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factorial with trailing zeros, but without calculating factorial

I'm calculating the trailing zeros of a factorial. My solution is to calculate the factorial then determine how many trailing zeros it has. As you can imagine this isn't very scalable. How can I solve this without calculating the factorial?
I've found these pages on SO:
Trailing zeroes in a Factorial
Calculating the factorial without trailing zeros efficiently?
However, neither are in Javascript. If you downvote this question please let me know why. Thank-you for your time and feedback.
My solution:
function zeros(n) {
var result = [];
var count = 0;
for (var i = 1; i <= n; i++) {
result.push(i);
} //generating range for factorial function
var factorial = result.reduce(function(acc, el) {
return acc * el;
}, 1); //calculating factorial
factorial = factorial.toString().split('');
for (var j = factorial.length - 1; j > 0; j--) {
if (parseInt(factorial[j]) === 0) {
count += 1;
} else {
break;
}
} //counting trailing zeros
return count;
}

Knowing the number of trailing zeroes in a number comes down to knowing how many times it can be divided by 10, i.e. by both 5 and 2.
With factorial numbers that is quite easy to count:
f! = 1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16. ... .f
^ ^ ^
The places where a factor 5 gets into the final product are marked. It is clear that factors of 2 occur more often, so the count of factors of 5 are determining the number of trailing zeroes.
Now, when the factor 25 occurs, it should be counted for 2; likewise 125 should count for 3 factors of 5, etc.
You can cover for that with a loop like this:
function zeros(n) {
var result = 0;
while (n = Math.floor(n / 5)) result += n;
return result;
}

public static void main(String[] args) {
int n=23;
String fact= factorial(BigInteger.valueOf(23)).toString();
System.out.format("Factorial value of %d is %s\n", n,fact);
int len=fact.length();
//Check end with zeros
if(fact.matches(".*0*$")){
String[] su=fact.split("0*$");
//Split the pattern from whole string
System.out.println(Arrays.toString(fact.split("0*$")));
//Subtract from the total length
System.out.println("Count of trailing zeros "+(len-su[0].length()));
}
}
public static BigInteger factorial(BigInteger n) {
if (n.equals(BigInteger.ONE) || n.equals(BigInteger.ZERO)) {
return BigInteger.ONE;
}
return n.multiply(factorial(n.subtract(BigInteger.ONE)));
}

You don't really need to calculate the factorial product to count the trailing zeroes.
Here a sample to count the number of trailing zeroes in n!
temp = 5;
zeroes = 0;
//counting the sum of multiples of 5,5^2,5^3....present in n!
while(n>=temp){
fives = n/temp;
zeroes = zeroes + fives;
temp = temp*5;
}
printf("%d",zeroes);
Note that each multiple of 5 in the factorial product will contribute 1 to the number of trailing zeros. On top of this, each multiple of 25 will contribute an additional 1 to the number of trailing zeros. Then, each multiple of 125 will contribute another 1 to the number of trailing zeros, and so on.
Here's a great link to understand the concept behind this: https://brilliant.org/wiki/trailing-number-of-zeros/

I came across this algorithm somewhere on here can not remember now, but it looks like this,
def zeros(n)
return 0 if n.zero?
k = (Math.log(n)/Math.log(5)).to_i
m = 5**k
n*(m-1)/(4*m)
end
This very effiecient as it does not need a loop.
You can further optimize it to look like this.
def zeros(n)
return 0 if n.zero?
n*(n-1)/(4*n)
end
A javascript translation of this will be.
function zeros(n) {
if (n == 0) return 0;
return n * (n-1)/(4*n);
}
Note that this algorithm is correct till about n >= 1000000000, in which case the return value has an error margin of +1, and this error margin increases by +1 every n * 10000.

Determine number of leading zeros in a floating point number

How can I calculate how many zeros come after the decimal point but before the first non-zero in a floating point number. Examples:
0 -> 0
1 -> 0
1.0 -> 0
1.1 -> 0
1.01 -> 1
1.00003456 ->4
Intuitively I assume there is a math function that provides this, or at least does the main part. But I can neither recall nor figure out which one.
I know it can be done by first converting the number to a string, as long as the number isn't in scientific notation, but I want a pure math solution.
In my case I don't need something that works for negative numbers if that's a complication.
I'd like to know what the general ways to do it are, irrespective of language.
But if there is a pretty standard math function for this, I would also like to know if JavaScript has this function.
As a sidenote, I wonder if this calculation is related to the method for determining how many digits are required for the decimal representation of an integer.

Let x be a non-whole number that can be written as n digits of the whole part, then the decimal point, then m zeroes, then the rest of the fractional part.
x = [a1a2...an] . [0102...0m][b1b2...bm]
This means that the fractional part of x is larger than or equal to 10–m, and smaller than 10–m+1.
In other words, the decimal logarithm of the fractional part of x is larger than or equal to –m, and smaller than –m+1.
Which, in turn, means that the whole part of the decimal logarithm of the fractional part of x equals –m.
function numZeroesAfterPoint(x) {
if (x % 1 == 0) {
return 0;
} else {
return -1 - Math.floor(Math.log10(x % 1));
}
}
console.log(numZeroesAfterPoint(0));
console.log(numZeroesAfterPoint(1));
console.log(numZeroesAfterPoint(1.0));
console.log(numZeroesAfterPoint(1.1));
console.log(numZeroesAfterPoint(1.01));
console.log(numZeroesAfterPoint(1.00003456));
As a sidenote, I wonder if this calculation is related to the method for determining how many digits are required for the decimal representation of an integer.
In the same manner, a positive integer x takes n decimal digits to represent it if and only if n - 1 <= log10(x) < n.
So the number of digits in the decimal representation of x is floor(log10(x)) + 1.
That said, I wouldn't recommend using this method of determining the number of digits in practice. log10 is not guaranteed to give the exact value of the logarithm (not even as exact as IEEE 754 permits), which may lead to incorrect results in some edge cases.

