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Why doesn't my function correctly replace when using some regex pattern

This is an extension of this SO question
I made a function to see if i can correctly format any number. The answers below work on tools like https://regex101.com and https://regexr.com/, but not within my function(tried in node and browser):
const
const format = (num, regex) => String(num).replace(regex, '$1')
Basically given any whole number, it should not exceed 15 significant digits. Given any decimal, it should not exceed 2 decimal points.
so...
Now
format(0.12345678901234567890, /^\d{1,13}(\.\d{1,2}|\d{0,2})$/)
returns 0.123456789012345678 instead of 0.123456789012345
but
format(0.123456789012345,/^-?(\d*\.?\d{0,2}).*/)
returns number formatted to 2 deimal points as expected.

Let me try to explain what's going on.
For the given input 0.12345678901234567890 and the regex /^\d{1,13}(\.\d{1,2}|\d{0,2})$/, let's go step by step and see what's happening.
^\d{1,13} Does indeed match the start of the string 0
(\. Now you've opened a new group, and it does match .
\d{1,2} It does find the digits 1 and 2
|\d{0,2} So this part is skipped
) So this is the end of your capture group.
$ This indicates the end of the string, but it won't match, because you've still got 345678901234567890 remaining.
Javascript returns the whole string because the match failed in the end.
Let's try removing $ at the end, to become /^\d{1,13}(\.\d{1,2}|\d{0,2})/
You'd get back ".12345678901234567890". This generates a couple of questions.
Why did the preceding 0 get removed?
Because it was not part of your matching group, enclosed with ().
Why did we not get only two decimal places, i.e. .12?
Remember that you're doing a replace. Which means that by default, the original string will be kept in place, only the parts that match will get replaced. Since 345678901234567890 was not part of the match, it was left intact. The only part that matched was 0.12.

Answer to title question: your function doesn't replace, because there's nothing to replace - the regex doesn't match anything in the string. csb's answer explains that in all details.
But that's perhaps not the answer you really need.
Now, it seems like you have an XY problem. You ask why your call to .replace() doesn't work, but .replace() is definitely not a function you should use. Role of .replace() is replacing parts of string, while you actually want to create a different string. Moreover, in the comments you suggest that your formatting is not only for presenting data to user, but you also intend to use it in some further computation. You also mention cryptocurriencies.
Let's cope with these problems one-by-one.
What to do instead of replace?
Well, just produce the string you need instead of replacing something in the string you don't like. There are some edge cases. Instead of writing all-in-one regex, just handle them one-by-one.
The following code is definitely not best possible, but it's main aim is to be simple and show exactly what is going on.
function format(n) {
const max_significant_digits = 15;
const max_precision = 2;
let digits_before_decimal_point;
if (n < 0) {
// Don't count minus sign.
digits_before_decimal_point = n.toFixed(0).length - 1;
} else {
digits_before_decimal_point = n.toFixed(0).length;
}
if (digits_before_decimal_point > max_significant_digits) {
throw new Error('No good representation for this number');
}
const available_significant_digits_for_precision =
Math.max(0, max_significant_digits - digits_before_decimal_point);
const effective_max_precision =
Math.min(max_precision, available_significant_digits_for_precision);
const with_trailing_zeroes = n.toFixed(effective_max_precision);
// I want to keep the string and change just matching part,
// so here .replace() is a proper method to use.
const withouth_trailing_zeroes = with_trailing_zeroes.replace(/\.?0*$/, '');
return withouth_trailing_zeroes;
}
So, you got the number formatted the way you want. What now?
What can you use this string for?
Well, you can display it to the user. And that's mostly it. The value was rounded to (1) represent it in a different base and (2) fit in limited precision, so it's pretty much useless for any computation. And, BTW, why would you convert it to String in the first place, if what you want is a number?
Was the value you are trying to print ever useful in the first place?
Well, that's the most serious question here. Because, you know, floating point numbers are tricky. And they are absolutely abysmal for representing money. So, most likely the number you are trying to format is already a wrong number.
What to use instead?
Fixed-point arithmetic is the most obvious answer. Works most of the time. However, it's pretty tricky in JS, where number may slip into floating-point representation almost any time. So, it's better to use decimal arithmetic library. Optionally, switch to a language that has built-in bignums and decimals, like Python.

