(6 answers)
Closed last year.
The community reviewed whether to reopen this question last year and left it closed:
Original close reason(s) were not resolved
So i want to be able to convert any decimal number into fraction. In both forms such as one without remainder like this: 3/5 or with remainder: 3 1/4.
what i was doing is this..
lets say i have number .3435.
Calculate amount of digits after decimals.
multiply by 10 with power of the amount before number.
then somehow find greatest common factor.
Now i don't know how to find GCF. And nor i know how to implement logic to find fraction that represents a number closely or in remainder form if exact fraction doesn't exists.
code i have so far: (testing)
x = 34/35;
a = x - x.toFixed();
tens = (10).pow(a.toString().length - 2);
numerator = tens * x;
denominator = tens;

Your first 2 steps are reasonable.
But what you should do is for the numerator and denominator calculate the Greatest Common Divisor (GCD) and then divide the numerator and denominator with that divisor to get the fraction you want.
GCD is rather easy to calculate. Here is Euclid's algorithm:
var gcd = function(a, b) {
if (!b) return a;
return gcd(b, a % b);
};
Edit
I've added a fully working JSFiddle.

Unless you are willing to work on developing something yourself then I would suggest using a library that someone has already put effort into, like fraction.js
Javascript
var frac = new Fraction(0.3435);
console.log(frac.toString());
Output
687/2000
On jsFiddle

You can use brute force test on different denominators and retain the result that has least error.
The algorithm below is an example of how you might go about this, but, suffers from being inefficient and limited to searching for denominators up to 10000.
function find_rational( value, maxdenom ) {
console.clear();
console.log( "Looking up: " + value );
let best = { numerator: 1, denominator: 1, error: Math.abs(value - 1) }
if ( !maxdenom ) maxdenom = 10000;
for ( let denominator = 1; best.error > 0 && denominator <= maxdenom; denominator++ ) {
let numerator = Math.round( value * denominator );
let error = Math.abs( value - numerator / denominator );
if ( error >= best.error ) continue;
best.numerator = numerator;
best.denominator = denominator;
best.error = error;
console.log( "Intermediate result: "
+ best.numerator + "/" + best.denominator
+ " (" + ( best.numerator/best.denominator)
+ " error " + best.error + " )" );
}
console.log( "Final result: " + JSON.stringify( best ) );
return best;
}
function calc() {
const value = parseFloat( $("#myInput").val() );
if ( isNaN(value) ) {
$( "#myResult" ).val( "NaN" );
return;
}
const rational = find_rational( value, 10000 );
$("#myResult").val( rational.numerator
+ " / " + rational.denominator
+ " ( Error: " + rational.error + " )" );
}
calc();
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<P>
Enter a decimal number:<BR/>
<INPUT type="text" name="myInput" id="myInput" value=".3435" onkeyup="calc()"/><BR/>
</P>
<P>
Resulting Rational:<BR/>
<INPUT name="myResult" id="myResult" value=""/><BR/>
</P>
The above determines the .3435 as a fraction is 687 / 2000.
Also, had you gave it PI (e.g. 3.1415926) it produces good looking fractions like 22/7 and 355/113.

One quick and easy way of doing it is
getFraction = (decimal) => {
for(var denominator = 1; (decimal * denominator) % 1 !== 0; denominator++);
return {numerator: decimal * denominator, denominator: denominator};
}

I get very poor results using the GCD approach. I got much better results using an iterative approach.
For example, here is a very crude approach that zeros in on a fraction from a decimal:
function toFraction(x, tolerance) {
if (x == 0) return [0, 1];
if (x < 0) x = -x;
if (!tolerance) tolerance = 0.0001;
var num = 1, den = 1;
function iterate() {
var R = num/den;
if (Math.abs((R-x)/x) < tolerance) return;
if (R < x) num++;
else den++;
iterate();
}
iterate();
return [num, den];
}
The idea is you increment the numerator if you are below the value, and increment the denominator if you are above the value.