You can do it with a simple while loop:
function CountZeros(Num) {
var Dec = Num % 1;
var Counter = -1;
while ((Dec < 1) && (Dec > 0)) {
Dec = Dec * 10;
Counter++;
}
Counter = Math.max(0, Counter); // In case there were no numbers at all after the decimal point.
console.log("There is: " + Counter + " zeros");
}
Then just pass the number you want to check into the function:
CountZeros(1.0034);

My approach is using a while() loop that compares the .floor(n) value with the n.toFixed(x) value of it while incrementing x until the two are not equal:
console.log(getZeros(0)); //0
console.log(getZeros(1)); //0
console.log(getZeros(1.0)); //0
console.log(getZeros(1.1)); //0
console.log(getZeros(1.01)); //1
console.log(getZeros(1.00003456)); //4
function getZeros(num) {
var x = 0;
if(num % 1 === 0) return x;
while(Math.floor(num)==num.toFixed(x)) {x++;}
return(x-1);
}

You can do it with toFixed() method, but there is only one flaw in my code, you need to specify the length of the numbers that comes after the point . It is because of the way the method is used.
NOTE:
The max length for toFixed() method is 20, so don't enter more than 20 numbers after . as said in the docs
var num = 12.0003400;
var lengthAfterThePoint = 7;
var l = num.toFixed(lengthAfterThePoint);
var pointFound = false;
var totalZeros = 0;
for(var i = 0; i < l.length; i++){
if(pointFound == false){
if(l[i] == '.'){
pointFound = true;
}
}else{
if(l[i] != 0){
break;
}else{
totalZeros++;
}
}
}
console.log(totalZeros);
Extra Answer
This is my extra answer, in this function, the program counts all the zeros until the last non-zero. So it ignores all the zeros at the end.
var num = 12.034000005608000;
var lengthAfterThePoint = 15;
var l = num.toFixed(lengthAfterThePoint);
var pointFound = false;
var theArr = [];
for(var i = 0; i < l.length; i++){
if(pointFound == false){
if(l[i] == '.'){
pointFound = true;
}
}else{
theArr.push(l[i]);
}
}
var firstNumFound = false;
var totalZeros = 0;
for(var j = 0; j < theArr.length; j++){
if(firstNumFound == false){
if(theArr[j] != 0){
firstNumFound = true;
totalZeros = totalZeros + j;
}
}else{
if(theArr[j] == 0){
totalZeros++;
}
}
}
var totalZerosLeft = 0;
for (var k = theArr.length; k > 0; k--) {
if(theArr[k -1] == 0){
totalZerosLeft++;
}else{
break;
}
}
console.log(totalZeros - totalZerosLeft);

What is the fastest way to count the number of significant digits of a number?

What is the fastest way to count the number of significant digits of a number?
I have the following function, which works, but is quite slow due to string operations.
/**
* Count the number of significant digits of a number.
*
* For example:
* 2.34 returns 3
* 0.0034 returns 2
* 120.5e+3 returns 4
*
* #param {Number} value
* #return {Number} The number of significant digits
*/
function digits (value) {
return value
.toExponential()
.replace(/e[\+\-0-9]*$/, '') // remove exponential notation
.replace( /^0\.?0*|\./, '') // remove decimal point and leading zeros
.length
};
Is there a faster way?
Update: here a list of assertions to test correct functioning:
assert.equal(digits(0), 0);
assert.equal(digits(2), 1);
assert.equal(digits(1234), 4);
assert.equal(digits(2.34), 3);
assert.equal(digits(3000), 1);
assert.equal(digits(0.0034), 2);
assert.equal(digits(120.5e50), 4);
assert.equal(digits(1120.5e+50), 5);
assert.equal(digits(120.52e-50), 5);
assert.equal(digits(Math.PI), 16);
My own method failed for digits(0), I fixed that by adding a ? to the second regexp.

Here's a more mathematical way of doing the same operation (which appears to be significantly faster)
JSPerf comparing the three implementations
Accurate for integer n < +-(2^53) per http://ecma262-5.com/ELS5_HTML.htm#Section_8.5
Floats are converted to a string and then coerced to an int (by removing the decimal so similar rules apply)
var log10 = Math.log(10);
function getSignificantDigitCount(n) {
n = Math.abs(String(n).replace(".", "")); //remove decimal and make positive
if (n == 0) return 0;
while (n != 0 && n % 10 == 0) n /= 10; //kill the 0s at the end of n
return Math.floor(Math.log(n) / log10) + 1; //get number of digits
}

Slight improvement of regular expression
function digits (value) {
return value
.toExponential()
.replace(/^([0-9]+)\.?([0-9]+)?e[\+\-0-9]*$/g, "$1$2")
.length
};