how to have one decimal point for numbers using angularJS

i've seen a few things for regExp but for some reason i'm thinking this should be more simple. Maybe not. Maybe regExp fits this perfectly and i don't understand the concept. Which, i do not. I don't understand regExp and thats why i haven't used it.
but what i am trying to do is build a basic calculator using AngularJS. My calculator has an ng-disable feature for my equals sign. So its not allowed to be pressed and won't work until a leftOperand, operator, and rightOperand are clicked.
Now i don't want to disable the decimal when its clicked, because i might need it for my rightOperand.
But in essence i want my code to reflect "if there is an operator, then the rightOperand is rightOperand + any other Operand entered. Otherwise, the leftOperand is going to equal the leftOperand + any other Operand Entered. (which this is currently working ex. 333 + 45948594 works. The decimal works too. But! it can allow 333.464.6454 and i don't want that).
Trying to implement:
Also/And/Or/If a decimal is put into the leftOperand, then no other decimals are allowed for the leftOperand. If the rightOperand has a decimal, then no other decimals are allowed for the rightOperand.
what is the simplest way to go about this?
https://codepen.io/tevon/pen/Moewba
this is my setDecimal method, but its allowing more than 1 decimal for both the right and left operands
$scope.setDecimal = function (decimalEntered) {
if ($scope.operator){
$scope.rightOperand += decimalEntered;
};
else {$scope.leftOperand += decimalEntered;
};
i was givin something like this: if string.indexOf('.') > -1 // then do something
but honestly, i'm not sure how to about the 'do something' part works, nor the -1 portion
not using Jquery.

So you're building your operand as a string. That means you can test the operand you currently have on your scope to see if it has a point using .indexOf.
if (myVariable.indexOf('.') === -1) myVariable += '.';

You were right when you thought about indexOf - you just need to stop execution of this method if there's already dot there - and easiest way is return
if ($scope.rightOperand.indexOf('.')>-1) return;
LINK

Is there a way to distinguish integers from very near decimals in Javascript?

Look at those evaluations (actual dump from node 0.10.33)
> parseFloat(2.1e-17) === parseInt(2.1e-17)
false
> parseFloat(2.1e-17 + 2) === parseInt(2.1e-17 + 2)
true
> parseFloat(2.000000000000000000000000000000000009) === parseInt(2.00000000000000000000000000000000000009)
true
How can I tell integers from decimals very near to integers?
It seems that JS (or at least V8) doesn't care about digits smaller than 10^-16 when doing calculations, even if the 64bit representation used by the language (reference) should handle it.

Your examples are pretty much straight forward to explain. First thing to note is, that parseInt() and parseFloat() take a string as an input. So you inputs first get converted to string, before actually getting parsed.
The first is easy to see:
> parseFloat(2.1e-17) === parseInt(2.1e-17)
false
// look at the result of each side
parseFloat(2.1e-17) == 2.1e-17
parseInt(2.1e-17) == 2
When parsing the string "2.1e-17" as integer, the parse will stop at the dot as that is no valid digit and return everything it found until then, which is just 2.
> parseFloat(2.1e-17 + 2) === parseInt(2.1e-17 + 2)
true
// look at the result of each side
parseFloat(2.1e-17 + 2) == 2
parseInt(2.1e-17 + 2) == 2
Here the formula in the parameter will be evaluated first. Due to the limitations of floating point math (we just have 52bit for the mantissa and can't represent something like 2.000000000000000021), this will result in just 2. So both parseX() function get the same integer parameter, which will result in the same parsed number.
> parseFloat(2.000000000000000000000000000000000009) === parseInt(2.00000000000000000000000000000000000009)
true
Same argument as for the second case. The only difference is, that instead of a formula, that gets evaluated, this time it is the JavaScript parser, which converts your numbers just to 2.
So to sum up: From JavaScript's point of view, your numbers are just the same. If you need more precision, you will have to use some library for arbitrary precision.

This is something I learned from ReSharper
instead of using expressions like
if (2.00001 == 2) {}
try
if (Math.abs(2.00001 - 2) < tolerance) {}
where tolerance should be an aceptable value for you for example .001
so all values wich difference is less than .001 will be equals
Do you really need 10^-16 precision I mean that is why 1000 meter = 1 kilometer, just change the unit of the output so you dont have to work with all those decimals

Working with string (array?) of bits of an unspecified length

I'm a javascript code monkey, so this is virgin territory for me.
I have two "strings" that are just zeros and ones:
var first = "00110101011101010010101110100101010101010101010";
var second = "11001010100010101101010001011010101010101010101";
I want to perform a bitwise & (which I've never before worked with) to determine if there's any index where 1 appears in both strings.
These could potentially be VERY long strings (in the thousands of characters). I thought about adding them together as numbers, then converting to strings and checking for a 2, but javascript can't hold precision in large intervals and I get back numbers as strings like "1.1111111118215729e+95", which doesn't really do me much good.
Can I take two strings of unspecified length (they may not be the same length either) and somehow use a bitwise & to compare them?
I've already built the loop-through-each-character solution, but 1001^0110 would strike me as a major performance upgrade. Please do not give the javascript looping solution as an answer, this question is about using bitwise operators.

