Develop Reference

How to encode integer to Uint8Array and back to integer in JavaScript?

I need to add compression to my project and I decided to use the LZJB algorithm that is fast and the code is small. Found this library https://github.com/copy/jslzjb-k
But the API is not very nice because to decompress the file you need input buffer length (because Uint8Array is not dynamic you need to allocate some data). So I want to save the length of the input buffer as the first few bytes of Uint8Array so I can extract that value and create output Uint8Array based on that integer value.
I want the function that returns Uint8Array from integer to be generic, maybe save the length of the bytes into the first byte so you know how much data you need to extract to read the integer. I guess I need to extract those bytes and use some bit shifting to get the original number. But I'm not exactly sure how to do this.
So how can I write a generic function that converts an integer into Uint8Array that can be embedded into a bigger array and then extract that number?

Here are working functions (based on Converting javascript Integer to byte array and back)
function numberToBytes(number) {
// you can use constant number of bytes by using 8 or 4
const len = Math.ceil(Math.log2(number) / 8);
const byteArray = new Uint8Array(len);
for (let index = 0; index < byteArray.length; index++) {
const byte = number & 0xff;
byteArray[index] = byte;
number = (number - byte) / 256;
}
return byteArray;
}
function bytesToNumber(byteArray) {
let result = 0;
for (let i = byteArray.length - 1; i >= 0; i--) {
result = (result * 256) + byteArray[i];
}
return result;
}
by using const len = Math.ceil(Math.log2(number) / 8); the array have only bytes needed. If you want a fixed size you can use a constant 8 or 4.
In my case, I just saved the length of the bytes in the first byte.

General answer
These functions allow any integer (it uses BigInts internally, but can accept Number arguments) to be encoded into, and decoded from, any part of a Uint8Array. It is somewhat overkill, but I wanted to learn how to work with arbitrary-sized integers in JS.
// n can be a bigint or a number
// bs is an optional Uint8Array of sufficient size
// if unspecified, a large-enough Uint8Array will be allocated
// start (optional) is the offset
// where the length-prefixed number will be written
// returns the resulting Uint8Array
function writePrefixedNum(n, bs, start) {
start = start || 0;
let len = start+2; // start, length, and 1 byte min
for (let i=0x100n; i<n; i<<=8n, len ++) /* increment length */;
if (bs === undefined) {
bs = new Uint8Array(len);
} else if (bs.length < len) {
throw `byte array too small; ${bs.length} < ${len}`;
}
let r = BigInt(n);
for (let pos = start+1; pos < len; pos++) {
bs[pos] = Number(r & 0xffn);
r >>= 8n;
}
bs[start] = len-start-1; // write byte-count to start byte
return bs;
}
// bs must be a Uint8Array from where the number will be read
// start (optional, defaults to 0)
// is where the length-prefixed number can be found
// returns a bigint, which can be coerced to int using Number()
function readPrefixedNum(bs, start) {
start = start || 0;
let size = bs[start]; // read byte-count from start byte
let n = 0n;
if (bs.length < start+size) {
throw `byte array too small; ${bs.length} < ${start+size}`;
}
for (let pos = start+size; pos >= start+1; pos --) {
n <<= 8n;
n |= BigInt(bs[pos])
}
return n;
}
function test(n) {
const array = undefined;
const offset = 2;
let bs = writePrefixedNum(n, undefined, offset);
console.log(bs);
let result = readPrefixedNum(bs, offset);
console.log(n, result, "correct?", n == result)
}
test(0)
test(0x1020304050607080n)
test(0x0807060504030201n)
Simple 4-byte answer
This answer encodes 4-byte integers to and from Uint8Arrays.
function intToArray(i) {
return Uint8Array.of(
(i&0xff000000)>>24,
(i&0x00ff0000)>>16,
(i&0x0000ff00)>> 8,
(i&0x000000ff)>> 0);
}
function arrayToInt(bs, start) {
start = start || 0;
const bytes = bs.subarray(start, start+4);
let n = 0;
for (const byte of bytes.values()) {
n = (n<<8)|byte;
}
return n;
}
for (let v of [123, 123<<8, 123<<16, 123<<24]) {
let a = intToArray(v);
let r = arrayToInt(a, 0);
console.log(v, a, r);
}

Posting this one-liner in case it is useful to anyone who is looking to work with numbers below 2^53. This strictly uses bitwise operations and has no need for constants or values other than the input to be defined.
export const encodeUvarint = (n: number): Uint8Array => n >= 0x80
? Uint8Array.from([(n & 0x7f) | 0x80, ...encodeUvarint(n >> 7)])
: Uint8Array.from([n & 0xff]);

How to implement toString to convert a number to a string?