And yet another approach, that uses string operations and handles some special cases for better performance:
function digits(value) {
if (value === 0) {
return 0;
}
//create absolute value and
var t1 = ("" + Math.abs(value));
//remove decimal point
var t2 = t1.replace(".","");
//if number is represented by scientific notation,
//the places before "e" (minus "-" and ".") are the
//significant digits. So here we can just return the index
//"-234.3e+50" -> "2343e+50" -> indexOf("e") === 4
var i = t2.indexOf("e");
if (i > -1) {
return i;
}
//if the original number had a decimal point,
//trailing zeros are already removed, since irrelevant
//0.001230000.toString() -> "0.00123" -> "000123"
if (t2.length < t1.length) {
// -> remove only leading zeros
return t2.replace(/^0+/,'').length;
}
//if number did not contain decimal point,
//leading zeros are already removed
//000123000.toString() -> "123000"
// -> remove only trailing zeros
return t2.replace(/0+$/,'').length;
}

You can directly examine the bytes of a floating-point value by using typed arrays. The advantages of doing this are that it's fast, and it doesn't require any math to be done. You can look directly at the bits of the mantissa.
You can start with this:
var n = yourFloatingPointValue;
var f64 = new Float64Array(1);
var dv = new DataView(f64.buffer);
dv.setFloat64(0, n, false); // false -> big-endian
var bytes = [];
for (var i = 0; i < 8; i++)
bytes.push(dv.getUint8(i));
Now the bytes array contains integers representing the 8-bit values of the floating point value as it looks in memory. The first byte contains the sign bit in the top bit position, and the first 7 bits of the exponent in the rest. The second byte contains the 5 least-significant bits of the exponent and the first three bits of the mantissa. The rest of the bytes are all mantissa.

Regular string checking. A slight of improvement though.
function digits(value) {
value = "" + value;
var res = 0;
for (var i = 0, len = value.length; i < len; i++){
if (value[i]==="e")break;
if (+value[i]>=0)
res++;
}
return res;
};
jsperf Benchmark testing result as compared to the OP's and other answers code.
Update
function digits(value) {
console.log(value);
value = "" + (+value);
var res = 0;
for (var i = 0, len = value.length; i < len; i++) {
if (value[i] === "e")
{
break;
}
if (+value[i] >= 0)
{
res++;
}
}
console.log(value);
return res;
}
function check(val1, val2) {
console.log( val1+"==="+val2 +" = "+ (val1 === val2));
return val1 === val2;
}
check(digits(0), 1);
check(digits(2), 1);
check(digits(1234), 4);
check(digits("0012003400"), 8);
check(digits("0022.002200"), 6);
check(digits(2.34), 3);
check(digits(3000), 4);
check(digits(0.0034), 2);
check(digits(12003), 5);
check(digits(1.23e+50), 3);
check(digits("1.23e+50"), 3);
check(digits(120.5e51), 4);
check(digits(1120.5e+52), 5);
check(digits(120.52e-53), 5);
check(digits(Math.PI), 16);

There is a faster and indirect way to do it, which is converting it to a string and finding the length of it.
a = 2.303
sig_fig = len(str(a))-len(str(int(a)))-1
The extra -1 is for the "."

Fix Javascript Value to 0 decimals

I have this script which is working in formatting my currency but now its not fixing my value to 0 decimal places, Im new to javascript so could anybody explain where and hwo I'd do this? Thanks
function FormatNumberBy3(num, decpoint, sep) {
// check for missing parameters and use defaults if so
if (arguments.length == 2) {
sep = ",";
}
if (arguments.length == 1) {
sep = ",";
decpoint = ".";
}
// need a string for operations
num = num.toString();
// separate the whole number and the fraction if possible
a = num.split(decpoint);
x = a[0]; // decimal
y = a[1]; // fraction
z = "";
if (typeof(x) != "undefined") {
// reverse the digits. regexp works from left to right.
for (i=x.length-1;i>=0;i--)
z += x.charAt(i);
// add seperators. but undo the trailing one, if there
z = z.replace(/(\d{3})/g, "$1" + sep);
if (z.slice(-sep.length) == sep)
z = z.slice(0, -sep.length);
x = "";
// reverse again to get back the number
for (i=z.length-1;i>=0;i--)
x += z.charAt(i);
// add the fraction back in, if it was there
if (typeof(y) != "undefined" && y.length > 0)
x += decpoint + y;
}
return x;
}

I may be misunderstanding your question, but it sounds like you want Math.floor()
http://www.javascripter.net/faq/mathfunc.htm

You can round numbers mathematically, which is much, much faster and much, much, much simpler. There are literally hundreds of examples of this algorithm all over this site alone.
function round(number, places)
{
var multiplicator = Math.pow(10, places);
return Math.round(number * multiplicator) / multiplicator;
}
alert(round(123.456, 0)); // alerts "123"
This function lets you round to an arbitrary decimal place. If you want to round to the nearest integer, you may simply use Math.round(x).
Math.round rounds a decimal number to the nearest integer. Assuming you want to keep 3 decimal places, all you have to do is multiply the decimal number by 1000 (which is 10 power 3), round to the nearest integer, then divide by 1000. This is what this snippet does.

