Develop Reference

Can someone explain this base conversion code

var ShortURL = new function() {
var _alphabet = '23456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ-_',
_base = _alphabet.length;
this.encode = function(num) {
var str = '';
while (num > 0) {
str = _alphabet.charAt(num % _base) + str;
num = Math.floor(num / _base);
}
return str;
};
this.decode = function(str) {
var num = 0;
for (var i = 0; i < str.length; i++) {
num = num * _base + _alphabet.indexOf(str.charAt(i));
}
return num;
};
};
I understand encode works by converting from decimal to custom base (custom alphabet/numbers in this case)
I am not quite sure how decode works.
Why do we multiply base by a current number and then add the position number of the alphabet? I know that to convert 010 base 2 to decimal, we would do
(2 * 0^2) + (2 * 1^1) + (2 * 0 ^ 0) = 2
Not sure how it is represented in that decode algorithm
EDIT:
My own decode version
this.decode2 = function (str) {
var result = 0;
var position = str.length - 1;
var value;
for (var i = 0; i < str.length; i++) {
value = _alphabet.indexOf(str[i]);
result += value * Math.pow(_base, position--);
}
return result;
}
This is how I wrote my own decode version (Just like I want convert this on paper. I would like someone to explain more in detail how the first version of decode works. Still don't get why we multiply num * base and start num with 0.

OK, so what does 376 mean as a base-10 output of your encode() function? It means:
1 * 100 +
5 * 10 +
4 * 1
Why? Because in encode(), you divide by the base on every iteration. That means that, implicitly, the characters pushed onto the string on the earlier iterations gain in significance by a factor of the base each time through the loop.
The decode() function, therefore, multiplies by the base each time it sees a new character. That way, the first digit is multiplied by the base once for every digit position past the first that it represents, and so on for the rest of the digits.
Note that in the explanation above, the 1, 5, and 4 come from the positions of the characters 3, 7, and 6 in the "alphabet" list. That's how your encoding/decoding mechanism works. If you feed your decode() function a numeric string encoded by something trying to produce normal base-10 numbers, then of course you'll get a weird result; that's probably obvious.
edit To further elaborate on the decode() function: forget (for now) about the special base and encoding alphabet. The process is basically the same regardless of the base involved. So, let's look at a function that interprets a base-10 string of numeric digits as a number:
function decode10(str) {
var num = 0, zero = '0'.charCodeAt(0);
for (var i = 0; i < str.length; ++i) {
num = (num * 10) + (str[i] - zero);
}
return num;
}
The accumulator variable num is initialized to 0 first, because before examining any characters of the input numeric string the only value that makes sense to start with is 0.
The function then iterates through each character of the input string from left to right. On each iteration, the accumulator is multiplied by the base, and the digit value at the current string position is added.
If the input string is "214", then, the iteration will proceed as follows:
num is set to 0
First iteration: str[i] is 2, so (num * 10) + 2 is 2
Second iteration: str[i] is 1, so (num * 10) + 1 is 21
Third iteration: str[i] is 4, so (num * 10) + 4 is 214
The successive multiplications by 10 achieve what the call to Math.pow() does in your code. Note that 2 is multiplied by 10 twice, which effectively multiplies it by 100.
The decode() routine in your original code does the same thing, only instead of a simple character code computation to get the numeric value of a digit, it performs a lookup in the alphabet string.