Use the Euclidean algorithm to find the greatest common divisor.
function reduce(numerator,denominator){
var gcd = function gcd(a,b){
return b ? gcd(b, a%b) : a;
};
gcd = gcd(numerator,denominator);
return [numerator/gcd, denominator/gcd];
}
This will provide you with the following results on your console
reduce(2,4);
// [1,2]
reduce(13427,3413358);
// [463,117702]
So by continuing from already what you have,
var x = 34/35;
var a = x - x.toFixed();
var tens = Math.pow(10,a.toString().length - 2);
var numerator = tens * x;
var denominator = tens;
reduce(numerator,denominator);
Source: https://stackoverflow.com/a/4652513/1998725

I had researched all over the website and I did combine all code into one, Here you go!
function fra_to_dec(num){
var test=(String(num).split('.')[1] || []).length;
var num=(num*(10**Number(test)))
var den=(10**Number(test))
function reduce(numerator,denominator){
var gcd = function gcd(a,b) {
return b ? gcd(b, a%b) : a;
};
gcd = gcd(numerator,denominator);
return [numerator/gcd, denominator/gcd];
}
return (reduce(num,den)[0]+"/"+reduce(num,den)[1])
}
This code is very easy to use! You can even put number in this function!

The tricky bit is not letting floating points get carried away.
Converting a number to a string restrains the trailing digits,
especially when you have a decimal with an integer, like 1.0625.
You can round off clumsy fractions, by passing a precision parameter.
Often you want to force a rounded value up, so a third parameter can specify that.
(e.g.; If you are using a precision of 1/64, the smallest return for a non-zero number will be 1/64, and not 0.)
Math.gcd= function(a, b){
if(b) return Math.gcd(b, a%b);
return Math.abs(a);
}
Math.fraction= function(n, prec, up){
var s= String(n),
p= s.indexOf('.');
if(p== -1) return s;
var i= Math.floor(n) || '',
dec= s.substring(p),
m= prec || Math.pow(10, dec.length-1),
num= up=== 1? Math.ceil(dec*m): Math.round(dec*m),
den= m,
g= Math.gcd(num, den);
if(den/g==1) return String(i+(num/g));
if(i) i= i+' and ';
return i+ String(num/g)+'/'+String(den/g);
}
Math.roundFraction(.3435,64); value: (String) 11/32

Inspired by #chowey answer, which contained recursive implementation of finding close fraction for a decimal value within given tolerance, here is better (see benchmark), iterative version of it.
function toFractionIterative(x, epsilon = 0.0001) {
if (x == 0) return [0, 1];
const a = Math.abs(x);
let n = 0;
let d = 1;
let r;
while (true) {
r = n / d;
if (Math.abs((r - a) / a) < epsilon) {
break;
}
if (r < a) {
n++;
}
else {
d++;
}
}
return [x < 0 ? -n : n, d];
}
Benchmark (tl;dr: recursive 1,589 ops/s, iterative 5,955 ops/s; use iterative approach)

let v = 3.141592;
document.write(d2f(v)); // 392699/125000
function d2f(v) // decimal to fraction
{
if (Math.floor(v) == v) return v + '/' + 1;
v = Math.abs(v);
let ret = .01, // rounding error tolerance
td = v-Math.floor(v), // trailing digits
r = 1/td, // reciprocal
d = r, // start building denominator
lim = 20; // max loop limit
for (let i = 0; i < lim; i++)
{
td = r-Math.floor(r);
if (Math.abs(r-Math.round(r)) < ret) break;
r = 1/td;
d *= r;
}
return Math.round(d*v) + '/' + Math.round(d);
}

I came up with this for 16ths
function getfract(theNum){
var input=theNum.toString();
var whole = input.split(".")[0];
var rem = input.split(".")[1] * .1;
return(whole + " " + Math.round(rem * 16) + "/16");
}

function decimalToFraction(num) {
let numsAfterDecPoint = num.toString().split('.')[1] ? num.toString().split('.')[1].length : 0;
let numerator = num * Math.pow(10, numsAfterDecPoint);
let denominator = Math.pow(10, numsAfterDecPoint);
console.log(numerator + " / " + denominator)
let d = GCD(numerator,denominator)
return numerator / d + " / " + denominator / d
}
console.log(decimalToFraction(0.5)); // 5 / 10 => 1 / 2
console.log(decimalToFraction(178.45)); // 17845 / 100 => 3569 / 20
function GCD(a,b) {
let r = 0;
while(b != 0) {
r = a % b
a = b;
b = r;
}
return a;
}

