In modern JavaScript, there are two types of numbers:
Regular numbers in JavaScript are stored in 64-bit format IEEE-754, also known as “double precision floating point numbers”. These are numbers that we’re using most of the time, and we’ll talk about them in this chapter.
BigInt numbers represent integers of arbitrary length. They are sometimes needed because a regular integer number can’t safely exceed (253-1) or be less than -(253-1), as we mentioned earlier in the chapter Data types. As bigints are used in a few special areas, we devote them to a special chapter BigInt.
So here we’ll talk about regular numbers. Let’s expand our knowledge of them.
Imagine we need to write 1 billion. The obvious way is:
let billion = 1000000000;
We also can use underscore _ as the separator:
let billion = 1_000_000_000;
Here the underscore _ plays the role of the “syntactic sugar”, it makes the number more readable. The JavaScript engine simply ignores _ between digits, so it’s exactly the same one billion as above.
In real life though, we try to avoid writing long sequences of zeroes. We’re too lazy for that. We’ll try to write something like "1bn" for a billion or "7.3bn" for 7 billion 300 million. The same is true for most large numbers.
In JavaScript, we can shorten a number by appending the letter "e" to it and specifying the zeroes count:
let billion = 1e9; // 1 billion, literally: 1 and 9 zeroes alert( 7.3e9 ); // 7.3 billions (same as 7300000000 or 7_300_000_000)
In other words, e multiplies the number by 1 with the given zeroes count.
1e3 === 1 * 1000; // e3 means *1000 1.23e6 === 1.23 * 1000000; // e6 means *1000000
Now let’s write something very small. Say, 1 microsecond (one-millionth of a second):
let mсs = 0.000001;
Just like before, using "e" can help. If we’d like to avoid writing the zeroes explicitly, we could write the same as:
let mcs = 1e-6; // five zeroes to the left from 1
If we count the zeroes in 0.000001, there are 6 of them. So naturally it’s 1e-6.
In other words, a negative number after "e" means a division by 1 with the given number of zeroes:
// -3 divides by 1 with 3 zeroes 1e-3 === 1 / 1000; // 0.001 // -6 divides by 1 with 6 zeroes 1.23e-6 === 1.23 / 1000000; // 0.00000123 // an example with a bigger number 1234e-2 === 1234 / 100; // 12.34, decimal point moves 2 times
Hexadecimal numbers are widely used in JavaScript to represent colors, encode characters, and for many other things. So naturally, there exists a shorter way to write them: 0x and then the number.
For instance:
alert( 0xff ); // 255 alert( 0xFF ); // 255 (the same, case doesn't matter)
Binary and octal numeral systems are rarely used, but also supported using the 0b and 0o prefixes:
let a = 0b11111111; // binary form of 255 let b = 0o377; // octal form of 255 alert( a == b ); // true, the same number 255 at both sides
There are only 3 numeral systems with such support. For other numeral systems, we should use the function parseInt (which we will see later in this chapter).
The method num.toString(base) returns a string representation of num in the numeral system with the given base.
For example:
let num = 255; alert( num.toString(16) ); // ff alert( num.toString(2) ); // 11111111
The base can vary from 2 to 36. By default, it’s 10.
Common use cases for this are:
base=16 is used for hex colors, character encodings etc, digits can be 0..9 or A..F.
base=2 is mostly for debugging bitwise operations, digits can be 0 or 1.
base=36 is the maximum, digits can be 0..9 or A..Z. The whole Latin alphabet is used to represent a number. A funny, but useful case for 36 is when we need to turn a long numeric identifier into something shorter, for example, to make a short url. Can simply represent it in the numeral system with base 36:
alert( 123456..toString(36) ); // 2n9c
Two dots to call a method
Please note that two dots in 123456..toString(36) is not a typo. If we want to call a method directly on a number, like toString in the example above, then we need to place two dots .. after it.
If we placed a single dot: 123456.toString(36), then there would be an error, because JavaScript syntax implies the decimal part after the first dot. And if we place one more dot, then JavaScript knows that the decimal part is empty and now goes the method.
Also could write (123456).toString(36).
One of the most used operations when working with numbers is rounding.
There are several built-in functions for rounding:
Math.floor
Rounds down: 3.1 becomes 3, and -1.1 becomes -2.
Math.ceil
Rounds up: 3.1 becomes 4, and -1.1 becomes -1.
Math.round
Rounds to the nearest integer: 3.1 becomes 3, 3.6 becomes 4. In the middle cases 3.5 rounds up to 4, and -3.5 rounds up to -3.
Math.trunc (not supported by Internet Explorer)
Removes anything after the decimal point without rounding: 3.1 becomes 3, -1.1 becomes -1.
