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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)

JavaScript factorial prevent infinity

I have been using this function for calculating factorial numbers in JavaScript:
var f = [];
function factorial (n) {
if (n == 0 || n == 1)
return 1;
if (f[n] > 0)
return f[n];
return f[n] = factorial(n-1) * n;
}
All seemed to be going well until I tried the number 500. It returned infinity.
Is there a way that I can prevent infinity as an answer?
Thank you.

You indeed need to use bignumbers. With math.js you can do:
// configure math.js to work with enough precision to do our calculation
math.config({precision: 2000});
// evaluate the factorial using a bignumber value
var value = math.bignumber(500);
var result = math.factorial(value);
// output the results
console.log(math.format(result, {notation: 'fixed'}));
This will output:
1220136825991110068701238785423046926253574342803192842192413588385845373153881997605496447502203281863013616477148203584163378722078177200480785205159329285477907571939330603772960859086270429174547882424912726344305670173270769461062802310452644218878789465754777149863494367781037644274033827365397471386477878495438489595537537990423241061271326984327745715546309977202781014561081188373709531016356324432987029563896628911658974769572087926928871281780070265174507768410719624390394322536422605234945850129918571501248706961568141625359056693423813008856249246891564126775654481886506593847951775360894005745238940335798476363944905313062323749066445048824665075946735862074637925184200459369692981022263971952597190945217823331756934581508552332820762820023402626907898342451712006207714640979456116127629145951237229913340169552363850942885592018727433795173014586357570828355780158735432768888680120399882384702151467605445407663535984174430480128938313896881639487469658817504506926365338175055478128640000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

500! is, for lack of a better term, "[bleep]ing huge".
It is far, far beyond what can be stored in a double-precision float, which is what JavaScript uses for numbers.
There's no way to prevent this, other than use numbers that are reasonable :p
EDIT: To show you just how huge it is, here's the answer:
500! = 1220136825991110068701238785423046926253574342803192842192413588385845373153881997605496447502203281863013616477148203584163378722078177200480785205159329285477907571939330603772960859086270429174547882424912726344305670173270769461062802310452644218878789465754777149863494367781037644274033827365397471386477878495438489595537537990423241061271326984327745715546309977202781014561081188373709531016356324432987029563896628911658974769572087926928871281780070265174507768410719624390394322536422605234945850129918571501248706961568141625359056693423813008856249246891564126775654481886506593847951775360894005745238940335798476363944905313062323749066445048824665075946735862074637925184200459369692981022263971952597190945217823331756934581508552332820762820023402626907898342451712006207714640979456116127629145951237229913340169552363850942885592018727433795173014586357570828355780158735432768888680120399882384702151467605445407663535984174430480128938313896881639487469658817504506926365338175055478128640000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
That right there is a 1,135-digit number. For comparison, double-precision floats can handle about 15 digits of precision.

You could consider using an arbitrary precision numeric library. This is a question of its own, though. Here's one related question: https://stackoverflow.com/questions/744099/is-there-a-good-javascript-bigdecimal-library.

I dont know if anyone has solved this elsewise...
I'm a novice beginner in coding and dont know all the aspects. But after I faced this factorial problem myself, i came here when searching for the answer. I solved the 'infinity' display problem in another way. I dont know if its very efficient or not. But it does show the results of even verry high intergers.
Sorry for any redundancy or untidiness in the code.
<!DOCTYPE html>
<html>
<head>
<title>Factorial</title>
<script src='http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js'></script>
</head>
<body>
<input type='text' id='number' />
<input type='button' value='!Factorial!' id='btn' />
<script>
var reslt=1;
var counter=0;
var mantissa=0; //stores the seperated matissa
var exponent=0; //stores the seperated exponent
$(document).ready(function (){
$('#btn').click(function (){
var num=parseFloat($('#number').val()); //number input by user
for(i=1;i<=num;i++){
reslt=reslt*i;
//when the result becomes so high that the exponent reaches 306, the number is divided by 1e300
if((parseFloat(reslt.toExponential().toString().split("e")[1]))>=300){
reslt=reslt/1e300; //the result becomes small again to be able to be iterated without becoming infinity
counter+=1; //the number of times that the number is divided in such manner is recorded by counter
}
}
//the mantissa of the final result is seperated first
mantissa=parseFloat(reslt.toExponential().toString().split("e")[0]);
//the exponent of the final result is obtained by adding the remaining exponent with the previously dropped exponents (1e300)
exponent=parseFloat(reslt.toExponential().toString().split("e")[1])+300*counter;
alert(mantissa+"e+"+exponent); //displays the result as a string by concatenating
//resets the variables and fields for the next input if any
$('#number').val('');
reslt=1;
mantissa=0;
exponent=0;
counter=0;
});
});
</script>
</body>
</html>