Your function appears to work for me... I assume the result of passing "1000" as an argument should be "1,000"? I'm not sure I understand what you mean by "fixing [your] value to 0 decimal places".
Here's a function I wrote to format numbers (like your above function does) for an API once:
function num_format(str) {
if (typeof str !== 'string' || /[^\d,\.e+-]/.test(str)) {
if (typeof str === 'number') {
str = str.toString();
} else {
throw new TypeError("Argument is not a string with a number in it");
}
}
var reg = /(\d+)(\d{3})/;
str = str.split(".");
while (reg.test(str[0])) {
str[0] = str[0].replace(reg, "$1,$2");
}
return str.join(".");
}
Remove the error checking code and it becomes quite a short little function (five lines). Pass it any number and it will format it correctly, although I didn't allow the option of specifying the separators.
As for rounding numbers to arbitrary decimal places, I suggest seeing zneak's answer above. Then would simply do:
num_format(round("100000.12341234", 2));
Which would give a result of "100,000.12".

This function should be capable of all various scenarios:
function formatNumber(num, precision, sep, thousand, addTrailing0) {
if (isNaN(num))
return "";
if (isNaN(precision))
precision = 2;
if (typeof addTrailing0 == "undefined")
addTrailing0 = true;
var factor = Math.pow(10, precision);
num = String(Math.round(num * factor) / factor);
if (addTrailing0 && precision > 0) {
if (num.indexOf(".") < 0)
num += ".";
for (var length = num.substr(num.indexOf(".")+1).length; length < precision; ++length)
num += "0";
}
var parts = num.split(".");
parts[0] = parts[0].split("").reverse().join("").replace(/(\d{3})(?=\d)/g, "$1" + (thousand || ",")).split("").reverse().join("");
num = parts[0] + (parts.length > 1 ? (sep || ".") + parts[1] : "");
//debug:
document.write(num + "<br />");
return num;
}
Examples:
formatNumber(32432342342.3574, 0); // 32,432,342,342
formatNumber(32432342342.3574, 2); // 32,432,342,342.36
formatNumber(1342.525423, 4, ".", ","); // 1,342.5254
formatNumber(1342.525423, 2, ",", "."); // 1.342,53
formatNumber(1342.525423); // 1,342.53
formatNumber(1342.525423, 0); // 1,343
formatNumber(342.525423, 8); // 342.52542300
formatNumber(42.525423, 8, null, null, false); // 42.525423
formatNumber(2.5, 3, ",", "."); // 2,500
Working Live Demo at http://jsfiddle.net/roberkules/xCqqh/

How to convert decimal to hexadecimal in JavaScript

How do you convert decimal values to their hexadecimal equivalent in JavaScript?

Convert a number to a hexadecimal string with:
hexString = yourNumber.toString(16);
And reverse the process with:
yourNumber = parseInt(hexString, 16);

If you need to handle things like bit fields or 32-bit colors, then you need to deal with signed numbers. The JavaScript function toString(16) will return a negative hexadecimal number which is usually not what you want. This function does some crazy addition to make it a positive number.
function decimalToHexString(number)
{
if (number < 0)
{
number = 0xFFFFFFFF + number + 1;
}
return number.toString(16).toUpperCase();
}
console.log(decimalToHexString(27));
console.log(decimalToHexString(48.6));

The code below will convert the decimal value d to hexadecimal. It also allows you to add padding to the hexadecimal result. So 0 will become 00 by default.
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}

function toHex(d) {
return ("0"+(Number(d).toString(16))).slice(-2).toUpperCase()
}

For completeness, if you want the two's-complement hexadecimal representation of a negative number, you can use the zero-fill-right shift >>> operator. For instance:
> (-1).toString(16)
"-1"
> ((-2)>>>0).toString(16)
"fffffffe"
There is however one limitation: JavaScript bitwise operators treat their operands as a sequence of 32 bits, that is, you get the 32-bits two's complement.

With padding:
function dec2hex(i) {
return (i+0x10000).toString(16).substr(-4).toUpperCase();
}