I was asked during an interview to implement toString() to convert a number into a string.
toString()
n => s
123 => "123"
Aside from:
converting the number by concatenating an empty string
123+""
using the native toString() function
(123).toString()
creating a new string
String(123)
How else could toString() be implemented in javascript?

You can use it as the property name of an object.
function toString(value) {
// Coerces value to a primitive string (or symbol)
var obj = {};
obj[value] = true;
return Object.getOwnPropertyNames(obj)[0];
}
console.log(toString(123)); // 123 -> "123"
console.log(toString(1.23)); // 1.23 -> "1.23"
console.log(toString(NaN)); // NaN -> "NaN"
console.log(Infinity); // Infinity -> "Infinity"
console.log(toString(-0)); // -0 -> "0"
console.log(toString(1e99)); // 1e99 -> "1e+99"
You can also use DOM attributes:
var obj = document.createElement('div');
obj.setAttribute('data-toString', value);
return obj.getAttribute('data-toString');
Or join an array
return [value].join();
And a big etcetera. There are lots of things which internally use the ToString abstract operation.

This works for integers. It takes the number modulo 10 and divides it by 10 repeatedly, then adds 48 to the digits and uses String.fromCharCode to get a string value of the digits, then joins everything.
function toString(n){
var minus = (n < 0
? "-"
: ""),
result = [];
n = Math.abs(n);
while(n > 0){
result.unshift(n % 10);
n = Math.floor(n / 10);
}
return minus + (result.map(function(d){
return String.fromCharCode(d + 48);
})
.join("") || "0");
}
console.log(toString(123123));
console.log(toString(999));
console.log(toString(0));
console.log(toString(-1));

The trick here is to consider a number as a series of digits. This is not an inherent property of numbers, since the base-10 representation that we use is quite arbitrary. But once a number is represented as a series of digits, it is quite easy to convert each digit individually to a string, and concatenate all such strings.
EDIT: As pointed out, this only takes integers into consideration (which is probably acceptable for an interview question).
var intToDigits = function(n) {
var highestPow = 1;
while (highestPow < n) highestPow *= 10;
var div = highestPow / 10;
// div is now the largest multiple of 10 smaller than n
var digits = [];
do {
var digit = Math.floor(n / div);
n = n - (digit * div);
div /= 10;
digits.push(digit);
} while (n > 0);
return digits;
};
var toString = function(n) {
var digitArr = intToDigits(n);
return digitArr.map(function(n) {
return "0123456789"[n];
}).join('');
};
Usage:
>> toString(678)
"678"

If you're using ES6 you could use template literals.
var a = 5;
console.log(`${a}`);

Converting a base 10 number to other bases 2 without built in Javascript functions

I am new to coding and javascript and was asked, for an assignment, to convert base 10 numbers to a binary base without using specific Javascript built in methods (like alert(a.toString(16))), and I am only allowed to use loops,arrays and functions. This is what i have so far:
var number = prompt("Enter an unsigned base 10 number");
if (number>=0) {
var base = prompt("Enter b for binary, o for octal, or h for hexadecimal");
if (base=="h"||base=="H") {
;
}
So as you can see, I don't have much to go on. I was curious as to what equation or formula I would use to convert the base 10 number, as well as how i'm supposed to show A=10, B=11, C=12 and so forth for a hexadecimal base. Any help would be greatly appreciated!

edit: This is a rather complicated way to do it,
as Alnitak showed me (see discussion below).
It is more a scibble, or the long way by foot.
Short explanation:
If we want to get the binary of the decimal number 10,
we have to try 2^n so that 2^n is still smaller than 10.
For example 2^3 = 8 (that is OK). But 2^4 = 16 (thats too big).
So we have 2^3 and store a 1 for that in an array at index 3.
Now we have to get the rest of 10-2^3, which is 2, and have to
make the same calculation again until we get a difference of zero.
At last we have to reverse the array because its the other way arround.
var a = prompt("Enter an unsigned base 10 number");
var arr = [];
var i = 0;
function decToBin(x) {
y = Math.pow(2, i);
if (y < x) {
arr[i] = 0;
i++;
decToBin(x);
} else if (y > x) {
i--;
newX = (x - Math.pow(2, i));
arr[i] = 1;
i = 0;
decToBin(newX)
} else if (y == x) {
arr[i] = 1;
result = arr.reverse().join();
}
return result;
}
var b = decToBin(a); // var b holds the result
document.write(b);

How can I find the length of a number?