Here’s the table to summarize the differences between them:
Math.floor | Math.ceil | Math.round | Math.trunc | |
|---|---|---|---|---|
3.1 | 3 | 4 | 3 | 3 |
3.5 | 3 | 4 | 4 | 3 |
3.6 | 3 | 4 | 4 | 3 |
-1.1 | -2 | -1 | -1 | -1 |
-1.5 | -2 | -1 | -1 | -1 |
-1.6 | -2 | -1 | -2 | -1 |
These functions cover all of the possible ways to deal with the decimal part of a number. But what if we’d like to round the number to n-th digit after the decimal?
For instance, we have 1.2345 and want to round it to 2 digits, getting only 1.23.
There are two ways to do so:
Multiply-and-divide.
For example, to round the number to the 2nd digit after the decimal, we can multiply the number by 100, call the rounding function and then divide it back.
let num = 1.23456; alert( Math.round(num * 100) / 100 ); // 1.23456 -> 123.456 -> 123 -> 1.23
The method toFixed(n) rounds the number to n digits after the point and returns a string representation of the result.
let num = 12.34; alert( num.toFixed(1) ); // "12.3"
This rounds up or down to the nearest value, similar to Math.round:
let num = 12.36; alert( num.toFixed(1) ); // "12.4"
Please note that the result of toFixed is a string. If the decimal part is shorter than required, zeroes are appended to the end:
let num = 12.34; alert( num.toFixed(5) ); // "12.34000", added zeroes to make exactly 5 digits
We can convert it to a number using the unary plus or a Number() call, e.g. write +num.toFixed(5).
Internally, a number is represented in 64-bit format IEEE-754, so there are exactly 64 bits to store a number: 52 of them are used to store the digits, 11 of them store the position of the decimal point, and 1 bit is for the sign.
If a number is really huge, it may overflow the 64-bit storage and become a special numeric value Infinity:
alert( 1e500 ); // Infinity
What may be a little less obvious, but happens quite often, is the loss of precision.
Consider this (falsy!) equality test:
alert( 0.1 + 0.2 == 0.3 ); // false
That’s right, if we check whether the sum of 0.1 and 0.2 is 0.3, we get false.
Strange! What is it then if not 0.3?
alert( 0.1 + 0.2 ); // 0.30000000000000004
Ouch! Imagine you’re making an e-shopping site and the visitor puts $0.10 and $0.20 goods into their cart. The order total will be $0.30000000000000004. That would surprise anyone.
But why does this happen?
A number is stored in memory in its binary form, a sequence of bits – ones and zeroes. But fractions like 0.1, 0.2 that look simple in the decimal numeric system are actually unending fractions in their binary form.
alert(0.1.toString(2)); // 0.0001100110011001100110011001100110011001100110011001101 alert(0.2.toString(2)); // 0.001100110011001100110011001100110011001100110011001101 alert((0.1 + 0.2).toString(2)); // 0.0100110011001100110011001100110011001100110011001101
What is 0.1? It is one divided by ten 1/10, one-tenth. In the decimal numeral system, such numbers are easily representable. Compare it to one-third: 1/3. It becomes an endless fraction 0.33333(3).
So, division by powers 10 is guaranteed to work well in the decimal system, but division by 3 is not. For the same reason, in the binary numeral system, the division by powers of 2 is guaranteed to work, but 1/10 becomes an endless binary fraction.
There’s just no way to store exactly 0.1 or exactly 0.2 using the binary system, just like there is no way to store one-third as a decimal fraction.
The numeric format IEEE-754 solves this by rounding to the nearest possible number. These rounding rules normally don’t allow us to see that “tiny precision loss”, but it exists.
We can see this in action:
alert( 0.1.toFixed(20) ); // 0.10000000000000000555
And when we sum two numbers, their “precision losses” add up.
That’s why 0.1 + 0.2 is not exactly 0.3.
Not only JavaScript
The same issue exists in many other programming languages.
PHP, Java, C, Perl, and Ruby give exactly the same result, because they are based on the same numeric format.
Can we work around the problem? Sure, the most reliable method is to round the result with the help of a method toFixed(n):
let sum = 0.1 + 0.2; alert( sum.toFixed(2) ); // "0.30"
Please note that toFixed always returns a string. It ensures that it has 2 digits after the decimal point. That’s actually convenient if we have an e-shopping and need to show $0.30. For other cases, we can use the unary plus to coerce it into a number:
let sum = 0.1 + 0.2; alert( +sum.toFixed(2) ); // 0.3
We also can temporarily multiply the numbers by 100 (or a bigger number) to turn them into integers, do the maths, and then divide back. Then, as we’re doing maths with integers, the error somewhat decreases, but we still get it on division:
alert( (0.1 * 10 + 0.2 * 10) / 10 ); // 0.3 alert( (0.28 * 100 + 0.14 * 100) / 100); // 0.4200000000000001
So, the multiply/divide approach reduces the error, but doesn’t remove it totally.