Both the original and your own version of the decode function achieve the same thing, but the original version does it more efficiently.
In the following assignment:
num = num * _base + _alphabet.indexOf(str.charAt(i));
... there are two parts:
_alphabet.indexOf(str.charAt(i))
The indexOf returns the value of a digit in base _base. You have this part in your own algorithm, so that should be clear.
num * _base
This multiplies the so-far accumulated result. The rest of my answer is about that part:
In the first iteration this has no effect, as num is still 0 at that point. But at the end of the first iteration, num contains the value as if the str only had its left most character. It is the base-51 digit value of the left most digit.
From the next iteration onwards, the result is multiplied by the base, which makes room for the next value to be added to it. It functions like a digit shift.
Take this example input to decode:
bd35
The individual characters represent value 8, 10, 1 and 3. As there are 51 characters in the alphabet, we're in base 51. So bd35 this represents value:
8*51³ + 10*51² + 1*51 + 3
Here is a table with the value of num after each iteration:
8
8*51 + 10
8*51² + 10*51 + 1
8*51³ + 10*51² + 1*51 + 3
Just to make the visualisation cleaner, let's put the power of 51 in a column header, and remove that from the rows:
3 2 1 0
----------------------------
8
8 10
8 10 1
8 10 1 3
Note how the 8 shifts to the left at each iteration and gets multiplied with the base (51). The same happens with 10, as soon as it is shifted in from the right, and the same with the 1, and 3, although that is the last one and doesn't shift any more.
The multiplication num * _base represents thus a shift of base-digits to the left, making room for a new digit to shift in from the right (through simple addition).
At the last iteration all digits have shifted in their correct position, i.e. they have been multiplied by the base just enough times.
Putting your own algorithm in the same scheme, you'd have this table:
3 2 1 0
----------------------------
8
8 10
8 10 1
8 10 1 3
Here, there is no shifting: the digits are immediately put in the right position, i.e. they are multiplied with the correct power of 51 immediately.

You ask
I would like to understand how the decode function works from logical perspective. Why are we using num * base and starting with num = 0.
and write that
I am not quite sure how decode works. Why do we multiply base by a
current number and then add the position number of the alphabet? I
know that to convert 010 base 2 to decimal, we would do
(2 * 0^2) + (2 * 1^1) + (2 * 0 ^ 0) = 2
The decode function uses an approach to base conversion known as Horner's rule, used because it is computationally efficient:
start with a variable set to 0, num = 0
multiply the variable num by the base
take the value of the most significant digit (the leftmost digit) and add it to num,
repeat step 2 and 3 for as long as there are digits left to convert,
the variable num now contains the converted value (in base 10)
Using an example of a hexadecimal number A5D:
start with a variable set to 0, num = 0
multiply by the base (16), num is now still 0
take the value of the most significant digit (the A has a digit value of 10) and add it to num, num is now 10
repeat step 2, multiply the variable num by the base (16), num is now 160
repeat step 3, add the hexadecimal digit 5 to num, num is now 165
repeat step 2, multiply the variable num by the base (16), num is now 2640
repeat step 3, add the hexadecimal digit D to num (add 13)
there are no digits left to convert, the variable num now contains the converted value (in base 10), which is 2653
Compare the expression of the standard approach:
(10 × 162) + (5 × 161) + (13 × 160) = 2653
to the use of Horner's rule:
(((10 × 16) + 5) × 16) + 13 = 2653
which is exactly the same computation, but rearranged in a form making it easier to compute. This is how the decode function works.
Why are we using num * base and starting with num = 0.
The conversion algorithm needs a start value, therefore num is set to 0. For each repetition (each loop iteration), num is multiplied by base. This only has any effect on the second iteration, but is written like this to make it easier to write the conversion as a for loop.

Check if a string starts with a number in range

I have a string that can be of random length and I need to check if it satisfies the following conditions:
starts with 4 digits
the first 4 digits form a number in range 2221-2720 (credit card number)
Right now I have the following code:
var isCardNumber = function (myString)
{
var num = parseInt(myString.substr(0, 4));
if(num) {
return num >= 2221 && num <= 2720;
};
return false;
};
Can this be done simpler/shorter? Probably with a regexp?

var isCardNumber = function ( cardNumber ) {
var n = +cardNumber.substr( 0, 4 )
return n && n > 2220 && n < 2721
}

Here is a regex for your general culture. As other said, I would not recommend using it. You could do performance tests but I'd expect it to be slower than the integer comparison test.
^2(2[2-9]\d|[3-6]\d\d|7([0-1]\d|2[0-1]))
To construct this kind of regex you need to break your range into easily reprensentable strings ; 2221-2720 is :
2220 to 2299 (22[2-9]\d)
2300 to 2699 (2[3-6]\d\d)
2700 to 2719 (27[0-1]\d)
2720 to 2721 (272[0-1])
Then you can factorise it as I did or just use it as is, with each fragment separated by a |.

How do I convert an integer to binary in JavaScript?

