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Javascript Word To Binary to Array of Bits

I have this code // Fiddle : https://jsfiddle.net/7buscnhw/
// Word to bits func
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
// binary to bit array
function bin2array(bin) {
let Bitarr = []
for(let i = 0; i < bin.length; ++i)
Bitarr[i] = (bin >> i) & 1;
return Bitarr;
}
R24011 = dec2bin(10);
Bits = bin2array(R24011);
msg = {
R24011: R24011,
BitArray: Bits
}
console.log(msg)
It correctly outputs 1010 for Ten in binary, but when I push it to an array I get [0,1,0,0]
I'm sure it'll be something stupid but I cant figure out what I've done wrong.

You pass bin (which is a string "1010") and apply >> to it, which makes Javascript convert it to a number in base 10, resulting in 1010 (one thousand and ten), which is binary 1111110010. Then, you convert that one to binary once again, using only four bits and in reversed order, which results in [0,1,0,0]
If you want to convert a number (not a string) to an array bits (and avoid built-ins), you can do that like this:
function num2bits(number) {
let Bitarr = []
while (number) {
Bitarr.unshift(number & 1);
number >>= 1
}
return Bitarr;
}
console.log(num2bits(10))
With built-ins this is as simple as
bits = [...yourNumber.toString(2)].map(Number)

You can directly use spread syntax to convert the binary string to an array.
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
function bin2array(bin) {
return [...bin].map(Number);
}
R24011 = dec2bin(10);
console.log(R24011)
Bits = bin2array(R24011);
console.log(Bits)

How to convert number start with 0 to string equivalent of the value?

I want to convert a number start with 0 to string equivalent of the value.
If I run
var num = 12;
var int = num.toString();
console.log(int);
it logs 12 as expected but if I apply the toString() to a number start with 0 like,
var num = 012;
var int = num.toString();
console.log(int);
it logs 10, why?

Number starting with 0 is interpreted as octal (base-8).

In sloppy mode (the default) numbers starting with 0 are interpreted as being written in octal (base 8) instead of decimal (base 10). If has been like that from the first released version of Javascript, and has this syntax in common with other programming languages. It is confusing, and have lead to many hard to detect buggs.
You can enable strict mode by adding "use strict" as the first non-comment in your script or function. It removes some of the quirks. It is still possible to write octal numbers in strict mode, but you have to use the same scheme as with hexadecimal and binary: 0o20 is the octal representation of 16 decimal.
The same problem can be found with the function paseInt, that takes up to two parameters, where the second is the radix. If not specified, numbers starting with 0 will be treated as octal up to ECMAScript 5, where it was changed to decimal. So if you use parseInt, specify the radix to be sure that you get what you expected.
"use strict";
// Diffrent ways to write the same number:
const values = [
0b10000, // binary
0o20, // octal
16, // decimal,
0x10 // hexadecimal
];
console.log("As binary:", values.map( value => value.toString(2)).join());
console.log("As decimal:", values.join());
console.log("As ocal", values.map( value => value.toString(8)).join());
console.log("As hexadecimal:", values.map( value => value.toString(16)).join());
console.log("As base36:", values.map( value => value.toString(36)).join());

All you have to do is add String to the front of the number that is
var num = 12;
var int = String(num);
console.log(int);
And if you want it to look like this 0012 all you have to do is
var num = 12;
var int = String(num).padStart(4, '0');
console.log(int);

toFixed function not working properly ( please give a reason not an alternative)

toFixed() function responding differently for float values.
For Example:
var a = 2.555;
var b = 5.555;
console.log(a.toFixed(2)); /* output is 2.56 */
console.log(b.toFixed(2)); /* output is 5.55 */
For 2.555/3.555 results are (2.56/3.56)
and
For other values(not sure for all values) it is showing #.55 (# refers to any number)
I am confused can any one help me out.
Thanks in advance.

Javascript uses a binary floating point representation for numbers (IEEE754).
Using this representation the only numbers that can be represented exactly are in the form n/2m where both n and m are integers.
Any number that is not a rational where the denominator is an integral power of two is impossible to represent exactly because in binary it is a periodic number (it has infinite binary digits after the point).
The number 0.5 (i.e. 1/2) is fine, (in binary is just 0.1₂) but for example 0.55 (i.e. 11/20) cannot be represented exactly (in binary it's 0.100011001100110011₂… i.e. 0.10(0011)₂ with the last part 0011₂ repeating infinite times).
If you need to do any computation in which the result depends on exact decimal numbers you need to use an exact decimal representation. A simple solution if the number of decimals is fixed (e.g. 3) is to keep all values as integers by multiplying them by 1000...
2.555 --> 2555
5.555 --> 5555
3.7 --> 3700
and adjusting your computation when doing multiplications and divisions accordingly (e.g. after multiplying two numbers you need to divide the result by 1000).
The IEEE754 double-precision format is accurate with integers up to 9,007,199,254,740,992 and this is often enough for prices/values (where the rounding is most often an issue).

