Avoid exponential notation and string - javascript

I would like to convert 8.8e-7 to the number 0.00000088, and not a string version of "0.00000088".
So far I've tried:
number.toFixed(8) // it returns a string "0.00000088"
BigNumber(number).toNumber() // it returns a number 8.8e-7
I need that value in Number because I'm using that data in a graph, and it doesn't take Strings.
Any idea how to force convert it?
Thank you in advance!

If you want to convert a string containing a number in exponential notation to a number(javascript Number type) you can use the Number constructor.
var number = Number("8.8e-7")
Then you can use number for calculations or pass it to functions that take numbers.

I recently had a problem that I needed to avoid exponential notation.
const avoidNotation = (numIn: string) => {
if (!numIn) return '';
if (numIn === 'NaN' || numIn === 'undefined') return numIn;
numIn += ''; // To cater to numric entries
let sign = ''; // To remember the number sign
numIn.charAt(0) === '-' && ((numIn = numIn.substring(1)), (sign = '-')); // remove - sign & remember it
let str = numIn.split(/[eE]/g); // Split numberic string at e or E
if (str.length < 2) return sign + numIn; // Not an Exponent Number? Exit with orginal Num back
const power = Number(str[1]); // Get Exponent (Power) (could be + or -)
const deciSp = (1.1).toLocaleString().substring(1, 2); // Get Deciaml Separator
str = str[0].split(deciSp); // Split the Base Number into LH and RH at the decimal point
let baseRH = str[1] || '', // RH Base part. Make sure we have a RH fraction else ""
baseLH = str[0]; // LH base part.
if (power >= 0) {
// ------- Positive Exponents (Process the RH Base Part)
if (power > baseRH.length) baseRH += '0'.repeat(power - baseRH.length); // Pad with "0" at RH
baseRH = baseRH.slice(0, power) + deciSp + baseRH.slice(power); // Insert decSep at the correct place into RH base
if (baseRH.charAt(baseRH.length - 1) === deciSp)
baseRH = baseRH.slice(0, -1); // If decSep at RH end? => remove it
} else {
// ------- Negative exponents (Process the LH Base Part)
const num = Math.abs(power) - baseLH.length; // Delta necessary 0's
if (num > 0) baseLH = '0'.repeat(num) + baseLH; // Pad with "0" at LH
baseLH = baseLH.slice(0, power) + deciSp + baseLH.slice(power); // Insert "." at the correct place into LH base
if (baseLH.charAt(0) === deciSp) baseLH = '0' + baseLH; // If decSep at LH most? => add "0"
}
// Rremove leading and trailing 0's and Return the long number (with sign)
return sign + (baseLH + baseRH).replace(/^0*(\d+|\d+\.\d+?)\.?0*$/, '$1');
};
It worked for me.

Related

How to format number to have n whole and decimal digits?

Say I have a float x that has a whole part a and decimal part b i.e. x = a.b. How can I format x in JavaScript such that it'll have at most n digits, as shown below:
const n = 5
format(123.0) == '123'
format(123.01) == '123.01'
format(123.012) == '123.01'
format(0.0123) == '0.0123'
format(0.01234) == '0.0123'
Based on a recently deleted answer that had the logic but didn't really work, I came up with this solution:
const format = (number, digits = 5) => {
const str = number + ''; // convert to string
let [integer, decimal] = str.split('.'); // split by dot delimiter
const integerDigits = integer.length;
if (integerDigits >= digits) {
return integer;
} else {
let decimalDigits = digits - integerDigits;
decimal = Number('0.' + decimal.substring(0, decimalDigits)).toString();
return integer + '.' + decimal.split('.')[1];
}
}

Alternative way to padStart [duplicate]

I am in need of a JavaScript function which can take a value and pad it to a given length (I need spaces, but anything would do). I found this, but I have no idea what the heck it is doing and it doesn't seem to work for me.
String.prototype.pad = function(l, s, t) {
return s || (s = " "),
(l -= this.length) > 0 ?
(s = new Array(Math.ceil(l / s.length) + 1).join(s))
.substr(0, t = !t ? l : t == 1 ?