As you already noticed yourself, javascript has limited capabilities if it's about integer values. You'll have to chop your strings into "edible" portions and work your way through them. Since the parseInt() function accepts a base, you could convert 64 characters to an 8 byte int (or 32 to a 4 byte int) and use an and-operator to test for set bits (if (a & b != 0))

var first = "00110101011101010010101110100101010101010101010010001001010001010100011111",
second = "10110101011101010010101110100101010101010101010010001001010001010100011100",
firstInt = parseInt(first, 2),
secondInt = parseInt(second, 2),
xorResult = firstInt ^ secondInt, //524288
xorString = xorResult.toString(2); //"10000000000000000000"

JavaScript Math.floor: how guarantee number will round down?

I want to normalize an array so that each value is
in [0-1) .. i.e. "the max will never be 1 but the min can be 0."
This is not unlike the random function returning numbers in the same range.
While looking at this, I found that .99999999999999999===1 is true!
Ditto (1-Number.MIN_VALUE) === 1 But Math.ceil(Number.MIN_VALUE) is 1, as it should be.
Some others: Math.floor(.999999999999) is 0
while Math.floor(.99999999999999999) is 1
OK so there are rounding problems in JS.
Is there any way I can normalize a set of numbers to lie in the range [0,1)?

It may help to examine the steps that JavaScript performs of each of your expressions.
In .99999999999999999===1:
The source text .99999999999999999 is converted to a Number. The closest Number is 1, so that is the result. (The next closest Number is 0.99999999999999988897769753748434595763683319091796875, which is 1–2–53.)
Then 1 is compared to 1. The result is true.
In (1-Number.MIN_VALUE) === 1:
Number.MIN_VALUE is 2–1074, about 5e–304.
1–2–1074 is extremely close to one. The exact value cannot be represented as a Number, so the nearest value is used. Again, the nearest value is 1.
Then 1 is compared to 1. The result is true.
In Math.ceil(Number.MIN_VALUE):
Number.MIN_VALUE is 2–1074, about 5e–304.
The ceiling function of that value is 1.
In Math.floor(.999999999999):
The source text .999999999999 is converted to a Number. The closest Number is 0.99999999999900002212172012150404043495655059814453125, so that is the result.
The floor function of that value is 0.
In Math.floor(.99999999999999999):
The source text .99999999999999999 is converted to a Number. The closest Number is 1, so that is the result.
The floor function of 1 is 1.
There are only two surprising things here, at most. One is that the numerals in the source text are converted to internal Number values. But this should not be surprising. Of course text has to be converted to internal representations of numbers, and the Number type cannot perfectly store all the infinitely many numbers. So it has to round. And of course numbers very near 1 round to 1.
The other possibly surprising thing is that 1-Number.MIN_VALUE is 1. But this is actually the same issue: The exact result is not representable, but it is very near 1, so 1 is used.
The Math.floor function works correctly. It never introduces any error, and you do not have to do anything to guarantee that it will round down. It always does.
However, since you want to normalize numbers, it seems likely you are going to divide numbers at some point. When you divide, there may be rounding problems, because many results of division are not exactly representable, so they must be rounded.
However, that is a separate problem, and you have not given enough information in this question to address the specific calculations you plan to do. You should open a separate question for it.

Javascript will treat any number between 0.999999999999999994 and 1 as 1, so just subtract .000000000000000006.
Of course that's not as easy as it sounds, since .000000000000000006 is evaluated as 0 in Javascript, so you could do something like:
function trueFloor(x)
{
x = x * 100;
if(x > .0000000000000006)
x = x - .0000000000000006;
x = Math.floor(x/100);
return x;
}
EDIT: Or at least you'd think you could. Apparently JS casts .99999999999999999 to 1 before passing it to a function, so you'd have to try something like:
trueFloor("0.99999999999999999")
function trueFloor(str)
{
x=str.substring(0,9) + 0;
return Math.floor(x); //=> 0
}
Not sure why you'd need that level of precision, but in theory, I guess it works. You can see a working fiddle here
As long as you cast your insanely precise float as a string, that's probably your best bet.

Please understand one thing: this...
.999999999999999999
... is just a Number literal. Just as
.999999999999999998
.999999999999999997
.999999999999999996
...
... you see the pattern.
How JavaScript treats these literals is completely another story. And yes, this treatment is limited by the number of bits that can be used to store a Number value.
The number of possible floating point literals is infinite by definition - no matter how small is the range set for them. For example, take the ones shown above: how many of numbers very close to 1 you may express? Right, it's infinite: just keep appending 9 to the line.
But the container for each Number value is quite finite: it has 64 bits. That means, it can store 2^64 different values (Infinite, -Infinite and NaN among them) - and that's all.
You want to work with such literals anyway? Use Strings to store them, not Numbers - and some BigMath JS library (take your pick) to work with those values - as Strings, again.
But from your question it looks like you're not, as you talked about array of Numbers - Number values, that is. And in no way there can be .999999999999999999 stored there, as there is no such Number value in JavaScript.