Change base of a number in JavaScript using a given digits array

I was required to make a method to convert integer from base ten to some another base in JavaScript, and it should also support providing your custom digits array. For example,
toBase(10, 2 ["A","B"])// returns 'BABA'
and if digits array is not provided, it should work as JavaScript 'toString' method
var a = 10;
a.toString(2);//returns '1010'
I have wrote a function to convert an integer to another base from base 10 number, with an option of providing digits array -
function toBase(number, radix, digits)
{
radix = radix || 10;
digits = digits || "0123456789abcdefghijklmnopqrstuvwxyz".split("").slice(0, radix)
if (radix > digits.length) {
var msg = "Not enough digits to represent the number '" + number + "' in base " + radix;
throw Error(msg);
}
if (number === 0) return digits[0];
var a = []
while (number) {
a.splice(0, 0, digits[number % radix])
number = parseInt(number / radix);
}
return a.join("");
}
This function works fine for me, but I want to know if is there any better way to do it? Thanks.

You can just use the native toString method and then replace the output with those from the digits array:
function toBase(number, radix, digits) {
if (digits && digits.length >= radix)
return number.toString(radix).replace(/./g, function(d) {
return digits[ parseInt(d, radix) ];
});
else
return number.toString(radix);
}

A method that might be slightly faster than the way you have is to bit shift. This works most easily when radix is a power of 2, here is an example
function toBase(x, radix, A) {
var r = 1, i = 0, s = '';
radix || (radix = 10); // case no radix
A || (A = '0123456789abcdefghijklmnopqrstuvwxyz'.split('')); // case no alphabet
if (A.length < radix) throw new RangeError('alphabet smaller than radix');
if (radix < 2) throw new RangeError('radix argument must be at least 2');
if (radix < 37) return useBergisMethod(x, radix, A); // this is arguably one of the fastest ways as it uses native `.toString`
if (x === 0) return A[0]; // short circuit 0
// test if radix is a power of 2
while (radix > r) {
r = r * 2;
i = i + 1;
}
if (r === radix) { // radix = 2 ^ i; fast method
r = r - 1; // Math.pow(2, i) - 1;
while (x > 0) {
s = A[x & r] + s;
x >>= i; // shift binary
}
return s; // done
}
return methodInOriginalQuestion(x, radix, A); // else not a power of 2, slower method
}
/*
toBase(74651278, 64, '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyzáé');
"4SnQE"
// check reverse
var i, j = 0, s = '4SnQE', a = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyzáé';
for (i = 0; i < s.length; ++i) j *= 64, j += a.indexOf(s[i]);
j; // 74651278, correct
*/

Converting hexadecimal to float in JavaScript

I would like to convert a number in base 10 with fraction to a number in base 16.
var myno = 28.5;
var convno = myno.toString(16);
alert(convno);
All is well there. Now I want to convert it back to decimal.
But now I cannot write:
var orgno = parseInt(convno, 16);
alert(orgno);
As it doesn't return the decimal part.
And I cannot use parseFloat, since per MDC, the syntax of parseFloat is
parseFloat(str);
It wouldn't have been a problem if I had to convert back to int, since parseInt's syntax is
parseInt(str [, radix]);
So what is an alternative for this?
Disclaimer: I thought it was a trivial question, but googling didn't give me any answers.
This question made me ask the above question.

