Develop Reference

Javascript: how to get micronumbers in decimal format

I have simple function that calculates number of decimals,
eg. _d(0.01) = 2, _d(0.001) = 3 and so on.
We added some new coins to our system that have 0.00000001 quantity and function broke.
Here is why:
0.00000001.toString() = 1e-8, so I cant split it it by '.' and calculate length of second part as I did before.
So the question is - how to get string '0.00000001' out of 0.00000001 number easiest way.
EDIT
I didnt mean exactly '0.00000001', I meant any micronumber to decimal without exp. Some function _d(x) that would work _d(0.000000000012) = '0.000000000012'and so on. What usually toString() does to large (but not too large) numbers.

Use toFixed() with a large number of digits, then count the number of zeroes after the decimal point.
function _d(num) {
var str = num.toFixed(100);
var fraction = str.split('.')[1];
var zeros = fraction.match(/^0*/)[0].length;
return zeros + 1;
}
console.log(_d(0.1));
console.log(_d(0.01));
console.log(_d(0.000000001));

Do you want some thing like this
function decimalPlaces(num) {
var match = (''+num).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
if (!match) { return 0; }
return Math.max(
0,
// Number of digits right of decimal point.
(match[1] ? match[1].length : 0)
// Adjust for scientific notation.
- (match[2] ? +match[2] : 0));
}
console.log(decimalPlaces(0.000000001))

First off, I got some inspiration for this answer from here:
How to avoid scientific notation for large numbers in JavaScript?
You can convert the number to a strong and then check for str.indexOf("e"). If true, then just return the scientific notation part of the string. For example:
function _d() {
// your current function here
if (str.indexOf("e")) {
var something = str.split("-")[1];
return something;
}
}
EDIT: I was working on this before your last comment to me, so this returns a string of the number, which I thought was what you wanted.
Leaving aside the point about significant digits, which is meaningful and correct but does not solve your problem, try this. We take the number, convert to string, if that string is not scientific notation then the answer is trivial. If it is scientific notation, then split the string twice (once on "e-" and then split the zeroth array on "." Add str[1]-1 zeroes to the lead of the number and add the digits to the end.
function _d(arg) {
var str = arg.toString();
if (str.indexOf("e-")) {
var digits = str.split("e-")[0];
var zeroes = str.split("e-")[1];
var zero = Number(zeroes);
var each = digits.split(".");
var something = "0.";
for (var i = 0; i < zeroes-1; i++) {
something += "0";
}
for (var j = 0; j < each.length; j++) {
something = something + each[j];
}
return something;
}
}
This won't work with very large numbers or very small negative numbers. And its pretty convoluted.

The other way is to use .toString() and then look for .length-2(2 characters - '0.'. It should give you the number of zeros.
The advantage of this method is you don't need to know the number of maximum decimals in the number.

To keep it as the full decimal:
Number(0.000001)
// 0.000001
To show it as a string:
0.000001.toFixed(6)
// "0.000001"

JavaScript displaying a float to 2 decimal places

I wanted to display a number to 2 decimal places.
I thought I could use toPrecision(2) in JavaScript .
However, if the number is 0.05, I get 0.0500. I'd rather it stay the same.
See it on JSbin.
What is the best way to do this?
I can think of coding a few solutions, but I'd imagine (I hope) something like this is built in?

float_num.toFixed(2);
Note:toFixed() will round or pad with zeros if necessary to meet the specified length.

You could do it with the toFixed function, but it's buggy in IE. If you want a reliable solution, look at my answer here.

number.parseFloat(2) works but it returns a string.
If you'd like to preserve it as a number type you can use:
Math.round(number * 100) / 100

Don't know how I got to this question, but even if it's many years since this has been asked, I would like to add a quick and simple method I follow and it has never let me down:
var num = response_from_a_function_or_something();
var fixedNum = parseFloat(num).toFixed( 2 );

with toFixed you can set length of decimal points like this:
let number = 6.1234
number.toFixed(2) // '6.12'
but toFixed returns a string and also if number doesn't have decimal point at all it will add redundant zeros.
let number = 6
number.toFixed(2) // '6.00'
to avoid this you have to convert the result to a number. you can do this with these two methods:
let number1 = 6
let number2 = 6.1234
// method 1
parseFloat(number1.toFixed(2)) // 6
parseFloat(number2.toFixed(2)) // 6.12
// method 2
+number1.toFixed(2) // 6
+number2.toFixed(2) // 6.12

Try toFixed instead of toPrecision.