Javascript numbers can only get so big before they just become "Infinity". If you want to support bigger numbers, you'll have to use BigInt.
Examples:
// Without BigInt
console.log(100 ** 1000) // Infinity
// With BigInt
// (stackOverflow doesn't seem to print the result,
// unless I turn it into a string first)
console.log(String(100n ** 1000n)) // A really big number
So, for your specific bit of code, all you need to do is turn your numeric literals into BigInt literals, like this:
var f = [];
function factorial (n) {
if (n == 0n || n == 1n)
return 1n;
if (f[n] > 0n)
return f[n];
return f[n] = factorial(n-1n) * n;
}
console.log(String(factorial(500n)));
You'll find that you computer can run that piece of code in a snap.

Hi this is due to the nature of java script as it can't represents number above 253-1 reference so to solve this either wrap the number with BigInt(n) or add to the number >> 3n
const factorial = (n) => {
n = BigInt(n)
if ( n < 1 ) return 1n
return factorial(n - 1n) * n
}

Something not working in this C code

I want to solve this problem and because I know JavaScript better than any other language I wrote it firstly in JavaScript and now I'm trying to translate it to C.
This is my JavaScript code
var SUM=0,RES=1,n=prompt('Enter n'),i=1,j=1;
for(var i=1;i<=n;i++){
SUM = 0;
for(var j=1;j<=i;j++){
SUM = SUM+ i/(2*j);
}
RES = RES * SUM
}
console.log(RES)
I think this is working well. I translated it to C and the result is:
#include "stdio.h"
int main(void)
{
int n,i,j;
float SUM=0,RES=1;
printf("n equals to ");
scanf("%d", &n);
for(i=1;i<=n;i++){
SUM = 0;
for(j=1;j<=i;j++){
SUM = SUM + i/(2*j);
}
RES = RES * SUM;
}
return RES;
}
The JavaScript at least returns a number. The C code always returns 0. Where is the problem?

Tricked by integer division (a very common problem)! If you change your one line to:
SUM = SUM + (float)i/(2*j);
The reason is because the result of integer division is another integer. So whenever
i/2*j < 1 (for positive integers)
your result gets truncated to 0. In your case that inequality is always true, so you're just adding up a whole lot of zeros. Just cast i to a float first and then divide to get a double result.

As noted in the comments you are doing integer devision in this line
SUM = SUM + i/(2*j);
which will return the result of an integer (like using the floor function)
you need to convert implicit variables in 1/(2*j) to a float or a double.
an integer/float is a float a integer/integer will return an integer.
a float times an integer is a float.
SUM = SUM + i/(2.0f*j);
should fix your problem by converting the denominator to a floating point number,

using the digits of a number as an array

var number = 342345820139586830203845861938475676
var output = []
var sum = 0;
while (number) {
output.push(number % 10);
number = Math.floor(number/10);
}
output = output.reverse();
function addTerms () {
for (i = 0; i < output.length; i=i+2) {
var term = Math.pow(output[i], output[i+1]);
sum += term;
}
return sum;
}
document.write(output);
document.write("<br>");
document.write(addTerms());
I am trying to take that large number and split it into its digits. Then, find the sum of the the first digit raised to the power of the 2nd, 3rd digit raiseed to the 4th, 5th raised to the 6th and so on. for some reason, my array is returning weird digits, causing my sum to be off. the correct answer is 2517052. Thanks

You're running into precision issues within JavaScript. Just evaluate the current value of number before you start doing anything, and the results may surprise you:
>>> var number = 342345820139586830203845861938475676; number;
3.423458201395868e+35
See also: What is JavaScript's highest integer value that a Number can go to without losing precision?
To resolve your issue, I'd store your input number as an array (or maybe even a string), then pull the digits off of that.
This will solve your calculation with the expected result of 2517052:
var number = "342345820139586830203845861938475676";
var sum = 0;
for(var i=0; i<number.length; i=i+2){
sum += Math.pow(number.charAt(i), number.charAt(i+1));
}
sum;

JavaScript stores numbers in floating point format (commonly double). double can store precisely only 15 digits.
You can use string to store this large number.

