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Using arrays to do basic calculations with negative exponent

I'm trying to write a function which outputs the correct result when multiplying a number by a negative power of ten using arrays and split() method. For example the following expressions get the right result: 1x10^-2 = 0.01 1x10^-4 = 0.0001.
Problem comes when the number's length is superior to the exponent value (note that my code treats num as a string to split it in an array as shown in code bellow :
//var num is treated as a string to be splited inside get_results() function
//exponent is a number
//Try different values for exponent and different lengths for num to reproduce the problem
//for example var num = 1234 and var exponent = 2 will output 1.234 instead of 12.34
var num = '1';
var sign = '-';
var exponent = 2;
var op = 'x10^'+sign+exponent;
var re = get_result(num);
console.log(num+op +' = '+ re);
function get_result(thisNum) {
if (sign == '-') {
var arr = [];
var splitNum = thisNum.split('');
for (var i = 0; i <= exponent-splitNum.length; i++) {
arr.push('0');
}
for (var j = 0; j < splitNum.length; j++) {
arr.push(splitNum[j]);
}
if (exponent > 0) {
arr.splice(1, 0, '.');
}
arr.join('');
}
return arr.join('');
}
Demo here : https://jsfiddle.net/Hal_9100/c7nobmnj/
I tried different approaches to get the right results with different num lengths and exponent values, but nothing I came with worked and I came to the point where I can't think of anything else.
You can see my latest try here : https://jsfiddle.net/Hal_9100/vq1hrru5/
Any idea how I could solve this problem ?
PS: I know most of the rounding errors due to javascript floating point conversion are pretty harmless and can be fixed using toFixed(n) or by using specialized third-party librairies, but my only goal here is to get better at writing pure javascript functions.

I am not sure if you want to keep going with the array approach to a solution, but it seems like this could be solved with using the Math.pow() method that already exists.
function computeExponentExpression ( test ) {
var base;
var multiplier;
var exponent;
test.replace(/^(\d+)(x)(\d+)([^])([-]?\d+)$/, function() {
base = parseInt(arguments[1], 10);
multiplier = parseInt(arguments[3], 10);
exponent = parseInt(arguments[5], 10);
return '';
} );
console.log( base * Math.pow(multiplier, exponent));
}
computeExponentExpression('1x10^-4');
computeExponentExpression('1x10^2');
computeExponentExpression('4x5^3');

The problem is where you push the decimal point .
instead of
arr.splice(1, 0, '.');
try this:
arr.splice(-exponent, 0, '.');
See fiddle: https://jsfiddle.net/free_soul/c7nobmnj/1/

Codefights: Correct solution but system does not accept it

Experienced codefighters, i have just started using Codefight website to learn Javascript. I have solved their task but system does not accept it. The task is to sum all integers (inidividual digit) in a number. For example sumDigit(111) = 3. What is wrong with my code? Please help me.
Code
function digitSum(n) {
var emptyArray = [];
var total = 0;
var number = n.toString();
var res = number.split("");
for (var i=0; i<res.length; i++) {
var numberInd = Number(res[i]);
emptyArray.push(numberInd);
}
var finalSum = emptyArray.reduce(add,total);
function add(a,b) {
return a + b;
}
console.log(finalSum);
//console.log(emptyArray);
//console.log(res);
}

Here's a faster trick for summing the individual digits of a number using only arithmetic:
var digitSum = function(n) {
var sum = 0;
while (n > 0) {
sum += n % 10;
n = Math.floor(n / 10);
}
return sum;
};
n % 10 is the remainder when you divide n by 10. Effectively, this retrieves the ones-digit of a number. Math.floor(n / 10) is the integer division of n by 10. You can think of it as chopping off the ones-digit of a number. That means that this code adds the ones digit to sum, chops off the ones digit (moving the tens digit down to where the ones-digit was) and repeats this process until the number is equal to zero (i.e. there are no digits left).
The reason why this is more efficient than your method is that it doesn't require converting the integer to a string, which is a potentially costly operation. Since CodeFights is mainly a test of algorithmic ability, they are most likely looking for the more algorithmic answer, which is the one I explained above.

Var overflow in JS and mongo [duplicate]

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);

Need help making an approximation of Euler's constant

It's very close but just one number off. If you can change anything here to make it better it'd be appreciated. I'm comparing my number with Math.E to see if I'm close.
var e = (function() {
var factorial = function(n) {
var a = 1;
for (var i = 1; i <= n; i++) {
a = a * i;
}
return a;
};
for (var k = 0, b = []; k < 18; k++) {
b.push(b.length ? b[k - 1] + 1 / factorial(k) : 1 / factorial(k));
}
return b[b.length - 1];
})();
document.write(e);document.write('<br />'+ Math.E);​
My number: 2.7182818284590455
Math.E: 2.718281828459045

Work from higher numbers to lower numbers to minimize cancellation:
var e = 1;
for(var k = 17; k > 0; --k) {
e = 1 + e/k;
}
return e;
Evaluating the Taylor polynomial by Horner's rule even avoids the factorial and allows you to use more terms (won't make a difference beyond 17, though).