function round(value, decimals) {
return Number(Math.round(value+'e'+decimals)+'e-'+decimals);
}
round(1.005, 2); // return 1.01
round(1.004, 2); // return 1 instead of 1.00
The answer is following this link: http://www.jacklmoore.com/notes/rounding-in-javascript/

I used this way if you need 2 digits and not string type.
const exFloat = 3.14159265359;
console.log(parseFloat(exFloat.toFixed(2)));

You could try mixing Number() and toFixed().
Have your target number converted to a nice string with X digits then convert the formated string to a number.
Number( (myVar).toFixed(2) )
See example below:
var myNumber = 5.01;
var multiplier = 5;
$('#actionButton').on('click', function() {
$('#message').text( myNumber * multiplier );
});
$('#actionButton2').on('click', function() {
$('#message').text( Number( (myNumber * multiplier).toFixed(2) ) );
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<button id="actionButton">Weird numbers</button>
<button id="actionButton2">Nice numbers</button>
<div id="message"></div>

The toFixed() method formats a number using fixed-point notation.
and here is the syntax
numObj.toFixed([digits])
digits argument is optional and by default is 0. And the return type is string not number. But you can convert it to number using
numObj.toFixed([digits]) * 1
It also can throws exceptions like TypeError, RangeError
Here is the full detail and compatibility in the browser.

let a = 0.0500
a.toFixed(2);
//output
0.05

There's also the Intl API to format decimals according to your locale value. This is important specially if the decimal separator isn't a dot "." but a comma "," instead, like it is the case in Germany.
Intl.NumberFormat('de-DE').formatToParts(0.05).reduce((acc, {value}) => acc += value, '');
Note that this will round to a maximum of 3 decimal places, just like the round() function suggested above in the default case. If you want to customize that behavior to specify the number of decimal places, there're options for minimum and maximum fraction digits:
Intl.NumberFormat('de-DE', {minimumFractionDigits: 3}).formatToParts(0.05)

float_num = parseFloat(float_num.toFixed(2))

I have made this function. It works fine but returns string.
function show_float_val(val,upto = 2){
var val = parseFloat(val);
return val.toFixed(upto);
}

Is there "0b" or something similar to represent a binary number in Javascript

I know that 0x is a prefix for hexadecimal numbers in Javascript. For example, 0xFF stands for the number 255.
Is there something similar for binary numbers ? I would expect 0b1111 to represent the number 15, but this doesn't work for me.

Update:
Newer versions of JavaScript -- specifically ECMAScript 6 -- have added support for binary (prefix 0b), octal (prefix 0o) and hexadecimal (prefix: 0x) numeric literals:
var bin = 0b1111; // bin will be set to 15
var oct = 0o17; // oct will be set to 15
var oxx = 017; // oxx will be set to 15
var hex = 0xF; // hex will be set to 15
// note: bB oO xX are all valid
This feature is already available in Firefox and Chrome. It's not currently supported in IE, but apparently will be when Spartan arrives.
(Thanks to Semicolon's comment and urish's answer for pointing this out.)
Original Answer:
No, there isn't an equivalent for binary numbers. JavaScript only supports numeric literals in decimal (no prefix), hexadecimal (prefix 0x) and octal (prefix 0) formats.
One possible alternative is to pass a binary string to the parseInt method along with the radix:
var foo = parseInt('1111', 2); // foo will be set to 15

In ECMASCript 6 this will be supported as a part of the language, i.e. 0b1111 === 15 is true. You can also use an uppercase B (e.g. 0B1111).
Look for NumericLiterals in the ES6 Spec.

I know that people says that extending the prototypes is not a good idea, but been your script...
I do it this way:
Object.defineProperty(
Number.prototype, 'b', {
set:function(){
return false;
},
get:function(){
return parseInt(this, 2);
}
}
);
100..b // returns 4
11111111..b // returns 511
10..b+1 // returns 3
// and so on

If your primary concern is display rather than coding, there's a built-in conversion system you can use:
var num = 255;
document.writeln(num.toString(16)); // Outputs: "ff"
document.writeln(num.toString(8)); // Outputs: "377"
document.writeln(num.toString(2)); // Outputs: "11111111"
Ref: MDN on Number.prototype.toString