The accepted answer did not take into account single digit returned hexadecimal codes. This is easily adjusted by:
function numHex(s)
{
var a = s.toString(16);
if ((a.length % 2) > 0) {
a = "0" + a;
}
return a;
}
and
function strHex(s)
{
var a = "";
for (var i=0; i<s.length; i++) {
a = a + numHex(s.charCodeAt(i));
}
return a;
}
I believe the above answers have been posted numerous times by others in one form or another. I wrap these in a toHex() function like so:
function toHex(s)
{
var re = new RegExp(/^\s*(\+|-)?((\d+(\.\d+)?)|(\.\d+))\s*$/);
if (re.test(s)) {
return '#' + strHex( s.toString());
}
else {
return 'A' + strHex(s);
}
}
Note that the numeric regular expression came from 10+ Useful JavaScript Regular Expression Functions to improve your web applications efficiency.
Update: After testing this thing several times I found an error (double quotes in the RegExp), so I fixed that. HOWEVER! After quite a bit of testing and having read the post by almaz - I realized I could not get negative numbers to work.
Further - I did some reading up on this and since all JavaScript numbers are stored as 64 bit words no matter what - I tried modifying the numHex code to get the 64 bit word. But it turns out you can not do that. If you put "3.14159265" AS A NUMBER into a variable - all you will be able to get is the "3", because the fractional portion is only accessible by multiplying the number by ten(IE:10.0) repeatedly. Or to put that another way - the hexadecimal value of 0xF causes the floating point value to be translated into an integer before it is ANDed which removes everything behind the period. Rather than taking the value as a whole (i.e.: 3.14159265) and ANDing the floating point value against the 0xF value.
So the best thing to do, in this case, is to convert the 3.14159265 into a string and then just convert the string. Because of the above, it also makes it easy to convert negative numbers because the minus sign just becomes 0x26 on the front of the value.
So what I did was on determining that the variable contains a number - just convert it to a string and convert the string. This means to everyone that on the server side you will need to unhex the incoming string and then to determine the incoming information is numeric. You can do that easily by just adding a "#" to the front of numbers and "A" to the front of a character string coming back. See the toHex() function.
Have fun!
After another year and a lot of thinking, I decided that the "toHex" function (and I also have a "fromHex" function) really needed to be revamped. The whole question was "How can I do this more efficiently?" I decided that a to/from hexadecimal function should not care if something is a fractional part but at the same time it should ensure that fractional parts are included in the string.
So then the question became, "How do you know you are working with a hexadecimal string?". The answer is simple. Use the standard pre-string information that is already recognized around the world.
In other words - use "0x". So now my toHex function looks to see if that is already there and if it is - it just returns the string that was sent to it. Otherwise, it converts the string, number, whatever. Here is the revised toHex function:
/////////////////////////////////////////////////////////////////////////////
// toHex(). Convert an ASCII string to hexadecimal.
/////////////////////////////////////////////////////////////////////////////
toHex(s)
{
if (s.substr(0,2).toLowerCase() == "0x") {
return s;
}
var l = "0123456789ABCDEF";
var o = "";
if (typeof s != "string") {
s = s.toString();
}
for (var i=0; i<s.length; i++) {
var c = s.charCodeAt(i);
o = o + l.substr((c>>4),1) + l.substr((c & 0x0f),1);
}
return "0x" + o;
}
This is a very fast function that takes into account single digits, floating point numbers, and even checks to see if the person is sending a hex value over to be hexed again. It only uses four function calls and only two of those are in the loop. To un-hex the values you use:
/////////////////////////////////////////////////////////////////////////////
// fromHex(). Convert a hex string to ASCII text.
/////////////////////////////////////////////////////////////////////////////
fromHex(s)
{
var start = 0;
var o = "";
if (s.substr(0,2).toLowerCase() == "0x") {
start = 2;
}
if (typeof s != "string") {
s = s.toString();
}
for (var i=start; i<s.length; i+=2) {
var c = s.substr(i, 2);
o = o + String.fromCharCode(parseInt(c, 16));
}
return o;
}
Like the toHex() function, the fromHex() function first looks for the "0x" and then it translates the incoming information into a string if it isn't already a string. I don't know how it wouldn't be a string - but just in case - I check. The function then goes through, grabbing two characters and translating those in to ASCII characters. If you want it to translate Unicode, you will need to change the loop to going by four(4) characters at a time. But then you also need to ensure that the string is NOT divisible by four. If it is - then it is a standard hexadecimal string. (Remember the string has "0x" on the front of it.)
A simple test script to show that -3.14159265, when converted to a string, is still -3.14159265.
<?php
echo <<<EOD
<html>
<head><title>Test</title>
<script>
var a = -3.14159265;
alert( "A = " + a );
var b = a.toString();
alert( "B = " + b );
</script>
</head>
<body>
</body>
</html>
EOD;
?>
Because of how JavaScript works in respect to the toString() function, all of those problems can be eliminated which before were causing problems. Now all strings and numbers can be converted easily. Further, such things as objects will cause an error to be generated by JavaScript itself. I believe this is about as good as it gets. The only improvement left is for W3C to just include a toHex() and fromHex() function in JavaScript.

Without the loop:
function decimalToHex(d) {
var hex = Number(d).toString(16);
hex = "000000".substr(0, 6 - hex.length) + hex;
return hex;
}
// Or "#000000".substr(0, 7 - hex.length) + hex;
// Or whatever
// *Thanks to MSDN
Also isn't it better not to use loop tests that have to be evaluated?
For example, instead of:
for (var i = 0; i < hex.length; i++){}
have
for (var i = 0, var j = hex.length; i < j; i++){}

Combining some of these good ideas for an RGB-value-to-hexadecimal function (add the # elsewhere for HTML/CSS):
function rgb2hex(r,g,b) {
if (g !== undefined)
return Number(0x1000000 + r*0x10000 + g*0x100 + b).toString(16).substring(1);
else
return Number(0x1000000 + r[0]*0x10000 + r[1]*0x100 + r[2]).toString(16).substring(1);
}

Constrained/padded to a set number of characters:
function decimalToHex(decimal, chars) {
return (decimal + Math.pow(16, chars)).toString(16).slice(-chars).toUpperCase();
}

For anyone interested, here's a JSFiddle comparing most of the answers given to this question.
And here's the method I ended up going with:
function decToHex(dec) {
return (dec + Math.pow(16, 6)).toString(16).substr(-6)
}
Also, bear in mind that if you're looking to convert from decimal to hex for use in CSS as a color data type, you might instead prefer to extract the RGB values from the decimal and use rgb().
For example (JSFiddle):
let c = 4210330 // your color in decimal format
let rgb = [(c & 0xff0000) >> 16, (c & 0x00ff00) >> 8, (c & 0x0000ff)]
// Vanilla JS:
document.getElementById('some-element').style.color = 'rgb(' + rgb + ')'
// jQuery:
$('#some-element').css('color', 'rgb(' + rgb + ')')
This sets #some-element's CSS color property to rgb(64, 62, 154).