0 :
Math.ceil(l / 2)) + this + s.substr(0, l - t) :
this;
};
var s = "Jonas";
document.write(
'<h2>S = '.bold(), s, "</h2>",
'S.pad(20, "[]", 0) = '.bold(), s.pad(20, "[]", 0), "<br />",
'S.pad(20, "[====]", 1) = '.bold(), s.pad(20, "[====]", 1), "<br />",
'S.pad(20, "~", 2) = '.bold(), s.pad(20, "~", 2)
);
ECMAScript 2017 (ES8) added String.padStart (along with String.padEnd) for just this purpose:
"Jonas".padStart(10); // Default pad string is a space
"42".padStart(6, "0"); // Pad with "0"
"*".padStart(8, "-/|\\"); // produces '-/|\\-/|*'
If not present in the JavaScript host, String.padStart can be added as a polyfill.
Pre ES8
I found this solution here and this is for me much much simpler:
var n = 123
String("00000" + n).slice(-5); // returns 00123
("00000" + n).slice(-5); // returns 00123
(" " + n).slice(-5); // returns " 123" (with two spaces)
And here I made an extension to the string object:
String.prototype.paddingLeft = function (paddingValue) {
return String(paddingValue + this).slice(-paddingValue.length);
};
An example to use it:
function getFormattedTime(date) {
var hours = date.getHours();
var minutes = date.getMinutes();
hours = hours.toString().paddingLeft("00");
minutes = minutes.toString().paddingLeft("00");
return "{0}:{1}".format(hours, minutes);
};
String.prototype.format = function () {
var args = arguments;
return this.replace(/{(\d+)}/g, function (match, number) {
return typeof args[number] != 'undefined' ? args[number] : match;
});
};
This will return a time in the format "15:30".
A faster method
If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.
function pad(pad, str, padLeft) {
if (typeof str === 'undefined')
return pad;
if (padLeft) {
return (pad + str).slice(-pad.length);
} else {
return (str + pad).substring(0, pad.length);
}
}
For example, to zero pad a number to a length of 10 digits,
pad('0000000000',123,true);
To pad a string with whitespace, so the entire string is 255 characters,
var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);
Performance Test
See the jsPerf test here.
And this is faster than ES6 string.repeat by 2x as well, as shown by the revised JsPerf here
Please note that jsPerf is no longer online
Please note that the jsPerf site that we originally used to benchmark the various methods is no longer online. Unfortunately, this means we can't get to those test results. Sad but true.
String.prototype.padStart() and String.prototype.padEnd() are currently TC39 candidate proposals: see github.com/tc39/proposal-string-pad-start-end (only available in Firefox as of April 2016; a polyfill is available).
http://www.webtoolkit.info/javascript_pad.html
/**
*
* JavaScript string pad
* http://www.webtoolkit.info/
*
**/
var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;
function pad(str, len, pad, dir) {
if (typeof(len) == "undefined") { var len = 0; }
if (typeof(pad) == "undefined") { var pad = ' '; }
if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }
if (len + 1 >= str.length) {
switch (dir){
case STR_PAD_LEFT:
str = Array(len + 1 - str.length).join(pad) + str;
break;
case STR_PAD_BOTH:
var padlen = len - str.length;
var right = Math.ceil( padlen / 2 );
var left = padlen - right;
str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
break;
default:
str = str + Array(len + 1 - str.length).join(pad);
break;
} // switch
}
return str;
}
It's a lot more readable.
Here's a recursive approach to it.
function pad(width, string, padding) {
return (width <= string.length) ? string : pad(width, padding + string, padding)
}
An example...
pad(5, 'hi', '0')
=> "000hi"
ECMAScript 2017 adds a padStart method to the String prototype. This method will pad a string with spaces to a given length. This method also takes an optional string that will be used instead of spaces for padding.
'abc'.padStart(10); // " abc"
'abc'.padStart(10, "foo"); // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0"); // "00000abc"
'abc'.padStart(1); // "abc"
A padEnd method was also added that works in the same manner.
For browser compatibility (and a useful polyfill) see this link.