I'm looking to get the length of a number in JavaScript or jQuery?
I've tried value.length without any success, do I need to convert this to a string first?

var x = 1234567;
x.toString().length;
This process will also work forFloat Number and for Exponential number also.

Ok, so many answers, but this is a pure math one, just for the fun or for remembering that Math is Important:
var len = Math.ceil(Math.log(num + 1) / Math.LN10);
This actually gives the "length" of the number even if it's in exponential form. num is supposed to be a non negative integer here: if it's negative, take its absolute value and adjust the sign afterwards.
Update for ES2015
Now that Math.log10 is a thing, you can simply write
const len = Math.ceil(Math.log10(num + 1));

Could also use a template string:
const num = 123456
`${num}`.length // 6

You have to make the number to string in order to take length
var num = 123;
alert((num + "").length);
or
alert(num.toString().length);

I've been using this functionality in node.js, this is my fastest implementation so far:
var nLength = function(n) {
return (Math.log(Math.abs(n)+1) * 0.43429448190325176 | 0) + 1; 
}
It should handle positive and negative integers (also in exponential form) and should return the length of integer part in floats.
The following reference should provide some insight into the method:
Weisstein, Eric W. "Number Length." From MathWorld--A Wolfram Web Resource.
I believe that some bitwise operation can replace the Math.abs, but jsperf shows that Math.abs works just fine in the majority of js engines.
Update: As noted in the comments, this solution has some issues :(
Update2 (workaround) : I believe that at some point precision issues kick in and the Math.log(...)*0.434... just behaves unexpectedly. However, if Internet Explorer or Mobile devices are not your cup of tea, you can replace this operation with the Math.log10 function. In Node.js I wrote a quick basic test with the function nLength = (n) => 1 + Math.log10(Math.abs(n) + 1) | 0; and with Math.log10 it worked as expected. Please note that Math.log10 is not universally supported.

There are three way to do it.
var num = 123;
alert(num.toString().length);
better performance one (best performance in ie11)
var num = 123;
alert((num + '').length);
Math (best performance in Chrome, firefox but slowest in ie11)
var num = 123
alert(Math.floor( Math.log(num) / Math.LN10 ) + 1)
there is a jspref here
http://jsperf.com/fastest-way-to-get-the-first-in-a-number/2

You should go for the simplest one (stringLength), readability always beats speed. But if you care about speed here are some below.
Three different methods all with varying speed.
// 34ms
let weissteinLength = function(n) {
return (Math.log(Math.abs(n)+1) * 0.43429448190325176 | 0) + 1;
}
// 350ms
let stringLength = function(n) {
return n.toString().length;
}
// 58ms
let mathLength = function(n) {
return Math.ceil(Math.log(n + 1) / Math.LN10);
}
// Simple tests below if you care about performance.
let iterations = 1000000;
let maxSize = 10000;
// ------ Weisstein length.
console.log("Starting weissteinLength length.");
let startTime = Date.now();
for (let index = 0; index < iterations; index++) {
weissteinLength(Math.random() * maxSize);
}
console.log("Ended weissteinLength length. Took : " + (Date.now() - startTime ) + "ms");
// ------- String length slowest.
console.log("Starting string length.");
startTime = Date.now();
for (let index = 0; index < iterations; index++) {
stringLength(Math.random() * maxSize);
}
console.log("Ended string length. Took : " + (Date.now() - startTime ) + "ms");
// ------- Math length.
console.log("Starting math length.");
startTime = Date.now();
for (let index = 0; index < iterations; index++) {
mathLength(Math.random() * maxSize);
}

First convert it to a string:
var mynumber = 123;
alert((""+mynumber).length);
Adding an empty string to it will implicitly cause mynumber to turn into a string.