var number = 3200;
var hexString = number.toString(16);
The 16 is the radix and there are 16 values in a hexadecimal number :-)

function dec2hex(i)
{
var result = "0000";
if (i >= 0 && i <= 15) { result = "000" + i.toString(16); }
else if (i >= 16 && i <= 255) { result = "00" + i.toString(16); }
else if (i >= 256 && i <= 4095) { result = "0" + i.toString(16); }
else if (i >= 4096 && i <= 65535) { result = i.toString(16); }
return result
}

If you want to convert a number to a hexadecimal representation of an RGBA color value, I've found this to be the most useful combination of several tips from here:
function toHexString(n) {
if(n < 0) {
n = 0xFFFFFFFF + n + 1;
}
return "0x" + ("00000000" + n.toString(16).toUpperCase()).substr(-8);
}

AFAIK comment 57807 is wrong and should be something like:
var hex = Number(d).toString(16);
instead of
var hex = parseInt(d, 16);
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}

And if the number is negative?
Here is my version.
function hexdec (hex_string) {
hex_string=((hex_string.charAt(1)!='X' && hex_string.charAt(1)!='x')?hex_string='0X'+hex_string : hex_string);
hex_string=(hex_string.charAt(2)<8 ? hex_string =hex_string-0x00000000 : hex_string=hex_string-0xFFFFFFFF-1);
return parseInt(hex_string, 10);
}

As the accepted answer states, the easiest way to convert from decimal to hexadecimal is var hex = dec.toString(16). However, you may prefer to add a string conversion, as it ensures that string representations like "12".toString(16) work correctly.
// Avoids a hard-to-track-down bug by returning `c` instead of `12`
(+"12").toString(16);
To reverse the process you may also use the solution below, as it is even shorter.
var dec = +("0x" + hex);
It seems to be slower in Google Chrome and Firefox, but is significantly faster in Opera. See http://jsperf.com/hex-to-dec.

I'm doing conversion to hex string in a pretty large loop, so I tried several techniques in order to find the fastest one. My requirements were to have a fixed-length string as a result, and encode negative values properly (-1 => ff..f).
Simple .toString(16) didn't work for me since I needed negative values to be properly encoded. The following code is the quickest I've tested so far on 1-2 byte values (note that symbols defines the number of output symbols you want to get, that is for 4-byte integer it should be equal to 8):
var hex = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
function getHexRepresentation(num, symbols) {
var result = '';
while (symbols--) {
result = hex[num & 0xF] + result;
num >>= 4;
}
return result;
}
It performs faster than .toString(16) on 1-2 byte numbers and slower on larger numbers (when symbols >= 6), but still should outperform methods that encode negative values properly.

Converting hex color numbers to hex color strings:
A simple solution with toString and ES6 padStart for converting hex color numbers to hex color strings.
const string = `#${color.toString(16).padStart(6, '0')}`;
For example:
0x000000 will become #000000
0xFFFFFF will become #FFFFFF
Check this example in a fiddle here

How to convert decimal to hexadecimal in JavaScript
I wasn't able to find a brutally clean/simple decimal to hexadecimal conversion that didn't involve a mess of functions and arrays ... so I had to make this for myself.
function DecToHex(decimal) { // Data (decimal)
length = -1; // Base string length
string = ''; // Source 'string'
characters = [ '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' ]; // character array
do { // Grab each nibble in reverse order because JavaScript has no unsigned left shift
string += characters[decimal & 0xF]; // Mask byte, get that character
++length; // Increment to length of string
} while (decimal >>>= 4); // For next character shift right 4 bits, or break on 0
decimal += 'x'; // Convert that 0 into a hex prefix string -> '0x'
do
decimal += string[length];
while (length--); // Flip string forwards, with the prefixed '0x'
return (decimal); // return (hexadecimal);
}
/* Original: */
D = 3678; // Data (decimal)
C = 0xF; // Check
A = D; // Accumulate
B = -1; // Base string length
S = ''; // Source 'string'
H = '0x'; // Destination 'string'
do {
++B;
A& = C;
switch(A) {
case 0xA: A='A'
break;
case 0xB: A='B'
break;
case 0xC: A='C'
break;
case 0xD: A='D'
break;
case 0xE: A='E'
break;
case 0xF: A='F'
break;
A = (A);
}
S += A;
D >>>= 0x04;
A = D;
} while(D)
do
H += S[B];
while (B--)
S = B = A = C = D; // Zero out variables
alert(H); // H: holds hexadecimal equivalent

You can do something like this in ECMAScript 6:
const toHex = num => (num).toString(16).toUpperCase();

If you are looking for converting Large integers i.e. Numbers greater than Number.MAX_SAFE_INTEGER -- 9007199254740991, then you can use the following code
const hugeNumber = "9007199254740991873839" // Make sure its in String
const hexOfHugeNumber = BigInt(hugeNumber).toString(16);
console.log(hexOfHugeNumber)

To sum it all up;
function toHex(i, pad) {
if (typeof(pad) === 'undefined' || pad === null) {
pad = 2;
}
var strToParse = i.toString(16);
while (strToParse.length < pad) {
strToParse = "0" + strToParse;
}
var finalVal = parseInt(strToParse, 16);
if ( finalVal < 0 ) {
finalVal = 0xFFFFFFFF + finalVal + 1;
}
return finalVal;
}
However, if you don't need to convert it back to an integer at the end (i.e. for colors), then just making sure the values aren't negative should suffice.