I’d like to see integers, positive or negative, in binary.
Rather like this question, but for JavaScript.

function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
console.log(dec2bin(1)); // 1
console.log(dec2bin(-1)); // 11111111111111111111111111111111
console.log(dec2bin(256)); // 100000000
console.log(dec2bin(-256)); // 11111111111111111111111100000000
You can use Number.toString(2) function, but it has some problems when representing negative numbers. For example, (-1).toString(2) output is "-1".
To fix this issue, you can use the unsigned right shift bitwise operator (>>>) to coerce your number to an unsigned integer.
If you run (-1 >>> 0).toString(2) you will shift your number 0 bits to the right, which doesn't change the number itself but it will be represented as an unsigned integer. The code above will output "11111111111111111111111111111111" correctly.
This question has further explanation.
-3 >>> 0 (right logical shift) coerces its arguments to unsigned integers, which is why you get the 32-bit two's complement representation of -3.

Try
num.toString(2);
The 2 is the radix and can be any base between 2 and 36
source here
UPDATE:
This will only work for positive numbers, Javascript represents negative binary integers in two's-complement notation. I made this little function which should do the trick, I haven't tested it out properly:
function dec2Bin(dec)
{
if(dec >= 0) {
return dec.toString(2);
}
else {
/* Here you could represent the number in 2s compliment but this is not what
JS uses as its not sure how many bits are in your number range. There are
some suggestions https://stackoverflow.com/questions/10936600/javascript-decimal-to-binary-64-bit
*/
return (~dec).toString(2);
}
}
I had some help from here

A simple way is just...
Number(42).toString(2);
// "101010"

The binary in 'convert to binary' can refer to three main things. The positional number system, the binary representation in memory or 32bit bitstrings. (for 64bit bitstrings see Patrick Roberts' answer)
1. Number System
(123456).toString(2) will convert numbers to the base 2 positional numeral system. In this system negative numbers are written with minus signs just like in decimal.
2. Internal Representation
The internal representation of numbers is 64 bit floating point and some limitations are discussed in this answer. There is no easy way to create a bit-string representation of this in javascript nor access specific bits.
3. Masks & Bitwise Operators
MDN has a good overview of how bitwise operators work. Importantly:
Bitwise operators treat their operands as a sequence of 32 bits (zeros and ones)
Before operations are applied the 64 bit floating points numbers are cast to 32 bit signed integers. After they are converted back.
Here is the MDN example code for converting numbers into 32-bit strings.
function createBinaryString (nMask) {
// nMask must be between -2147483648 and 2147483647
for (var nFlag = 0, nShifted = nMask, sMask = ""; nFlag < 32;
nFlag++, sMask += String(nShifted >>> 31), nShifted <<= 1);
return sMask;
}
createBinaryString(0) //-> "00000000000000000000000000000000"
createBinaryString(123) //-> "00000000000000000000000001111011"
createBinaryString(-1) //-> "11111111111111111111111111111111"
createBinaryString(-1123456) //-> "11111111111011101101101110000000"
createBinaryString(0x7fffffff) //-> "01111111111111111111111111111111"