Using the ECMAScript 6 method String#repeat, a pad function is as simple as:
String.prototype.padLeft = function(char, length) {
return char.repeat(Math.max(0, length - this.length)) + this;
}
String#repeat is currently supported in Firefox and Chrome only. for other implementation, one might consider the following simple polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : (this + this.repeat(n-1));
}
Using the ECMAScript 6 method String#repeat and Arrow functions, a pad function is as simple as:
var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"
jsfiddle
edit:
suggestion from the comments:
const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;
this way, it wont throw an error when s.lengthis greater than n
edit2:
suggestion from the comments:
const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }
this way, you can use the function for strings and non-strings alike.
The key trick in both those solutions is to create an array instance with a given size (one more than the desired length), and then to immediately call the join() method to make a string. The join() method is passed the padding string (spaces probably). Since the array is empty, the empty cells will be rendered as empty strings during the process of joining the array into one result string, and only the padding will remain. It's a really nice technique.
With ES8, there are two options for padding.
You can check them in the documentation.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
Taking up Samuel's ideas, upward here. And remember an old SQL script, I tried with this:
a=1234;
'0000'.slice(a.toString().length)+a;
It works in all the cases I could imagine:
a= 1 result 0001
a= 12 result 0012
a= 123 result 0123
a= 1234 result 1234
a= 12345 result 12345
a= '12' result 0012
Pad with default values
I noticed that I mostly need the padLeft for time conversion / number padding.
So I wrote this function:
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + a).slice(-b)
}
This simple function supports Number or String as input.
The default pad is two characters.
The default char is 0.
So I can simply write:
padL(1);
// 01
If I add the second argument (pad width):
padL(1, 3);
// 001
The third parameter (pad character)
padL('zzz', 10, 'x');
// xxxxxxxzzz
#BananaAcid: If you pass a undefined value or a 0 length string, you get 0undefined, so:
As suggested
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array((b || 1) + 1).join(c || 0) + (a || '')).slice(-(b || 2))
}
But this can also be achieved in a shorter way.
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + (a || c || 0)).slice(-b)
}
It also works with:
padL(0)
padL(NaN)
padL('')
padL(undefined)
padL(false)
And if you want to be able to pad in both ways:
function pad(a, b, c, d) { // string/number, length=2, char=0, 0/false=Left-1/true=Right
return a = (a || c || 0), c = new Array(b || 2).join(c || 0), d ? (a + c).slice(0, b) : (c + a).slice(-b)
}
which can be written in a shorter way without using slice.
function pad(a, b, c, d) {
return a = (a || c || 0) + '', b = new Array((++b || 3) - a.length).join(c || 0), d ? a+b : b+a
}
/*
Usage:
pad(
input // (int or string) or undefined, NaN, false, empty string
// default:0 or PadCharacter
// Optional
,PadLength // (int) default:2
,PadCharacter // (string or int) default:'0'
,PadDirection // (bolean) default:0 (padLeft) - (true or 1) is padRight
)
*/
Now if you try to pad 'averylongword' with 2... that’s not my problem.
I said that I would give you a tip.
Most of the time, if you pad, you do it for the same value N times.
Using any type of function inside a loop slows down the loop!!!
So if you just want to pad left some numbers inside a long list, don't use functions to do this simple thing.
Use something like this:
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7],
paddedArray = [],
len = arrayOfNumbers.length;
while(len--) {
paddedArray[len] = ('0000' + arrayOfNumbers[len]).slice(-4);
}
If you don't know how the maximum padding size based on the numbers inside the array.
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7, 49095],
paddedArray = [],
len = arrayOfNumbers.length;
// Search the highest number
var arrayMax = Function.prototype.apply.bind(Math.max, null),
// Get that string length
padSize = (arrayMax(arrayOfNumbers) + '').length,
// Create a Padding string
padStr = new Array(padSize).join(0);
// And after you have all this static values cached start the loop.
while(len--) {
paddedArray[len] = (padStr + arrayOfNumbers[len]).slice(-padSize); // substr(-padSize)
}
console.log(paddedArray);
/*
0: "00001"
1: "00002"
2: "00003"
3: "00004"
4: "00005"
5: "00006"
6: "00007"
7: "49095"
*/
padding string has been inplemented in new javascript version.