As mentioned, this is a problem with numeric precision. It applies to all programming languages that use native numeric formats. Your problem works fine if you use a string instead
var number = '342345820139586830203845861938475676'
var digits = number.split('')
var total = 0
while (digits.length > 1) {
var [n, power] = digits.splice(0, 2)
total += Math.pow(n, power)
}
(the result is 2517052, byt the way!)

Cast the number as a string and then iterate through it doing your math.
var number = "342345820139586830203845861938475676";//number definition
var X = 0;//some iterator
var numberAtX = 0 + number.charAt(X);//number access

The greatest integer supported by Javascript is 9007199254740992. So that only your output is weird.
For Reference go through the link http://ecma262-5.com/ELS5_HTML.htm#Section_8.5

[edit] adjusted the answer based on Borodins comment.
Mmm, I think the result should be 2517052. I'd say this does the same:
var numbers = '342345820139586830203845861938475676'.split('')
,num = numbers.splice(0,2)
,result = Math.pow(num[0],num[1]);
while ( (num = numbers.splice(0,2)) && num.length ){
result += Math.pow(num[0],num[1]);
}
console.log(result); //=> 2517052

The array methods map and reduce are supported in modern browsers,
and could be worth defining in older browsers. This is a good opportunity,
if you haven't used them before.
If you are going to make an array of a string anyway,
match pairs of digits instead of splitting to single digits.
This example takes numbers or strings.
function sumPower(s){
return String(s).match(/\d{2}/g).map(function(itm){
return Math.pow(itm.charAt(0), itm.charAt(1));
}).reduce(function(a, b){
return a+b;
});
}
sumPower('342345820139586830203845861938475676');
alert(sumPower(s))
/*
returned value:(Number)
2517052
*/

Math.random() returns value greater than one?

While playing around with random numbers in JavaScript I discovered a surprising bug, presumably in the V8 JavaScript engine in Google Chrome. Consider:
// Generate a random number [1,5].
var rand5 = function() {
return parseInt(Math.random() * 5) + 1;
};
// Return a sample distribution over MAX times.
var testRand5 = function(dist, max) {
if (!dist) { dist = {}; }
if (!max) { max = 5000000; }
for (var i=0; i<max; i++) {
var r = rand5();
dist[r] = (dist[r] || 0) + 1;
}
return dist;
};
Now when I run testRand5() I get the following results (of course, differing slightly with each run, you might need to set "max" to a higher value to reveal the bug):
var d = testRand5();
d = {
1: 1002797,
2: 998803,
3: 999541,
4: 1000851,
5: 998007,
10: 1 // XXX: Math.random() returned 4.5?!
}
Interestingly, I see comparable results in node.js, leading me to believe it's not specific to Chrome. Sometimes there are different or multiple mystery values (7, 9, etc).
Can anyone explain why I might be getting the results I see? I'm guessing it has something to do with using parseInt (instead of Math.floor()) but I'm still not sure why it could happen.

The edge case occurs when you happen to generate a very small number, expressed with an exponent, like this for example 9.546056389808655e-8.
Combined with parseInt, which interprets the argument as a string, hell breaks loose. And as suggested before me, it can be solved using Math.floor.
Try it yourself with this piece of code:
var test = 9.546056389808655e-8;
console.log(test); // prints 9.546056389808655e-8
console.log(parseInt(test)); // prints 9 - oh noes!
console.log(Math.floor(test)) // prints 0 - this is better

Of course, it's a parseInt() gotcha. It converts its argument to a string first, and that can force scientific notation which will cause parseInt to do something like this:
var x = 0.000000004;
(x).toString(); // => "4e-9"
parseInt(x); // => 4
Silly me...

I would suggest changing your random number function to this:
var rand5 = function() {
return(Math.floor(Math.random() * 5) + 1);
};
This will reliably generate an integer value between 1 and 5 inclusive.
You can see your test function in action here: http://jsfiddle.net/jfriend00/FCzjF/.
In this case, parseInt isn't the best choice because it's going to convert your float to a string which can be a number of different formats (including scientific notation) and then try to parse an integer out of it. Much better to just operate on the float directly with Math.floor().