This answer attempts to address inputs with an absolute value in the range of 214748364810 (231) – 900719925474099110 (253-1).
In JavaScript, numbers are stored in 64-bit floating point representation, but bitwise operations coerce them to 32-bit integers in two's complement format, so any approach which uses bitwise operations restricts the range of output to -214748364810 (-231) – 214748364710 (231-1).
However, if bitwise operations are avoided and the 64-bit floating point representation is preserved by using only mathematical operations, we can reliably convert any safe integer to 64-bit two's complement binary notation by sign-extending the 53-bit twosComplement:
function toBinary (value) {
if (!Number.isSafeInteger(value)) {
throw new TypeError('value must be a safe integer');
}
const negative = value < 0;
const twosComplement = negative ? Number.MAX_SAFE_INTEGER + value + 1 : value;
const signExtend = negative ? '1' : '0';
return twosComplement.toString(2).padStart(53, '0').padStart(64, signExtend);
}
function format (value) {
console.log(value.toString().padStart(64));
console.log(value.toString(2).padStart(64));
console.log(toBinary(value));
}
format(8);
format(-8);
format(2**33-1);
format(-(2**33-1));
format(2**53-1);
format(-(2**53-1));
format(2**52);
format(-(2**52));
format(2**52+1);
format(-(2**52+1));
.as-console-wrapper{max-height:100%!important}
For older browsers, polyfills exist for the following functions and values:
Number.isSafeInteger()
Number.isInteger()
Number.MAX_SAFE_INTEGER
String.prototype.padStart()
As an added bonus, you can support any radix (2–36) if you perform the two's complement conversion for negative numbers in ⌈64 / log2(radix)⌉ digits by using BigInt:
function toRadix (value, radix) {
if (!Number.isSafeInteger(value)) {
throw new TypeError('value must be a safe integer');
}
const digits = Math.ceil(64 / Math.log2(radix));
const twosComplement = value < 0
? BigInt(radix) ** BigInt(digits) + BigInt(value)
: value;
return twosComplement.toString(radix).padStart(digits, '0');
}
console.log(toRadix(0xcba9876543210, 2));
console.log(toRadix(-0xcba9876543210, 2));
console.log(toRadix(0xcba9876543210, 16));
console.log(toRadix(-0xcba9876543210, 16));
console.log(toRadix(0x1032547698bac, 2));
console.log(toRadix(-0x1032547698bac, 2));
console.log(toRadix(0x1032547698bac, 16));
console.log(toRadix(-0x1032547698bac, 16));
.as-console-wrapper{max-height:100%!important}
If you are interested in my old answer that used an ArrayBuffer to create a union between a Float64Array and a Uint16Array, please refer to this answer's revision history.

A solution i'd go with that's fine for 32-bits, is the code the end of this answer, which is from developer.mozilla.org(MDN), but with some lines added for A)formatting and B)checking that the number is in range.
Some suggested x.toString(2) which doesn't work for negatives, it just sticks a minus sign in there for them, which is no good.
Fernando mentioned a simple solution of (x>>>0).toString(2); which is fine for negatives, but has a slight issue when x is positive. It has the output starting with 1, which for positive numbers isn't proper 2s complement.
Anybody that doesn't understand the fact of positive numbers starting with 0 and negative numbers with 1, in 2s complement, could check this SO QnA on 2s complement. What is “2's Complement”?
A solution could involve prepending a 0 for positive numbers, which I did in an earlier revision of this answer. And one could accept sometimes having a 33bit number, or one could make sure that the number to convert is within range -(2^31)<=x<2^31-1. So the number is always 32bits. But rather than do that, you can go with this solution on mozilla.org
Patrick's answer and code is long and apparently works for 64-bit, but had a bug that a commenter found, and the commenter fixed patrick's bug, but patrick has some "magic number" in his code that he didn't comment about and has forgotten about and patrick no longer fully understands his own code / why it works.
Annan had some incorrect and unclear terminology but mentioned a solution by developer.mozilla.org
Note- the old link https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators now redirects elsewhere and doesn't have that content but the proper old link , which comes up when archive.org retrieves pages!, is available here https://web.archive.org/web/20150315015832/https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators
The solution there works for 32-bit numbers.
The code is pretty compact, a function of three lines.
But I have added a regex to format the output in groups of 8 bits. Based on How to format a number with commas as thousands separators? (I just amended it from grouping it in 3s right to left and adding commas, to grouping in 8s right to left, and adding spaces)
And, while mozilla made a comment about the size of nMask(the number fed in)..that it has to be in range, they didn't test for or throw an error when the number is out of range, so i've added that.
I'm not sure why they named their parameter 'nMask' but i'll leave that as is.
https://web.archive.org/web/20150315015832/https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators
function createBinaryString(nMask) {
// nMask must be between -2147483648 and 2147483647
if (nMask > 2**31-1)
throw "number too large. number shouldn't be > 2**31-1"; //added
if (nMask < -1*(2**31))
throw "number too far negative, number shouldn't be < -(2**31)" //added
for (var nFlag = 0, nShifted = nMask, sMask = ''; nFlag < 32;
nFlag++, sMask += String(nShifted >>> 31), nShifted <<= 1);
sMask=sMask.replace(/\B(?=(.{8})+(?!.))/g, " ") // added
return sMask;
}
console.log(createBinaryString(-1)) // "11111111 11111111 11111111 11111111"
console.log(createBinaryString(1024)) // "00000000 00000000 00000100 00000000"
console.log(createBinaryString(-2)) // "11111111 11111111 11111111 11111110"
console.log(createBinaryString(-1024)) // "11111111 11111111 11111100 00000000"
//added further console.log example
console.log(createBinaryString(2**31 -1)) //"01111111 11111111 11111111 11111111"