As far as I know it is not possible to use a binary denoter in Javascript. I have three solutions for you, all of which have their issues. I think alternative 3 is the most "good looking" for readability, and it is possibly much faster than the rest - except for it's initial run time cost. The problem is it only supports values up to 255.
Alternative 1: "00001111".b()
String.prototype.b = function() { return parseInt(this,2); }
Alternative 2: b("00001111")
function b(i) { if(typeof i=='string') return parseInt(i,2); throw "Expects string"; }
Alternative 3: b00001111
This version allows you to type either 8 digit binary b00000000, 4 digit b0000 and variable digits b0. That is b01 is illegal, you have to use b0001 or b1.
String.prototype.lpad = function(padString, length) {
var str = this;
while (str.length < length)
str = padString + str;
return str;
}
for(var i = 0; i < 256; i++)
window['b' + i.toString(2)] = window['b' + i.toString(2).lpad('0', 8)] = window['b' + i.toString(2).lpad('0', 4)] = i;

May be this will usefull:
var bin = 1111;
var dec = parseInt(bin, 2);
// 15

No, but you can use parseInt and optionally omit the quotes.
parseInt(110, 2); // this is 6
parseInt("110", 2); // this is also 6
The only disadvantage of omitting the quotes is that, for very large numbers, you will overflow faster:
parseInt(10000000000000000000000, 2); // this gives 1
parseInt("10000000000000000000000", 2); // this gives 4194304

I know this does not actually answer the asked Q (which was already answered several times) as is, however I suggest that you (or others interested in this subject) consider the fact that the most readable & backwards/future/cross browser-compatible way would be to just use the hex representation.
From the phrasing of the Q it would seem that you are only talking about using binary literals in your code and not processing of binary representations of numeric values (for which parstInt is the way to go).
I doubt that there are many programmers that need to handle binary numbers that are not familiar with the mapping of 0-F to 0000-1111.
so basically make groups of four and use hex notation.
so instead of writing 101000000010 you would use 0xA02 which has exactly the same meaning and is far more readable and less less likely to have errors.
Just consider readability, Try comparing which of those is bigger:
10001000000010010 or 1001000000010010
and what if I write them like this:
0x11012 or 0x9012

Convert binary strings to numbers and visa-versa.
var b = function(n) {
if(typeof n === 'string')
return parseInt(n, 2);
else if (typeof n === 'number')
return n.toString(2);
throw "unknown input";
};

Using Number() function works...
// using Number()
var bin = Number('0b1111'); // bin will be set to 15
var oct = Number('0o17'); // oct will be set to 15
var oxx = Number('0xF'); // hex will be set to 15
// making function convTo
const convTo = (prefix,n) => {
return Number(`${prefix}${n}`) //Here put prefix 0b, 0x and num
}
console.log(bin)
console.log(oct)
console.log(oxx)
// Using convTo function
console.log(convTo('0b',1111))

bitwise AND in Javascript with a 64 bit integer

I am looking for a way of performing a bitwise AND on a 64 bit integer in JavaScript.
JavaScript will cast all of its double values into signed 32-bit integers to do the bitwise operations (details here).

Javascript represents all numbers as 64-bit double precision IEEE 754 floating point numbers (see the ECMAscript spec, section 8.5.) All positive integers up to 2^53 can be encoded precisely. Larger integers get their least significant bits clipped. This leaves the question of how can you even represent a 64-bit integer in Javascript -- the native number data type clearly can't precisely represent a 64-bit int.
The following illustrates this. Although javascript appears to be able to parse hexadecimal numbers representing 64-bit numbers, the underlying numeric representation does not hold 64 bits. Try the following in your browser:
<html>
<head>
<script language="javascript">
function showPrecisionLimits() {
document.getElementById("r50").innerHTML = 0x0004000000000001 - 0x0004000000000000;
document.getElementById("r51").innerHTML = 0x0008000000000001 - 0x0008000000000000;
document.getElementById("r52").innerHTML = 0x0010000000000001 - 0x0010000000000000;
document.getElementById("r53").innerHTML = 0x0020000000000001 - 0x0020000000000000;
document.getElementById("r54").innerHTML = 0x0040000000000001 - 0x0040000000000000;
}
</script>
</head>
<body onload="showPrecisionLimits()">
<p>(2^50+1) - (2^50) = <span id="r50"></span></p>
<p>(2^51+1) - (2^51) = <span id="r51"></span></p>
<p>(2^52+1) - (2^52) = <span id="r52"></span></p>
<p>(2^53+1) - (2^53) = <span id="r53"></span></p>
<p>(2^54+1) - (2^54) = <span id="r54"></span></p>
</body>
</html>
In Firefox, Chrome and IE I'm getting the following. If numbers were stored in their full 64-bit glory, the result should have been 1 for all the substractions. Instead, you can see how the difference between 2^53+1 and 2^53 is lost.
(2^50+1) - (2^50) = 1
(2^51+1) - (2^51) = 1
(2^52+1) - (2^52) = 1
(2^53+1) - (2^53) = 0
(2^54+1) - (2^54) = 0
So what can you do?
If you choose to represent a 64-bit integer as two 32-bit numbers, then applying a bitwise AND is as simple as applying 2 bitwise AND's, to the low and high 32-bit 'words'.
For example:
var a = [ 0x0000ffff, 0xffff0000 ];
var b = [ 0x00ffff00, 0x00ffff00 ];
var c = [ a[0] & b[0], a[1] & b[1] ];
document.body.innerHTML = c[0].toString(16) + ":" + c[1].toString(16);
gets you:
ff00:ff0000