Well without converting the integer to a string you could make a funky loop:
var number = 20000;
var length = 0;
for(i = number; i > 1; ++i){
++length;
i = Math.floor(i/10);
}
alert(length);​
Demo: http://jsfiddle.net/maniator/G8tQE/

I got asked a similar question in a test.
Find a number's length without converting to string
const numbers = [1, 10, 100, 12, 123, -1, -10, -100, -12, -123, 0, -0]
const numberLength = number => {
let length = 0
let n = Math.abs(number)
do {
n /= 10
length++
} while (n >= 1)
return length
}
console.log(numbers.map(numberLength)) // [ 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 1 ]
Negative numbers were added to complicate it a little more, hence the Math.abs().

I'm perplex about converting into a string the given number because such an algorithm won't be robust and will be prone to errors: it will show all its limitations especially in case it has to evaluate very long numbers. In fact before converting the long number into a string it will "collapse" into its exponential notation equivalent (example: 1.2345e4). This notation will be converted into a string and this resulting string will be evaluated for returning its length. All of this will give a wrong result. So I suggest not to use that approach.
Have a look at the following code and run the code snippet to compare the different behaviors:
let num = 116234567891011121415113441236542134465236441625344625344625623456723423523429798771121411511034412365421344652364416253446253446254461253446221314623879235441623683749283441136232514654296853446323214617456789101112141511344122354416236837492834411362325146542968534463232146172368374928344113623251465429685;
let lenFromMath;
let lenFromString;
// The suggested way:
lenFromMath = Math.ceil(Math.log10(num + 1)); // this works in fact returns 309
// The discouraged way:
lenFromString = String(num).split("").length; // this doesn't work in fact returns 23
/*It is also possible to modify the prototype of the primitive "Number" (but some programmer might suggest this is not a good practice). But this is will also work:*/
Number.prototype.lenght = () => {return Math.ceil(Math.log10(num + 1));}
lenFromPrototype = num.lenght();
console.log({lenFromMath, lenFromPrototype, lenFromString});

A way for integers or for length of the integer part without banal converting to string:
var num = 9999999999; // your number
if (num < 0) num = -num; // this string for negative numbers
var length = 1;
while (num >= 10) {
num /= 10;
length++;
}
alert(length);

I would like to correct the #Neal answer which was pretty good for integers, but the number 1 would return a length of 0 in the previous case.
function Longueur(numberlen)
{
var length = 0, i; //define `i` with `var` as not to clutter the global scope
numberlen = parseInt(numberlen);
for(i = numberlen; i >= 1; i)
{
++length;
i = Math.floor(i/10);
}
return length;
}

To get the number of relevant digits (if the leading decimal part is 0 then the whole part has a length of 0) of any number separated by whole part and decimal part I use:
function getNumberLength(x) {
let numberText = x.toString();
let exp = 0;
if (numberText.includes('e')) {
const [coefficient, base] = numberText.split('e');
exp = parseInt(base, 10);
numberText = coefficient;
}
const [whole, decimal] = numberText.split('.');
const wholeLength = whole === '0' ? 0 : whole.length;
const decimalLength = decimal ? decimal.length : 0;
return {
whole: wholeLength > -exp ? wholeLength + exp : 0,
decimal: decimalLength > exp ? decimalLength - exp : 0,
};
}

var x = 1234567;
String(x).length;
It is shorter than with .toString() (which in the accepted answer).

Try this:
$("#element").text().length;
Example of it in use

Yes you need to convert to string in order to find the length.For example
var x=100;// type of x is number
var x=100+"";// now the type of x is string
document.write(x.length);//which would output 3.

JavaScript summing large integers

In JavaScript I would like to create the binary hash of a large boolean array (54 elements) with the following method:
function bhash(arr) {
for (var i = 0, L = arr.length, sum = 0; i < L; sum += Math.pow(2,i)*arr[i++]);
return sum;
}
In short: it creates the smallest integer to store an array of booleans in. Now my problem is that javascript apparently uses floats as default. The maximum number I have to create is 2^54-1 but once javascript reaches 2^53 it starts doing weird things:
9007199254740992+1 = 9007199254740994
Is there any way of using integers instead of floats in javascript? Or large integer summations?