You can write your own function that returns an array of bits.
Example how to convert number to bits
Divisor| Dividend| bits/remainder
2 | 9 | 1
2 | 4 | 0
2 | 2 | 0
~ | 1 |~
example of above line: 2 * 4 = 8 and remainder is 1
so 9 = 1 0 0 1
function numToBit(num){
var number = num
var result = []
while(number >= 1 ){
result.unshift(Math.floor(number%2))
number = number/2
}
return result
}
Read remainders from bottom to top. Digit 1 in the middle to top.

This is how I manage to handle it:
const decbin = nbr => {
if(nbr < 0){
nbr = 0xFFFFFFFF + nbr + 1
}
return parseInt(nbr, 10).toString(2)
};
got it from this link: https://locutus.io/php/math/decbin/

we can also calculate the binary for positive or negative numbers as below:
function toBinary(n){
let binary = "";
if (n < 0) {
n = n >>> 0;
}
while(Math.ceil(n/2) > 0){
binary = n%2 + binary;
n = Math.floor(n/2);
}
return binary;
}
console.log(toBinary(7));
console.log(toBinary(-7));

You could use a recursive solution:
function intToBinary(number, res = "") {
if (number < 1)
if (res === "") return "0"
else
return res
else return intToBinary(Math.floor(number / 2), number % 2 + res)
}
console.log(intToBinary(12))
console.log(intToBinary(546))
console.log(intToBinary(0))
console.log(intToBinary(125))
Works only with positive numbers.

An actual solution that logic can be implemented by any programming language:
If you sure it is positive only:
var a = 0;
var n = 12; // your input
var m = 1;
while(n) {
a = a + n%2*m;
n = Math.floor(n/2);
m = m*10;
}
console.log(n, ':', a) // 12 : 1100
If can negative or positive -
(n >>> 0).toString(2)

I’d like to see integers, positive or negative, in binary.
This is an old question and I think there are very nice solutions here but there is no explanation about the use of these clever solutions.
First, we need to understand that a number can be positive or negative.
Also, JavaScript provides a MAX_SAFE_INTEGER constant that has a value of 9007199254740991. The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent integers between -(2^53 - 1) and 2^53 - 1.
So, now we know the range where numbers are "safe". Also, JavaScript ES6 has the built-in method Number.isSafeInteger() to check if a number is a safe integer.
Logically, if we want to represent a number n as binary, this number needs a length of 53 bits, but for better presentation lets use 7 groups of 8 bits = 56 bits and fill the left side with 0 or 1 based on its sign using the padStart function.
Next, we need to handle positive and negative numbers: positive numbers will add 0s to the left while negative numbers will add 1s. Also, negative numbers will need a two's-complement representation. We can easily fix this by adding Number.MAX_SAFE_INTEGER + 1 to the number.
For example, we want to represent -3 as binary, lets assume that Number.MAX_SAFE_INTEGER is 00000000 11111111 (255) then Number.MAX_SAFE_INTEGER + 1 will be 00000001 00000000 (256). Now lets add the number Number.MAX_SAFE_INTEGER + 1 - 3 this will be 00000000 11111101 (253) but as we said we will fill with the left side with 1 like this 11111111 11111101 (-3), this represent -3 in binary.
Another algorithm will be we add 1 to the number and invert the sign like this -(-3 + 1) = 2 this will be 00000000 00000010 (2). Now we invert every bit like this 11111111 11111101 (-3) again we have a binary representation of -3.
Here we have a working snippet of these algos:
function dec2binA(n) {
if (!Number.isSafeInteger(n)) throw new TypeError('n value must be a safe integer')
if (n > 2**31) throw 'number too large. number should not be greater than 2**31'
if (n < -1*(2**31)) throw 'number too far negative, number should not be lesser than 2**31'
const bin = n < 0 ? Number.MAX_SAFE_INTEGER + 1 + n : n
const signBit = n < 0 ? '1' : '0'
return parseInt(bin, 10).toString(2)
.padStart(56, signBit)
.replace(/\B(?=(.{8})+(?!.))/g, ' ')
}
function dec2binB(n) {
if (!Number.isSafeInteger(n)) throw new TypeError('n value must be a safe integer')
if (n > 2**31) throw 'number too large. number should not be greater than 2**31'
if (n < -1*(2**31)) throw 'number too far negative, number should not be lesser than 2**31'
const bin = n < 0 ? -(1 + n) : n
const signBit = n < 0 ? '1' : '0'
return parseInt(bin, 10).toString(2)
.replace(/[01]/g, d => +!+d)
.padStart(56, signBit)
.replace(/\B(?=(.{8})+(?!.))/g, ' ')
}
const a = -805306368
console.log(a)
console.log('dec2binA:', dec2binA(a))
console.log('dec2binB:', dec2binB(a))
const b = -3
console.log(b)
console.log('dec2binA:', dec2binA(b))
console.log('dec2binB:', dec2binB(b))