I haven't found a clear answer, without checks if it is negative or positive, that uses two's complement (negative numbers included). For that, I show my solution to one byte:
((0xFF + number +1) & 0x0FF).toString(16);
You can use this instruction to any number bytes, only you add FF in respective places. For example, to two bytes:
((0xFFFF + number +1) & 0x0FFFF).toString(16);
If you want cast an array integer to string hexadecimal:
s = "";
for(var i = 0; i < arrayNumber.length; ++i) {
s += ((0xFF + arrayNumber[i] +1) & 0x0FF).toString(16);
}

In case you're looking to convert to a 'full' JavaScript or CSS representation, you can use something like:
numToHex = function(num) {
var r=((0xff0000&num)>>16).toString(16),
g=((0x00ff00&num)>>8).toString(16),
b=(0x0000ff&num).toString(16);
if (r.length==1) { r = '0'+r; }
if (g.length==1) { g = '0'+g; }
if (b.length==1) { b = '0'+b; }
return '0x'+r+g+b; // ('#' instead of'0x' for CSS)
};
var dec = 5974678;
console.log( numToHex(dec) ); // 0x5b2a96

This is based on Prestaul and Tod's solutions. However, this is a generalisation that accounts for varying size of a variable (e.g. Parsing signed value from a microcontroller serial log).
function decimalToPaddedHexString(number, bitsize)
{
let byteCount = Math.ceil(bitsize/8);
let maxBinValue = Math.pow(2, bitsize)-1;
/* In node.js this function fails for bitsize above 32bits */
if (bitsize > 32)
throw "number above maximum value";
/* Conversion to unsigned form based on */
if (number < 0)
number = maxBinValue + number + 1;
return "0x"+(number >>> 0).toString(16).toUpperCase().padStart(byteCount*2, '0');
}
Test script:
for (let n = 0 ; n < 64 ; n++ ) {
let s=decimalToPaddedHexString(-1, n);
console.log(`decimalToPaddedHexString(-1,${(n+"").padStart(2)}) = ${s.padStart(10)} = ${("0b"+parseInt(s).toString(2)).padStart(34)}`);
}
Test results:
decimalToPaddedHexString(-1, 0) = 0x0 = 0b0
decimalToPaddedHexString(-1, 1) = 0x01 = 0b1
decimalToPaddedHexString(-1, 2) = 0x03 = 0b11
decimalToPaddedHexString(-1, 3) = 0x07 = 0b111
decimalToPaddedHexString(-1, 4) = 0x0F = 0b1111
decimalToPaddedHexString(-1, 5) = 0x1F = 0b11111
decimalToPaddedHexString(-1, 6) = 0x3F = 0b111111
decimalToPaddedHexString(-1, 7) = 0x7F = 0b1111111
decimalToPaddedHexString(-1, 8) = 0xFF = 0b11111111
decimalToPaddedHexString(-1, 9) = 0x01FF = 0b111111111
decimalToPaddedHexString(-1,10) = 0x03FF = 0b1111111111
decimalToPaddedHexString(-1,11) = 0x07FF = 0b11111111111
decimalToPaddedHexString(-1,12) = 0x0FFF = 0b111111111111
decimalToPaddedHexString(-1,13) = 0x1FFF = 0b1111111111111
decimalToPaddedHexString(-1,14) = 0x3FFF = 0b11111111111111
decimalToPaddedHexString(-1,15) = 0x7FFF = 0b111111111111111
decimalToPaddedHexString(-1,16) = 0xFFFF = 0b1111111111111111
decimalToPaddedHexString(-1,17) = 0x01FFFF = 0b11111111111111111
decimalToPaddedHexString(-1,18) = 0x03FFFF = 0b111111111111111111
decimalToPaddedHexString(-1,19) = 0x07FFFF = 0b1111111111111111111
decimalToPaddedHexString(-1,20) = 0x0FFFFF = 0b11111111111111111111
decimalToPaddedHexString(-1,21) = 0x1FFFFF = 0b111111111111111111111
decimalToPaddedHexString(-1,22) = 0x3FFFFF = 0b1111111111111111111111
decimalToPaddedHexString(-1,23) = 0x7FFFFF = 0b11111111111111111111111
decimalToPaddedHexString(-1,24) = 0xFFFFFF = 0b111111111111111111111111
decimalToPaddedHexString(-1,25) = 0x01FFFFFF = 0b1111111111111111111111111
decimalToPaddedHexString(-1,26) = 0x03FFFFFF = 0b11111111111111111111111111
decimalToPaddedHexString(-1,27) = 0x07FFFFFF = 0b111111111111111111111111111
decimalToPaddedHexString(-1,28) = 0x0FFFFFFF = 0b1111111111111111111111111111
decimalToPaddedHexString(-1,29) = 0x1FFFFFFF = 0b11111111111111111111111111111
decimalToPaddedHexString(-1,30) = 0x3FFFFFFF = 0b111111111111111111111111111111
decimalToPaddedHexString(-1,31) = 0x7FFFFFFF = 0b1111111111111111111111111111111
decimalToPaddedHexString(-1,32) = 0xFFFFFFFF = 0b11111111111111111111111111111111
Thrown: 'number above maximum value'
Note: Not too sure why it fails above 32 bitsize

rgb(255, 255, 255) // returns FFFFFF
rgb(255, 255, 300) // returns FFFFFF
rgb(0,0,0) // returns 000000
rgb(148, 0, 211) // returns 9400D3
function rgb(...values){
return values.reduce((acc, cur) => {
let val = cur >= 255 ? 'ff' : cur <= 0 ? '00' : Number(cur).toString(16);
return acc + (val.length === 1 ? '0'+val : val);
}, '').toUpperCase();
}