Another possibility is to parse the digits separately, splitting the string up in two and treating both parts as ints during the conversion and then add them back together.
function parseFloat(str, radix)
{
var parts = str.split(".");
if ( parts.length > 1 )
{
return parseInt(parts[0], radix) + parseInt(parts[1], radix) / Math.pow(radix, parts[1].length);
}
return parseInt(parts[0], radix);
}
var myno = 28.4382;
var convno = myno.toString(16);
var f = parseFloat(convno, 16);
console.log(myno + " -> " + convno + " -> " + f);

Try this.
The string may be raw data (simple text) with four characters (0 - 255) or
a hex string "0xFFFFFFFF" four bytes in length.
jsfiddle.net
var str = '0x3F160008';
function parseFloat(str) {
var float = 0, sign, order, mantissa, exp,
int = 0, multi = 1;
if (/^0x/.exec(str)) {
int = parseInt(str, 16);
}
else {
for (var i = str.length -1; i >=0; i -= 1) {
if (str.charCodeAt(i) > 255) {
console.log('Wrong string parameter');
return false;
}
int += str.charCodeAt(i) * multi;
multi *= 256;
}
}
sign = (int >>> 31) ? -1 : 1;
exp = (int >>> 23 & 0xff) - 127;
mantissa = ((int & 0x7fffff) + 0x800000).toString(2);
for (i=0; i<mantissa.length; i+=1) {
float += parseInt(mantissa[i]) ? Math.pow(2, exp) : 0;
exp--;
}
return float*sign;
}

Please try this:
function hex2dec(hex) {
hex = hex.split(/\./);
var len = hex[1].length;
hex[1] = parseInt(hex[1], 16);
hex[1] *= Math.pow(16, -len);
return parseInt(hex[0], 16) + hex[1];
}
function hex2dec(hex) {
hex = hex.split(/\./);
var len = hex[1].length;
hex[1] = parseInt(hex[1], 16);
hex[1] *= Math.pow(16, -len);
return parseInt(hex[0], 16) + hex[1];
}
// ----------
// TEST
// ----------
function calc(hex) {
let dec = hex2dec(hex);
msg.innerHTML = `dec: <b>${dec}</b><br>hex test: <b>${dec.toString(16)}</b>`
}
let init="bad.a55";
inp.value=init;
calc(init);
<input oninput="calc(this.value)" id="inp" /><div id="msg"></div>

I combined Mark's and Kent's answers to make an overloaded parseFloat function that takes an argument for the radix (much simpler and more versatile):
function parseFloat(string, radix)
{
// Split the string at the decimal point
string = string.split(/\./);
// If there is nothing before the decimal point, make it 0
if (string[0] == '') {
string[0] = "0";
}
// If there was a decimal point & something after it
if (string.length > 1 && string[1] != '') {
var fractionLength = string[1].length;
string[1] = parseInt(string[1], radix);
string[1] *= Math.pow(radix, -fractionLength);
return parseInt(string[0], radix) + string[1];
}
// If there wasn't a decimal point or there was but nothing was after it
return parseInt(string[0], radix);
}

Try this:
Decide how many digits of precision you need after the decimal point.
Multiply your original number by that power of 16 (e.g. 256 if you want two digits).
Convert it as an integer.
Put the decimal point in manually according to what you decided in step 1.
Reverse the steps to convert back.
Take out the decimal point, remembering where it was.
Convert the hex to decimal in integer form.
Divide the result by the the appropriate power of 16 (16^n, where n is the number of digits after the decimal point you took out in step 1).
A simple example:
Convert decimal 23.5 into hex, and want one digit after the decimal point after conversion.
23.5 x 16 = 376.
Converted to hex = 0x178.
Answer in base 16: 17.8
Now convert back to decimal:
Take out the decimal point: 0x178
Convert to decimal: 376
Divide by 16: 23.5

I'm not sure what hexadecimal format you wanted to parse there. Was this something like: "a1.2c"?
Floats are commonly stored in hexadecimal format using the IEEE 754 standard. That standard doesn't use any dots (which don't exist in pure hexadecimal alphabet). Instead of that there are three groups of bits of predefined length (1 + 8 + 23 = 32 bits in total ─ double uses 64 bits).
I've written the following function for parsing such a numbers into float:
function hex2float(num) {
var sign = (num & 0x80000000) ? -1 : 1;
var exponent = ((num >> 23) & 0xff) - 127;
var mantissa = 1 + ((num & 0x7fffff) / 0x7fffff);
return sign * mantissa * Math.pow(2, exponent);
}