One more alternative
const decToBin = dec => {
let bin = '';
let f = false;
while (!f) {
bin = bin + (dec % 2);
dec = Math.trunc(dec / 2);
if (dec === 0 ) f = true;
}
return bin.split("").reverse().join("");
}
console.log(decToBin(0));
console.log(decToBin(1));
console.log(decToBin(2));
console.log(decToBin(3));
console.log(decToBin(4));
console.log(decToBin(5));
console.log(decToBin(6));

I used a different approach to come up with something that does this. I've decided to not use this code in my project, but I thought I'd leave it somewhere relevant in case it is useful for someone.
Doesn't use bit-shifting or two's complement coercion.
You choose the number of bits that comes out (it checks for valid values of '8', '16', '32', but I suppose you could change that)
You choose whether to treat it as a signed or unsigned integer.
It will check for range issues given the combination of signed/unsigned and number of bits, though you'll want to improve the error handling.
It also has the "reverse" version of the function which converts the bits back to the int. You'll need that since there's probably nothing else that will interpret this output :D
function intToBitString(input, size, unsigned) {
if ([8, 16, 32].indexOf(size) == -1) {
throw "invalid params";
}
var min = unsigned ? 0 : - (2 ** size / 2);
var limit = unsigned ? 2 ** size : 2 ** size / 2;
if (!Number.isInteger(input) || input < min || input >= limit) {
throw "out of range or not an int";
}
if (!unsigned) {
input += limit;
}
var binary = input.toString(2).replace(/^-/, '');
return binary.padStart(size, '0');
}
function bitStringToInt(input, size, unsigned) {
if ([8, 16, 32].indexOf(size) == -1) {
throw "invalid params";
}
input = parseInt(input, 2);
if (!unsigned) {
input -= 2 ** size / 2;
}
return input;
}
// EXAMPLES
var res;
console.log("(uint8)10");
res = intToBitString(10, 8, true);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(uint8)127");
res = intToBitString(127, 8, true);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(int8)127");
res = intToBitString(127, 8, false);
console.log("intToBitString(res, 8, false)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, false));
console.log("---");
console.log("(int8)-128");
res = intToBitString(-128, 8, false);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(uint16)5000");
res = intToBitString(5000, 16, true);
console.log("intToBitString(res, 16, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 16, true));
console.log("---");
console.log("(uint32)5000");
res = intToBitString(5000, 32, true);
console.log("intToBitString(res, 32, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 32, true));
console.log("---");

This is a method that I use. It's a very fast and concise method that works for whole numbers.
If you want, this method also works with BigInts. You just have to change each 1 to 1n.
// Assuming {num} is a whole number
function toBin(num){
let str = "";
do {
str = `${num & 1}${str}`;
num >>= 1;
} while(num);
return str
}
Explanation
This method, in a way, goes through all the bits of the number as if it's already a binary number.
It starts with an empty string, and then it prepends the last bit. num & 1 will return the last bit of the number (1 or 0). num >>= 1 then removes the last bit and makes the second-to-last bit the new last bit. The process is repeated until all the bits have been read.
Of course, this is an extreme simplification of what's actually going on. But this is how I generalize it.