str.padStart(targetLength [, padString])
https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart
If you want your own function check this example:
const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house
Here is a build in method you can use -
str1.padStart(2, '0')
Here's a simple function that I use.
var pad=function(num,field){
var n = '' + num;
var w = n.length;
var l = field.length;
var pad = w < l ? l-w : 0;
return field.substr(0,pad) + n;
};
For example:
pad (20,' '); // 20
pad (321,' '); // 321
pad (12345,' '); //12345
pad ( 15,'00000'); //00015
pad ( 999,'*****'); //**999
pad ('cat','_____'); //__cat
A short way:
(x=>(new Array(int-x.length+1)).join(char)+x)(String)
Example:
(x=>(new Array(6-x.length+1)).join("0")+x)("1234")
return: "001234"
Here is a simple answer in basically one line of code.
var value = 35 // the numerical value
var x = 5 // the minimum length of the string
var padded = ("00000" + value).substr(-x);
Make sure the number of characters in you padding, zeros here, is at least as many as your intended minimum length. So really, to put it into one line, to get a result of "00035" in this case is:
var padded = ("00000" + 35).substr(-5);
ES7 is just drafts and proposals right now, but if you wanted to track compatibility with the specification, your pad functions need:
Multi-character pad support.
Don't truncate the input string
Pad defaults to space
From my polyfill library, but apply your own due diligence for prototype extensions.
// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === ' hello'
'hello'.rpad(10) === 'hello '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'
String.prototype.lpad || (String.prototype.lpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str = pad + str;
}
return str.substr( -length );
});
String.prototype.rpad || (String.prototype.rpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str += pad;
}
return str.substr(0, length);
});
Array manipulations are really slow compared to simple string concat. Of course, benchmark for your use case.
function(string, length, pad_char, append) {
string = string.toString();
length = parseInt(length) || 1;
pad_char = pad_char || ' ';
while (string.length < length) {
string = append ? string+pad_char : pad_char+string;
}
return string;
};
A variant of #Daniel LaFavers' answer.
var mask = function (background, foreground) {
bg = (new String(background));
fg = (new String(foreground));
bgl = bg.length;
fgl = fg.length;
bgs = bg.substring(0, Math.max(0, bgl - fgl));
fgs = fg.substring(Math.max(0, fgl - bgl));
return bgs + fgs;
};
For example:
mask('00000', 11 ); // '00011'
mask('00011','00' ); // '00000'
mask( 2 , 3 ); // '3'
mask('0' ,'111'); // '1'
mask('fork' ,'***'); // 'f***'
mask('_____','dog'); // '__dog'
If you don't mind including a utility library, lodash library has _.pad, _.padLeft and _.padRight functions.
I think its better to avoid recursion because its costly.
function padLeft(str,size,padwith) {
if(size <= str.length) {
// not padding is required.
return str;
} else {
// 1- take array of size equal to number of padding char + 1. suppose if string is 55 and we want 00055 it means we have 3 padding char so array size should be 3 + 1 (+1 will explain below)
// 2- now join this array with provided padding char (padwith) or default one ('0'). so it will produce '000'
// 3- now append '000' with orginal string (str = 55), will produce 00055
// why +1 in size of array?
// it is a trick, that we are joining an array of empty element with '0' (in our case)
// if we want to join items with '0' then we should have at least 2 items in the array to get joined (array with single item doesn't need to get joined).
// <item>0<item>0<item>0<item> to get 3 zero we need 4 (3+1) items in array
return Array(size-str.length+1).join(padwith||'0')+str
}
}
alert(padLeft("59",5) + "\n" +
padLeft("659",5) + "\n" +
padLeft("5919",5) + "\n" +
padLeft("59879",5) + "\n" +
padLeft("5437899",5));
It's 2014, and I suggest a JavaScript string-padding function. Ha!