Here is code for AND int64 numbers, you can replace AND with other bitwise operation
function and(v1, v2) {
var hi = 0x80000000;
var low = 0x7fffffff;
var hi1 = ~~(v1 / hi);
var hi2 = ~~(v2 / hi);
var low1 = v1 & low;
var low2 = v2 & low;
var h = hi1 & hi2;
var l = low1 & low2;
return h*hi + l;
}

This can now be done with the new BigInt built-in numeric type. BigInt is currently (July 2019) only available in certain browsers, see the following link for details:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/BigInt
I have tested bitwise operations using BigInts in Chrome 67 and can confirm that they work as expected with up to 64 bit values.

Javascript doesn't support 64 bit integers out of the box. This is what I ended up doing:
Found long.js, a self contained Long implementation on github.
Convert the string value representing the 64 bit number to a Long.
Extract the high and low 32 bit values
Do a 32 bit bitwise and between the high and low bits, separately
Initialise a new 64 bit Long from the low and high bit
If the number is > 0 then there is correlation between the two numbers
Note: for the code example below to work you need to load
long.js.
// Handy to output leading zeros to make it easier to compare the bits when outputting to the console
function zeroPad(num, places){
var zero = places - num.length + 1;
return Array(+(zero > 0 && zero)).join('0') + num;
}
// 2^3 = 8
var val1 = Long.fromString('8', 10);
var val1High = val1.getHighBitsUnsigned();
var val1Low = val1.getLowBitsUnsigned();
// 2^61 = 2305843009213693960
var val2 = Long.fromString('2305843009213693960', 10);
var val2High = val2.getHighBitsUnsigned();
var val2Low = val2.getLowBitsUnsigned();
console.log('2^3 & (2^3 + 2^63)')
console.log(zeroPad(val1.toString(2), 64));
console.log(zeroPad(val2.toString(2), 64));
var bitwiseAndResult = Long.fromBits(val1Low & val2Low, val1High & val2High, true);
console.log(bitwiseAndResult);
console.log(zeroPad(bitwiseAndResult.toString(2), 64));
console.log('Correlation betwen val1 and val2 ?');
console.log(bitwiseAndResult > 0);
Console output:
2^3
0000000000000000000000000000000000000000000000000000000000001000
2^3 + 2^63
0010000000000000000000000000000000000000000000000000000000001000
2^3 & (2^3 + 2^63)
0000000000000000000000000000000000000000000000000000000000001000
Correlation between val1 and val2?
true

The Closure library has goog.math.Long with a bitwise add() method.

Unfortunately, the accepted answer (and others) appears not to have been adequately tested. Confronted by this problem recently, I initially tried to split my 64-bit numbers into two 32-bit numbers as suggested, but there's another little wrinkle.
Open your JavaScript console and enter:
0x80000001
When you press Enter, you'll obtain 2147483649, the decimal equivalent. Next try:
0x80000001 & 0x80000003
This gives you -2147483647, not quite what you expected. It's clear that in performing the bitwise AND, the numbers are treated as signed 32-bit integers. And the result is wrong. Even if you negate it.
My solution was to apply ~~ to the 32-bit numbers after they were split off, check for a negative sign, and then deal with this appropriately.
This is clumsy. There may be a more elegant 'fix', but I can't see it on quick examination. There's a certain irony that something that can be accomplished by a couple of lines of assembly should require so much more labour in JavaScript.

How can I round a number in JavaScript? .toFixed() returns a string?

Am I missing something here?
var someNumber = 123.456;
someNumber = someNumber.toFixed(2);
alert(typeof(someNumber));
//alerts string
Why does .toFixed() return a string?
I want to round the number to 2 decimal digits.