This is my code:
var x = prompt("enter number", "7");
var i = 0;
var binaryvar = " ";
function add(n) {
if (n == 0) {
binaryvar = "0" + binaryvar;
}
else {
binaryvar = "1" + binaryvar;
}
}
function binary() {
while (i < 1) {
if (x == 1) {
add(1);
document.write(binaryvar);
break;
}
else {
if (x % 2 == 0) {
x = x / 2;
add(0);
}
else {
x = (x - 1) / 2;
add(1);
}
}
}
}
binary();

This is the solution . Its quite simple as a matter of fact
function binaries(num1){
var str = num1.toString(2)
return(console.log('The binary form of ' + num1 + ' is: ' + str))
}
binaries(3
)
/*
According to MDN, Number.prototype.toString() overrides
Object.prototype.toString() with the useful distinction that you can
pass in a single integer argument. This argument is an optional radix,
numbers 2 to 36 allowed.So in the example above, we’re passing in 2 to
get a string representation of the binary for the base 10 number 100,
i.e. 1100100.
*/

Javascript Brainteaser - Reverse Number Determining

Lets say I have a list of numbers in the following form(Ignore the | they are there for formating help).
00|00|xx
00|xx|00
xx|00|00
etc.
Rules: XX can be any number between 1 and 50. No XX values can be identical.
Now I select a random set of numbers(no duplicates) from a list qualifying the above format, and randomly add and subtract them. For example
000011 - 002400 - 230000 = -232389
How can I determine the original numbers and if they were added or subtracted solely from -232389? I'm stumped.
Thanks!
EDIT:
I was looking for a function so I ended up having to make one. Its just a proof of concept function so variables names are ugly http://jsfiddle.net/jPW8A/.
There are bugs in the following implementation, and it fails to work in a dozen of scenarios. Check the selected answer below.
function reverse_add_subtract(num){
var nums = [];
while(num != 0){
var str = num.toString(),
L = Math.abs(num).toString().length,
MA = str.match(/^(-?[0-9]?[0-9])([0-9][0-9])([0-9][0-9])*$/);
if(MA){
var num1 = MA[1],
num2 = MA[2];
}else{
var num1 = num,
num2 = 0;
}
if(L%2)L++;
if( num2 > 50){
if(num < 0) num1--;
else num1++;
}
nums.push(num1);
var add = parseInt(num1 + Array(--L).join(0),10);
num = (num-add);
}
return nums;
}
reverse_add_subtract(-122436);

First note that each xx group is constrained from [1, 50). This implies that each associated pair in the number that is in the range [50, 99) is really 100 - xx and this means that it "borrowed from" the group to the left. (It also means that there is only one set of normalized numbers and one solution, if any.)
So given the input 23|23|89 (the initial xx spots from -232389), normalize it -- that is, starting from the right, if the value is >= 50, get 100 - value and carry the 100 rightward (must balance). Example: (23 * 100) + 89 = 2300 * 89 = 2400 - 11 = 2389. And example that shows that it doesn't matter if it's negative as the only things that change is the signs: (-23 * 100) - 89 = -2300 - 89 = -2400 + 11 = -2389
(Notes: Remember, 1 is added to the 23 group to make it 24: the sign of the groups is not actually considered in this step, the math is just to show an example that it's okay to do! It may be possible to use this step to determine the sign and avoid extra math below, but this solution just tries to find the candidate numbers at this step. If there are any repeats of the number groups after this step then there is no solution; otherwise a solution exists.)
The candidate numbers after the normalization are then 23|24|11 (let's say this is aa|bb|cc, for below). All the xx values are now known and it is just a matter of finding the combination such that e * (aa * 10000) + f * (bb * 100) + g * (cc * 1) = -232389. The values aa, bb, cc are known from above and e, f, and g will be either 1 or -1, respectively.
Solution Warning: A method of finding the addition or subtraction given the determined numbers (determined above) is provided below the horizontal separator. Take a break and reflect on the above sections before deciding if the extra "hints" are required.
This can then be solved by utilizing the fact that all the xx groups are not dependent after the normalization. (At each step, try to make the input number for the next step approach zero.)
Example:
-232389 + (23 * 10000) = -2389 (e is -1 because that undoes the + we just did)
-2389 + (24 * 100) = 11 (likewise, f is -1)
11 - (11 * 1) = 0 (0 = win! g is 1 and solution is (-1 * 23 * 10000) + (-1 * 24 * 100) + (1 * 11 * 1) = -232389)
Happy homeworking.