Here is a size-improvement of Mark Eirich's answer:
function hex2dec(hex) {
let h = hex.split(/\./);
return ('0x'+h[1])*(16**-h[1].length)+ +('0x'+h[0]);
}
function hex2dec(hex) {
let h = hex.split(/\./);
return ('0x'+h[1])*(16**-h[1].length)+ +('0x'+h[0]);
}
function calc(hex) {
let dec = hex2dec(hex);
msg.innerHTML = `dec: <b>${dec}</b><br>hex test: <b>${dec.toString(16)}</b>`
}
let init = "bad.a55";
inp.value = init;
calc(init);
<input oninput="calc(this.value)" id="inp" /><div id="msg"></div>

private hexStringToFloat(hexString: string): number {
return Buffer.from(hexString, 'hex').readFloatBE(0);
}

Someone might find this useful.
bytes to Float32
function Int2Float32(bytes) {
var sign = (bytes & 0x80000000) ? -1 : 1;
var exponent = ((bytes >> 23) & 0xFF) - 127;
var significand = (bytes & ~(-1 << 23));
if (exponent == 128)
return sign * ((significand) ? Number.NaN : Number.POSITIVE_INFINITY);
if (exponent == -127) {
if (significand === 0) return sign * 0.0;
exponent = -126;
significand /= (1 << 22);
} else significand = (significand | (1 << 23)) / (1 << 23);
return sign * significand * Math.pow(2, exponent);
}

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

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;
}
}

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"

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'

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

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 x10y
// 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;
}
}

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!

This didn't help me:
console.log( myNumb.toLocaleString('fullwide', {useGrouping:false}) );
but this:
value.toLocaleString("fullwide", {
useGrouping: false,
maximumSignificantDigits: 20,
})

Try this:
Number.standardizenumber = function (number,n) {
var mantissa = number.toLocaleString(
'en-US', {
useGrouping: false,
signDisplay: "never",
notation: "scientific",
minimumFractionDigits: 16,
maximumFractionDigits: 16
}
).toLowerCase().split('e')[0].replace(/\./g,'');
var exponentNegative = "0".repeat(Math.max(+Math.abs(number).toExponential().toLowerCase().split('e-')[1]-1,0)) + mantissa;
var exponentPositive = Math.abs(number)<1E17?mantissa.slice(0,+Math.abs(number).toExponential().toLowerCase().split('e+')[1]+1):mantissa+(Math.abs(number).toExponential().toLowerCase().split('e+')[1]-16);
var decimalExpPositive = Math.abs(number)<1E17?mantissa.slice(0,Math.abs(number).toExponential().toLowerCase().split('e+')[0]-16):undefined;
var fullDec = number===0?(1/number<0?'-0':'0'):(1/Math.sign(number)<0?'-':'')+(Math.abs(number)>=1?[exponentPositive,(number%1===0?(decimalExpPositive.slice(+Math.abs(number).toExponential().toLowerCase().split('e+')[1]+1)): undefined)].join('.'):`.${exponentNegative}`);
return isNaN(number)===false&&Math.abs(number)<1E17?((number%1===0?number.toLocaleString('en-US', {useGrouping: false}):fullDec).includes('.')===false?fullDec.split('.')[0].replace(/\B(?=(\d{3})+(?!\d))/g, ","):fullDec.replace(/(\.[0-9]*[1-9])0+$|\.0*$/,'$1').replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",")):number.toLocaleString('en-US');
}
Number.standardizenumber(.0000001) // .0000001
Number.standardizenumber(1E21) // 1,000,000,000,000,000,000,000
Number.standardizenumber(1_234_567_890.123456) // 1,234,567,890.123456

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.

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.

If you want to convert scientific notation to integer :
parseInt("5.645656456454545e+23", 10)
result: 5

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