Bare-bones: right-pad with spaces
function pad (str, length) {
var padding = (new Array(Math.max(length - str.length + 1, 0))).join(" ");
return str + padding;
}
Fancy: pad with options
/**
* #param {*} str Input string, or any other type (will be converted to string)
* #param {number} length Desired length to pad the string to
* #param {Object} [opts]
* #param {string} [opts.padWith=" "] Character to use for padding
* #param {boolean} [opts.padLeft=false] Whether to pad on the left
* #param {boolean} [opts.collapseEmpty=false] Whether to return an empty string if the input was empty
* #returns {string}
*/
function pad(str, length, opts) {
var padding = (new Array(Math.max(length - (str + "").length + 1, 0))).join(opts && opts.padWith || " "),
collapse = opts && opts.collapseEmpty && !(str + "").length;
return collapse ? "" : opts && opts.padLeft ? padding + str : str + padding;
}
Usage (fancy):
pad("123", 5);
// Returns "123 "
pad(123, 5);
// Returns "123 " - non-string input
pad("123", 5, { padWith: "0", padLeft: true });
// Returns "00123"
pad("", 5);
// Returns " "
pad("", 5, { collapseEmpty: true });
// Returns ""
pad("1234567", 5);
// Returns "1234567"
/**************************************************************************************************
Pad a string to pad_length fillig it with pad_char.
By default the function performs a left pad, unless pad_right is set to true.
If the value of pad_length is negative, less than, or equal to the length of the input string, no padding takes place.
**************************************************************************************************/
if(!String.prototype.pad)
String.prototype.pad = function(pad_char, pad_length, pad_right)
{
var result = this;
if( (typeof pad_char === 'string') && (pad_char.length === 1) && (pad_length > this.length) )
{
var padding = new Array(pad_length - this.length + 1).join(pad_char); //thanks to http://stackoverflow.com/questions/202605/repeat-string-javascript/2433358#2433358
result = (pad_right ? result + padding : padding + result);
}
return result;
}
And then you can do:
alert( "3".pad("0", 3) ); //shows "003"
alert( "hi".pad(" ", 3) ); //shows " hi"
alert( "hi".pad(" ", 3, true) ); //shows "hi "
If you just want a very simple hacky one-liner to pad, just make a string of the desired padding character of the desired max padding length and then substring it to the length of what you want to pad.
Example: padding the string store in e with spaces to 25 characters long.
var e = "hello"; e = e + " ".substring(e.length)
Result: "hello "
If you want to do the same with a number as input just call .toString() on it before.
A friend asked about using a JavaScript function to pad left. It turned into a little bit of an endeavor between some of us in chat to code golf it. This was the result:
function l(p,t,v){
v+="";return v.length>=t?v:l(p,t,p+v);
}
It ensures that the value to be padded is a string, and then if it isn't the length of the total desired length it will pad it once and then recurse. Here is what it looks like with more logical naming and structure
function padLeft(pad, totalLength, value){
value = value.toString();
if( value.length >= totalLength ){
return value;
}else{
return padLeft(pad, totalLength, pad + value);
}
}
The example we were using was to ensure that numbers were padded with 0 to the left to make a max length of 6. Here is an example set:
function l(p,t,v){v+="";return v.length>=t?v:l(p,t,p+v);}
var vals = [6451,123,466750];
var pad = l(0,6,vals[0]);// pad with 0's, max length 6
var pads = vals.map(function(i){ return l(0,6,i) });
document.write(pads.join("<br />"));
A little late, but thought I might share anyway. I found it useful to add a prototype extension to Object. That way I can pad numbers and strings, left or right. I have a module with similar utilities I include in my scripts.
// include the module in your script, there is no need to export
var jsAddOns = require('<path to module>/jsAddOns');
~~~~~~~~~~~~ jsAddOns.js ~~~~~~~~~~~~
/*
* method prototype for any Object to pad it's toString()
* representation with additional characters to the specified length
*
* #param padToLength required int
* entire length of padded string (original + padding)
* #param padChar optional char
* character to use for padding, default is white space
* #param padLeft optional boolean
* if true padding added to left
* if omitted or false, padding added to right
*
* #return padded string or
* original string if length is >= padToLength
*/
Object.prototype.pad = function(padToLength, padChar, padLeft) {
// get the string value
s = this.toString()
// default padToLength to 0
// if omitted, original string is returned
padToLength = padToLength || 0;
// default padChar to empty space
padChar = padChar || ' ';
// ignore padding if string too long
if (s.length >= padToLength) {
return s;
}
// create the pad of appropriate length
var pad = Array(padToLength - s.length).join(padChar);
// add pad to right or left side
if (padLeft) {
return pad + s;
} else {
return s + pad;
}
};
Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end!
Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string.
Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).
if (!String.prototype.lpad) {
String.prototype.lpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return pad.substr(0, len-this.length) + this;
}
}
if (!String.prototype.rpad) {
String.prototype.rpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return this + pad.substr(0, len-this.length);
}
}
Here is a JavaScript function that adds a specified number of paddings with a custom symbol. The function takes three parameters.
padMe --> string or number to left pad
pads --> number of pads
padSymble --> custom symbol, default is "0"
function leftPad(padMe, pads, padSymble) {
if(typeof padMe === "undefined") {
padMe = "";
}
if (typeof pads === "undefined") {
pads = 0;
}
if (typeof padSymble === "undefined") {
padSymble = "0";
}
var symble = "";
var result = [];
for(var i=0; i < pads; i++) {
symble += padSymble;
}
var length = symble.length - padMe.toString().length;
result = symble.substring(0, length);
return result.concat(padMe.toString());
}
Here are some results:
> leftPad(1)
"1"
> leftPad(1, 4)
"0001"
> leftPad(1, 4, "0")
"0001"
> leftPad(1, 4, "#")
"###1"
Yet another take at with combination of a couple of solutions:
/**
* pad string on left
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingLeft = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this || c || 0) + '', b = new Array((++b || 3) - this.length).join(c || 0), b + this
};
/**
* pad string on right
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingRight = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this||c||0) + '', b = new Array((++b || 3) - this.length).join(c || 0), this + b
};

Wrong Convert amount

The following function works perfect, but when the amount over 1 million, the function don't work exactly.
Example:
AMOUNTPAID = 35555
The output is: 35.555,00 - work fine
But when the amount paid is for example: 1223578 (over 1 Million),
is the output the following output value: 1.223.235,00 (but it must be: 1.223.578,00) - there is a deviation of 343
Any ideas?
I call the function via HTML as follows:
<td class="tr1 td2"><p class="p2 ft4"><script type="text/javascript">document.write(ConvertBetrag('{{NETAMOUNT}}'))</script> €</P></TD>
#
Here ist the Javascript:
function Convertamount( amount ){
var number = amount;
number = Math.round(number * Math.pow(12, 2)) / Math.pow(12, 2);
number = number.toFixed(2);
number = number.toString();
var negative = false;
if (number.indexOf("-") == 0)
{
negative = true ;
number = number.replace("-","");
}
var str = number.toString();
str = str.replace(".", ",");
// number before decimal point
var intbeforedecimaln = str.length - (str.length - str.indexOf(","));
// number of delimiters
var intKTrenner = Math.floor((intbeforedecimaln - 1) / 3);
// Leading digits before the first dot
var intZiffern = (intbeforedecimaln % 3 == 0) ? 3 : (intbeforedecimaln % 3);
// Provided digits before the first thousand separator with point
strNew = str.substring(0, intZiffern);
// Auxiliary string without the previously treated digits
strHelp = str.substr(intZiffern, (str.length - intZiffern));
// Through thousands of remaining ...
for(var i=0; i<intKTrenner; i++)
{
// attach 3 digits of the nearest thousand group point to String
strNew += "." + strHelp.substring(0, 3);
// Post new auxiliary string without the 3 digits being treated
strHelp = strHelp.substr(intZiffern, (strHelp.length - intZiffern));
}
// attach a decimal
var szdecimal = str.substring(intbeforedecimaln, str.length);
if (szdecimal.length < 3 )
{
strNew += str.substring(intbeforedecimaln, str.length) + '0';
}
else
{
strNew += str.substring(intbeforedecimaln, str.length);
}
var number = strNew;
if (negative)
{
number = "- " + number ;
}
return number;
}
JavaScript's Math functions have a toLocaleString method. Why don't you just use this?
var n = (1223578.00).toLocaleString();
-> "1,223,578.00"
The locale you wish to use can be passed in as a parameter, for instance:
var n = (1223578.00).toLocaleString('de-DE');
-> "1.223.578,00"

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.

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