As far as I can see your number is the same as Math.E and even has a better precision.
2.7182818284590455
2.718281828459045
What is the problem after all?

With javascript, you cannot calculate e this way due to the level of precision of javascript computations. See http://www.javascripter.net/faq/accuracy.htm for more info.
To demonstrate this problem please take a look at the following fiddle which calculates e with n starting at 50000000, incrementing n by 1 every 10 milliseconds:
http://jsfiddle.net/q8xRs/1/

I like using integer values to approximate real ones.
Possible approximations of e in order of increasing accuracy are:
11/4
87/32
23225/8544
3442297523731/1266350489376
That last one is fairly accurate, equating to:
2.7182818284590452213260834432
which doesn't diverge from wikipedia's value till the 18th:
2.71828182845904523536028747135266249775724709369995
So there's that, if you're interested.

How to do bitwise AND in javascript on variables that are longer than 32 bit?

I have 2 numbers in javascript that I want to bit and. They both are 33bit long
in C#:
((4294967296 & 4294967296 )==0) is false
but in javascript:
((4294967296 & 4294967296 )==0) is true
4294967296 is ((long)1) << 32
As I understand it, it is due to the fact that javascript converts values to int32 when performing bit wise operations.
How do I work around this?
Any suggestions on how to replace bit and with a set of other math operations so that bits are not lost?

Here's a fun function for arbitrarily large integers:
function BitwiseAndLarge(val1, val2) {
var shift = 0, result = 0;
var mask = ~((~0) << 30); // Gives us a bit mask like 01111..1 (30 ones)
var divisor = 1 << 30; // To work with the bit mask, we need to clear bits at a time
while( (val1 != 0) && (val2 != 0) ) {
var rs = (mask & val1) & (mask & val2);
val1 = Math.floor(val1 / divisor); // val1 >>> 30
val2 = Math.floor(val2 / divisor); // val2 >>> 30
for(var i = shift++; i--;) {
rs *= divisor; // rs << 30
}
result += rs;
}
return result;
}
Assuming that the system handles at least 30-bit bitwise operations properly.

You could split each of the vars into 2 32-bit values (like a high word and low word), then do a bitwise operation on both pairs.
The script below runs as a Windows .js script. You can replace WScript.Echo() with alert() for Web.
var a = 4294967296;
var b = 4294967296;
var w = 4294967296; // 2^32
var aHI = a / w;
var aLO = a % w;
var bHI = b / w;
var bLO = b % w;
WScript.Echo((aHI & bHI) * w + (aLO & bLO));

There are several BigInteger librairy in Javascript, but none of them offer bitwise operation you need at the moment. If you are motivated and really need that functionality you can modify one of those librairy and add a method to do so. They already offer a good code base to work with huge number.
You can find a list of the BigInteger librairy in Javascript in this question :
Huge Integer JavaScript Library

The simplest bit-wise AND, that works up to JavaScript's maximum number
JavaScript's max integer value is 2^53 for internal reasons (it's a double float). If you need more there are good libraries for that big integer handling.
2^53 is 9,007,199,254,740,992, or about 9,000 trillion (~9 quadrillion).
// Works with values up to 2^53
function bitwiseAnd_53bit(value1, value2) {
const maxInt32Bits = 4294967296; // 2^32
const value1_highBits = value1 / maxInt32Bits;
const value1_lowBits = value1 % maxInt32Bits;
const value2_highBits = value2 / maxInt32Bits;
const value2_lowBits = value2 % maxInt32Bits;
return (value1_highBits & value2_highBits) * maxInt32Bits + (value1_lowBits & value2_lowBits)
}

Ran into this problem today and this is what I came up with:
function bitwiseAnd(firstNumber, secondNumber) {
let // convert the numbers to binary strings
firstBitArray = (firstNumber).toString(2),
secondBitArray = (secondNumber).toString(2),
resultedBitArray = [],
// get the length of the largest number
maxLength = Math.max(firstBitArray.length, secondBitArray.length);
//add zero fill ahead in case the binary strings have different lengths
//so we can have strings equal in length and compare bit by bit
firstBitArray = firstBitArray.padStart(maxLength,'0');
secondBitArray = secondBitArray.padStart(maxLength,'0');
// bit by bit comparison
for(let i = 0; i < maxLength; i++) {
resultedBitArray.push(parseInt(firstBitArray[i]) && secondBitArray[i]);
}
//concat the result array back to a string and parse the binary string back to an integer
return parseInt(resultedBitArray.join(''),2);
}
Hope this helps anyone else who runs into this problem.