First, your math is wrong. Your leading zeros are converting the first two numbers to octal. If that is the intent, the rest of this post doesn't exactly apply but may be able to be adapted.
11-2400-230000 = -232389
Now the last number is easy, it's always the first two digits, 23 in this case. Remove that:
-232389 + 230000 = -2389
Your 2nd number is the next 100 below this, -2400 in this case. And your final number is simply:
-2389 + 2400 = 11

Aww! Someone posted an answer saying "brute force it" that I was about to respond to with:
function find(num){for(var i=1;i<50;i++){for(var o1=0;o1<2;o1++){for(var j=1;j<50;j++){for(var o2=0;o2<2;o2++){for(var k=1;k<50;k++){var eq;if(eval(eq=(i+(o1?'+':'-')+j+'00'+(o2?'+':'-')+k+'0000'))==num){ return eq; }}}}}}}
they deleted it... :(
It was going to go in the comment, but here's a cleaner format:
function find(num){
for(var i=1;i<50;i++){
for(var o1=0;o1<2;o1++){
for(var j=1;j<50;j++){
for(var o2=0;o2<2;o2++){
for(var k=1;k<50;k++){
var eq;
if(eval(eq=(i+(o1?'+':'-')+j+'00'+(o2?'+':'-')+k+'0000'))==num){ return eq; }
}
}
}
}
}
}

get the number of n digit in a 2+ digit number

For example, getting "5" in "256". The closest I've gotten is Math.floor(256/10)), but that'll still return the numbers in front. Is there any simple way to get what I want or would I have to make a big function for it? Also, for clarity: "n digit" would be defined. Example, getDigit(2,256) would return 5 (second digit)

Math.floor((256 / 10) % 10)
or more generally:
Math.floor(N / (Math.pow(10, n)) % 10)
where N is the number to be extracted, and n is the position of the digit. Note that this counts from 0 starting from the right (i.e., the least significant digit = 0), and doesn't account for invalid values of n.

how about
(12345 + "")[3]
or
(12345 + "").charAt(3)
to count from the other end
[length of string - digit you want] so if you want the 2 it's:
5 - 4 = 1
(12345 + "")[1] = "2"
function getNumber (var num, var pos){
var sNum = num + "";
if(pos > sNum.length || pos <= 0){return "";}
return sNum[sNum.length - pos];
}

First, you need to cast the number to a string, then you can access the character as normal:
var num = 256;
var char = num.toString()[1]; // get the 2nd (0-based index) character from the stringified version of num
Edit: Note also that, if you want to access it without setting the number as a variable first, you need a double dot .. to access the function:
var char = 256..toString()[1];
The first dot tells the interpreter "this is a number"; the second accesses the function.

Convert to string and substring(2,2)?

This should do it:
function getDigit ( position, number ) {
number = number + ""; // convert number to string
return number.substr ( position + 1, 1 ); // I'm adding 1 to position, since 0 is the position of the first character and so on
}

Try this, last line is key:
var number = 12345;
var n = 2;
var nDigit = parseInt((number + '').substr(1,1));

If you want to try to do everything mathematically:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,digitNum-1))%10;
This code counts the digit from the right starting with 1, not 0. If you wish to change it to start at 0, delete the -1 portion in the call.
If you wish to count from the left, it gets more complicated and similar to other solutions:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,number.tostring().length-digitNum))%10;
edit:
Also, this assumes you want base 10 for your number system, but both of those will work with other bases. All you need to do is change instances of 10 in the final line of code to the number representing the base for the number system you'd like to use. (ie. hexadecimal =16, binary = 2)

// You do not say if you allow decimal fractions or negative numbers-
// the strings of those need adjusting.
Number.prototype.nthDigit= function(n){
var s= String(this).replace(/\D+/g,'');
if(s.length<=n) return null;
return Number(s.charAt(n))
}

use variable "count" to control loop
var count = 1; //starting 1
for(i=0; i<100; i++){
console.log(count);
if(i%10 == 0) count++;
}
output will fill
1
2
3
4
5
6
7
8
9