Arbitrary precision
This solution take on input decimal string, and return hex string. A decimal fractions are supported. Algorithm
split number to sign (s), integer part (i) and fractional part (f) e.g for -123.75 we have s=true, i=123, f=75
integer part to hex:
if i='0' stop
get modulo: m=i%16 (in arbitrary precision)
convert m to hex digit and put to result string
for next step calc integer part i=i/16 (in arbitrary precision)
fractional part
count fractional digits n
multiply k=f*16 (in arbitrary precision)
split k to right part with n digits and put them to f, and left part with rest of digits and put them to d
convert d to hex and add to result.
finish when number of result fractional digits is enough
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function dec2HexArbitrary(decStr, fracDigits=0) {
// Helper: divide arbitrary precision number by js number
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function arbDivision(decStr, divisor)
{
// algorithm https://www.geeksforgeeks.org/divide-large-number-represented-string/
let ans='';
let idx = 0;
let temp = +decStr[idx];
while (temp < divisor) temp = temp * 10 + +decStr[++idx];
while (decStr.length > idx) {
ans += (temp / divisor)|0 ;
temp = (temp % divisor) * 10 + +decStr[++idx];
}
if (ans.length == 0) return "0";
return ans;
}
// Helper: calc module of arbitrary precision number
// #param decStr - string with non-negative integer
// #param mod - positive integer
function arbMod(decStr, mod) {
// algorithm https://www.geeksforgeeks.org/how-to-compute-mod-of-a-big-number/
let res = 0;
for (let i = 0; i < decStr.length; i++)
res = (res * 10 + +decStr[i]) % mod;
return res;
}
// Helper: multiply arbitrary precision integer by js number
// #param decStr - string with non-negative integer
// #param mult - positive integer
function arbMultiply(decStr, mult) {
let r='';
let m=0;
for (let i = decStr.length-1; i >=0 ; i--) {
let n = m+mult*(+decStr[i]);
r= (i ? n%10 : n) + r
m= n/10|0;
}
return r;
}
// dec2hex algorithm starts here
let h= '0123456789abcdef'; // hex 'alphabet'
let m= decStr.match(/-?(.*?)\.(.*)?/) || decStr.match(/-?(.*)/); // separate sign,integer,ractional
let i= m[1].replace(/^0+/,'').replace(/^$/,'0'); // integer part (without sign and leading zeros)
let f= (m[2]||'0').replace(/0+$/,'').replace(/^$/,'0'); // fractional part (without last zeros)
let s= decStr[0]=='-'; // sign
let r=''; // result
if(i=='0') r='0';
while(i!='0') { // integer part
r=h[arbMod(i,16)]+r;
i=arbDivision(i,16);
}
if(fracDigits) r+=".";
let n = f.length;
for(let j=0; j<fracDigits; j++) { // frac part
let k= arbMultiply(f,16);
f = k.slice(-n);
let d= k.slice(0,k.length-n);
r+= d.length ? h[+d] : '0';
}
return (s?'-':'')+r;
}
// -----------
// TESTS
// -----------
let tests = [
["0",2],
["000",2],
["123",0],
["-123",0],
["00.000",2],
["255.75",5],
["-255.75",5],
["127.999",32],
];
console.log('Input Standard Abitrary');
tests.forEach(t=> {
let nonArb = (+t[0]).toString(16).padEnd(17,' ');
let arb = dec2HexArbitrary(t[0],t[1]);
console.log(t[0].padEnd(10,' '), nonArb, arb);
});
// Long Example (40 digits after dot)
let example = "123456789012345678901234567890.09876543210987654321"
console.log(`\nLong Example:`);
console.log('dec:',example);
console.log('hex: ',dec2HexArbitrary(example,40));

The problem basically how many padding zeros to expect.
If you expect string 01 and 11 from Number 1 and 17. it's better to use Buffer as a bridge, with which number is turn into bytes, and then the hex is just an output format of it. And the bytes organization is well controlled by Buffer functions, like writeUInt32BE, writeInt16LE, etc.
import { Buffer } from 'buffer';
function toHex(n) { // 4byte
const buff = Buffer.alloc(4);
buff.writeInt32BE(n);
return buff.toString('hex');
}
> toHex(1)
'00000001'
> toHex(17)
'00000011'
> toHex(-1)
'ffffffff'
> toHex(-1212)
'fffffb44'
> toHex(1212)
'000004bc'

Here's my solution:
hex = function(number) {
return '0x' + Math.abs(number).toString(16);
}
The question says: "How to convert decimal to hexadecimal in JavaScript". While, the question does not specify that the hexadecimal string should begin with a 0x prefix, anybody who writes code should know that 0x is added to hexadecimal codes to distinguish hexadecimal codes from programmatic identifiers and other numbers (1234 could be hexadecimal, decimal, or even octal).
Therefore, to correctly answer this question, for the purpose of script-writing, you must add the 0x prefix.
The Math.abs(N) function converts negatives to positives, and as a bonus, it doesn't look like somebody ran it through a wood-chipper.
The answer I wanted, would have had a field-width specifier, so we could for example show 8/16/32/64-bit values the way you would see them listed in a hexadecimal editing application. That, is the actual, correct answer.