JavaScript uses floating point internally.
What is JavaScript's highest integer value that a number can go to without losing precision?
In other words you can't use more than 53 bits. In some implementations you may be limited to 31.
Try storing the bits in more than one variable, use a string, or get a bignum library, or if you only need to deal with integers, a biginteger library.

BigInt is being added as a native feature of JavaScript.
typeof 123;
// → 'number'
typeof 123n;
// → 'bigint'
Example:
const max = BigInt(Number.MAX_SAFE_INTEGER);
const two = 2n;
const result = max + two;
console.log(result);
// → '9007199254740993'

javascript now has experimental support for BigInt.
At the time of writing only chrome supports this.
caniuse has no entry yet.
BigInt can be either used with a constructor, e.g. BigInt(20) or by appending n, e.g. 20n
Example:
const max = Number.MAX_SAFE_INTEGER;
console.log('javascript Number limit reached', max + 1 === max + 2) // true;
console.log('javascript BigInt limit reached', BigInt(max) + 1n === BigInt(max) + 2n); // false

No. Javascript only has one numeric type. You've to code yourself or use a large integer library (and you cannot even overload arithmetic operators).
Update
This was true in 2010... now (2019) a BigInt library is being standardized and will most probably soon arrive natively in Javascript and it will be the second numeric type present (there are typed arrays, but - at least formally - values extracted from them are still double-precision floating point numbers).

Another implementation of large integer arithmetic (also using BigInt.js) is available at www.javascripter.net/math/calculators/100digitbigintcalculator.htm. Supports the operations + - * / as well as remainder, GCD, LCM, factorial, primality test, next prime, previous prime.

So while attempting one of the leetcode problem I have written a function which takes two numbers in form of string and returns the sum of those numbers in form of string.
(This doesn't work with negative numbers though we can modify this function to cover that)
var addTwoStr = function (s1, s2) {
s1 = s1.split("").reverse().join("")
s2 = s2.split("").reverse().join("")
var carry = 0, rS = '', x = null
if (s1.length > s2.length) {
for (let i = 0; i < s1.length; i++) {
let s = s1[i]
if (i < s2.length) {
x = Number(s) + Number(s2[i]) + carry
rS += String((x % 10))
carry = parseInt(x/10)
} else {
if (carry) {
x = Number(s) + carry
rS += String((x % 10))
carry = parseInt(x/10)
} else {
rS += s
}
}
}
} else {
for (let i = 0; i < s2.length; i++) {
let s = s2[i]
if (i < s1.length) {
x = Number(s) + Number(s1[i]) + carry
rS += String((x % 10))
carry = parseInt(x/10)
} else {
if (carry) {
x = Number(s) + carry
rS += String((x % 10))
carry = parseInt(x/10)
} else {
rS += s
}
}
}
}
if (carry) {
rS += String(carry)
}
return rS.split("").reverse().join("")
}
Example: addTwoStr('120354566', '321442535')
Output: "441797101"

There are various BigInteger Javascript libraries that you can find through googling. e.g. http://www.leemon.com/crypto/BigInt.html

Here's (yet another) wrapper around Leemon Baird's BigInt.js
It is used in this online demo of a big integer calculator in JavaScript which implements the usual four operations + - * /, the modulus (%), and four builtin functions : the square root (sqrt), the power (pow), the recursive factorial (fact) and a memoizing Fibonacci (fibo).

You're probably running into a byte length limit on your system. I'd take the array of booleans, convert it to an array of binary digits ([true, false, true] => [1,0,1]), then join this array into a string "101", then use parseInt('101',2), and you'll have your answer.

/** --if you want to show a big int as your wish use install and require this module
* By using 'big-integer' module is easier to use and handling the big int numbers than regular javascript
* https://www.npmjs.com/package/big-integer
*/
let bigInt = require('big-integer');
//variable: get_bigInt
let get_bigInt = bigInt("999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999");
let arr = [1, 100000, 21, 30, 4, BigInt(999999999999), get_bigInt.value];
console.log(arr[6]); // Output: 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999n
//Calculation
console.log(arr[6] + 1n); // +1
console.log(arr[6] + 100n); // +100
console.log(arr[6] - 1n); // -1
console.log(arr[6] - 10245n); // -1000n
console.log((arr[6] * 10000n) + 145n - 435n);