Try this Demo Here
function roundToTwo(num) {
alert(+(Math.round(num + "e+2") + "e-2"));
}
roundToTwo(2.555);
roundToTwo(5.555);

toFixed() method depending on Browser rounds down or retain.
Here is the solution for this problem, check for "5" at the end
var num = 5.555;
var temp = num.toString();
if(temp .charAt(temp .length-1)==="5"){
temp = temp .slice(0,temp .length-1) + '6';
}
num = Number(temp);
Final = num.toFixed(2);
Or reusable function would be like
function toFixedCustom(num,upto){
var temp = num.toString();
if(temp .charAt(temp .length-1)==="5"){
temp = temp .slice(0,temp .length-1) + '6';
}
num = Number(temp);
Final = num.toFixed(upto);
return Final;
}
var a = 2.555;
var b = 5.555;
console.log(toFixedCustom(a,2));
console.log(toFixedCustom(b,2));

Convert A Large Integer To a Hex String In Javascript

I need to find a way to convert a large number into a hex string in javascript. Straight off the bat, I tried myBigNumber.toString(16) but if myBigNumber has a very large value (eg 1298925419114529174706173) then myBigNumber.toString(16) will return an erroneous result, which is just brilliant. I tried writing by own function as follows:
function (integer) {
var result = '';
while (integer) {
result = (integer % 16).toString(16) + result;
integer = Math.floor(integer / 16);
}
}
However, large numbers modulo 16 all return 0 (I think this fundamental issue is what is causing the problem with toString. I also tried replacing (integer % 16) with (integer - 16 * Math.floor(integer/16)) but that had the same issue.
I have also looked at the Big Integer Javascript library but that is a huge plugin for one, hopefully relatively straightforward problem.
Any thoughts as to how I can get a valid result? Maybe some sort of divide and conquer approach? I am really rather stuck here.

Assuming you have your integer stored as a decimal string like '1298925419114529174706173':
function dec2hex(str){ // .toString(16) only works up to 2^53
var dec = str.toString().split(''), sum = [], hex = [], i, s
while(dec.length){
s = 1 * dec.shift()
for(i = 0; s || i < sum.length; i++){
s += (sum[i] || 0) * 10
sum[i] = s % 16
s = (s - sum[i]) / 16
}
}
while(sum.length){
hex.push(sum.pop().toString(16))
}
return hex.join('')
}

The numbers in question are above javascript's largest integer. However, you can work with such large numbers by strings and there are some plugins which can help you do this. An example which is particularly useful in this circumstance is hex2dec

The approach I took was to use the bignumber.js library and create a BigNumber passing in the value as a string then just use toString to convert to hex:
const BigNumber = require('bignumber.js');
const lrgIntStr = '1298925419114529174706173';
const bn = new BigNumber(lrgIntStr);
const hex = bn.toString(16);

How do I round a number in JavaScript?

While working on a project, I came across a JS-script created by a former employee that basically creates a report in the form of
Name : Value
Name2 : Value2
etc.
The peoblem is that the values can sometimes be floats (with different precision), integers, or even in the form 2.20011E+17. What I want to output are pure integers. I don't know a lot of JavaScript, though. How would I go about writing a method that takes these sometimes-floats and makes them integers?

If you need to round to a certain number of digits use the following function
function roundNumber(number, digits) {
var multiple = Math.pow(10, digits);
var rndedNum = Math.round(number * multiple) / multiple;
return rndedNum;
}

You hav to convert your input into a number and then round them:
function toInteger(number){
return Math.round( // round to nearest integer
Number(number) // type cast your input
);
};
Or as a one liner:
function toInt(n){ return Math.round(Number(n)); };
Testing with different values:
toInteger(2.5); // 3
toInteger(1000); // 1000
toInteger("12345.12345"); // 12345
toInteger("2.20011E+17"); // 220011000000000000

According to the ECMAScript specification, numbers in JavaScript are represented only by the double-precision 64-bit format IEEE 754. Hence there is not really an integer type in JavaScript.
Regarding the rounding of these numbers, there are a number of ways you can achieve this. The Math object gives us three rounding methods wich we can use:
The Math.round() is most commonly used, it returns the value rounded to the nearest integer. Then there is the Math.floor() wich returns the largest integer less than or equal to a number. Lastly we have the Math.ceil() function that returns the smallest integer greater than or equal to a number.
There is also the toFixed() that returns a string representing the number using fixed-point notation.
Ps.: There is no 2nd argument in the Math.round() method. The toFixed() is not IE specific, its within the ECMAScript specification aswell

Here is a way to be able to use Math.round() with a second argument (number of decimals for rounding):
// 'improve' Math.round() to support a second argument
var _round = Math.round;
Math.round = function(number, decimals /* optional, default 0 */)
{
if (arguments.length == 1)
return _round(number);
var multiplier = Math.pow(10, decimals);
return _round(number * multiplier) / multiplier;
}
// examples
Math.round('123.4567', 2); // => 123.46
Math.round('123.4567'); // => 123

You can also use toFixed(x) or toPrecision(x) where x is the number of digits.
Both these methods are supported in all major browsers

You can use Math.round() for rounding numbers to the nearest integer.
Math.round(532.24) => 532
Also, you can use parseInt() and parseFloat() to cast a variable to a certain type, in this case integer and floating point.

A very good approximation for rounding:
function Rounding (number, precision){
var newNumber;
var sNumber = number.toString();
var increase = precision + sNumber.length - sNumber.indexOf('.') + 1;
if (number < 0)
newNumber = (number - 5 * Math.pow(10,-increase));
else
newNumber = (number + 5 * Math.pow(10,-increase));
var multiple = Math.pow(10,precision);
return Math.round(newNumber * multiple)/multiple;
}
Only in some cases when the length of the decimal part of the number is very long will it be incorrect.

Math.floor(19.5) = 19 should also work.