Number.prototype.toFixed is a function designed to format a number before printing it out. It's from the family of toString, toExponential and toPrecision.
To round a number, you would do this:
someNumber = 42.008;
someNumber = Math.round( someNumber * 1e2 ) / 1e2;
someNumber === 42.01;
// if you need 3 digits, replace 1e2 with 1e3 etc.
// or just copypaste this function to your code:
function toFixedNumber(num, digits, base){
var pow = Math.pow(base||10, digits);
return Math.round(num*pow) / pow;
}
.
Or if you want a “native-like” function, you can extend the prototype:
Number.prototype.toFixedNumber = function(digits, base){
var pow = Math.pow(base||10, digits);
return Math.round(this*pow) / pow;
}
someNumber = 42.008;
someNumber = someNumber.toFixedNumber(2);
someNumber === 42.01;
//or even hexadecimal
someNumber = 0xAF309/256; //which is af3.09
someNumber = someNumber.toFixedNumber(1, 16);
someNumber.toString(16) === "af3.1";
However, bear in mind that polluting the prototype is considered bad when you're writing a module, as modules shouldn't have any side effects. So, for a module, use the first function.

I've solved this problem by changing this:
someNumber = someNumber.toFixed(2)
...to this:
someNumber = +someNumber.toFixed(2);
However this will convert the number to a string and parse it again, which will have a significant impact on performance. If you care about performance or type safety, check the the other answers as well.

It returns a string because 0.1, and powers thereof (which are used to display decimal fractions), are not representable (at least not with full accuracy) in binary floating-point systems.
For example, 0.1 is really 0.1000000000000000055511151231257827021181583404541015625, and 0.01 is really 0.01000000000000000020816681711721685132943093776702880859375. (Thanks to BigDecimal for proving my point. :-P)
Therefore (absent a decimal floating point or rational number type), outputting it as a string is the only way to get it trimmed to exactly the precision required for display.

Why not use parseFloat?
var someNumber = 123.456;
someNumber = parseFloat(someNumber.toFixed(2));
alert(typeof(someNumber));
//alerts number

I solved it with converting it back to number using JavaScript's Number() function
var x = 2.2873424;
x = Number(x.toFixed(2));

Of course it returns a string. If you wanted to round the numeric variable you'd use Math.round() instead. The point of toFixed is to format the number with a fixed number of decimal places for display to the user.

You can simply use a '+' to convert the result to a number.
var x = 22.032423;
x = +x.toFixed(2); // x = 22.03

May be too late to answer but you can multiple the output with 1 to convert to number again, here is an example.
const x1 = 1211.1212121;
const x2 = x1.toFixed(2)*1;
console.log(typeof(x2));

What would you expect it to return when it's supposed to format a number ? If you have a number you can't pretty much do anything with it because e.g.2 == 2.0 == 2.00 etc. so it has to be a string.

Because its primary use is displaying numbers? If you want to round numbers, use Math.round() with apropriate factors.

To supply an example of why it has to be a string:
If you format 1.toFixed(2) you would get '1.00'.
This is not the same as 1, as 1 does not have 2 decimals.
I know JavaScript isn't exactly a performance language, but chances are you'd get better performance for a rounding if you use something like:
roundedValue = Math.round(value * 100) * 0.01

You should use it like below.
var someNumber: number = 0.000000;
someNumber = Number(someNumber.toFixed(2))

Why not * the result by 1 i.e
someNumber.toFixed(2) * 1

Here's a slightly more functional version of the answer m93a provided.
const toFixedNumber = (toFixTo = 2, base = 10) => num => {
const pow = Math.pow(base, toFixTo)
return +(Math.round(num * pow) / pow)
}
const oneNumber = 10.12323223
const result1 = toFixedNumber(2)(oneNumber) // 10.12
const result2 = toFixedNumber(3)(oneNumber) // 10.123
// or using pipeline-operator
const result3 = oneNumber |> toFixedNumber(2) // 10.12

For others like me that happen upon this very old question, a modern solution:
const roundValue = (num, decimals = 2) => {
let scaling = 10 ** decimals;
return Math.round((num + Number.EPSILON) * scaling) / scaling;
}
ref: https://stackoverflow.com/a/11832950

Be careful using toFixed() and Math.round(), they can produce unexpected results due to the floating point number system:
function toFixedNumber(num, digits, base){
var pow = Math.pow(base||10, digits);
return Math.round(num*pow) / pow;
}
console.log(toFixedNumber(130.795, 2, 10));
// 130.79 (incorrect)
console.log(toFixedNumber(100.795, 2, 10));
// 100.8
console.log(+130.795.toFixed(2));
// 130.79 (incorrect)
console.log(+100.795.toFixed(2));
// 100.8
I recommend using Lodash's _.round() function: https://lodash.com/docs/4.17.15#round
_.round(130.795, 2);
// 130.8