Sometimes we could try to evade fractions at all. Like if we’re dealing with a shop, then we can store prices in cents instead of dollars. But what if we apply a discount of 30%? In practice, totally evading fractions is rarely possible. Just round them to cut “tails” when needed.
The funny thing
Try running this:
// Hello! I'm a self-increasing number! alert( 9999999999999999 ); // shows 10000000000000000
This suffers from the same issue: a loss of precision. There are 64 bits for the number, 52 of them can be used to store digits, but that’s not enough. So the least significant digits disappear.
JavaScript doesn’t trigger an error in such events. It does its best to fit the number into the desired format, but unfortunately, this format is not big enough.
Two zeroes
Another funny consequence of the internal representation of numbers is the existence of two zeroes: 0 and -0.
That’s because a sign is represented by a single bit, so it can be set or not set for any number including a zero.
In most cases, the distinction is unnoticeable, because operators are suited to treat them as the same.
Remember these two special numeric values?
Infinity (and -Infinity) is a special numeric value that is greater (less) than anything.
NaN represents an error.
They belong to the type number, but are not “normal” numbers, so there are special functions to check for them:
isNaN(value) converts its argument to a number and then tests it for being NaN:
alert( isNaN(NaN) ); // true
alert( isNaN("str") ); // trueBut do we need this function? Can’t we just use the comparison === NaN? Unfortunately not. The value NaN is unique in that it does not equal anything, including itself:
alert( NaN === NaN ); // false
isFinite(value) converts its argument to a number and returns true if it’s a regular number, not NaN/Infinity/-Infinity:
alert( isFinite("15") ); // true
alert( isFinite("str") ); // false, because a special value: NaN
alert( isFinite(Infinity) ); // false, because a special value: InfinitySometimes isFinite is used to validate whether a string value is a regular number:
let num = +prompt("Enter a number", '');
// will be true unless you enter Infinity, -Infinity or not a number
alert( isFinite(num) );Please note that an empty or a space-only string is treated as 0 in all numeric functions including isFinite.
Number.isNaN and Number.isFinite
Number.isNaN and Number.isFinite methods are the more “strict” versions of isNaN and isFinite functions. They do not autoconvert their argument into a number, but check if it belongs to the number type instead.
Number.isNaN(value) returns true if the argument belongs to the number type and it is NaN. In any other case, it returns false.
alert( Number.isNaN(NaN) ); // true
alert( Number.isNaN("str" / 2) ); // true
// Note the difference:
alert( Number.isNaN("str") ); // false, because "str" belongs to the string type, not the number type
alert( isNaN("str") ); // true, because isNaN converts string "str" into a number and gets NaN as a result of this conversionNumber.isFinite(value) returns true if the argument belongs to the number type and it is not NaN/Infinity/-Infinity. In any other case, it returns false.
alert( Number.isFinite(123) ); // true
alert( Number.isFinite(Infinity) ); // false
alert( Number.isFinite(2 / 0) ); // false
// Note the difference:
alert( Number.isFinite("123") ); // false, because "123" belongs to the string type, not the number type
alert( isFinite("123") ); // true, because isFinite converts string "123" into a number 123In a way, Number.isNaN and Number.isFinite are simpler and more straightforward than isNaN and isFinite functions. In practice though, isNaN and isFinite are mostly used, as they’re shorter to write.
Comparison with Object.is
There is a special built-in method Object.is that compares values like ===, but is more reliable for two edge cases:
It works with NaN: Object.is(NaN, NaN) === true, that’s a good thing.
Values 0 and -0 are different: Object.is(0, -0) === false, technically that’s correct because internally the number has a sign bit that may be different even if all other bits are zeroes.
In all other cases, Object.is(a, b) is the same as a === b.
We mention Object.is here, because it’s often used in JavaScript specification. When an internal algorithm needs to compare two values for being exactly the same, it uses Object.is (internally called SameValue).
Numeric conversion using a plus + or Number() is strict. If a value is not exactly a number, it fails:
alert( +"100px" ); // NaN
The sole exception is spaces at the beginning or at the end of the string, as they are ignored.
But in real life, we often have values in units, like "100px" or "12pt" in CSS. Also in many countries, the currency symbol goes after the amount, so we have "19€" and would like to extract a numeric value out of that.
That’s what parseInt and parseFloat are for.
They “read” a number from a string until they can’t. In case of an error, the gathered number is returned. The function parseInt returns an integer, whilst parseFloat will return a floating-point number:
alert( parseInt('100px') ); // 100
alert( parseFloat('12.5em') ); // 12.5
alert( parseInt('12.3') ); // 12, only the integer part is returned
alert( parseFloat('12.3.4') ); // 12.3, the second point stops the readingThere are situations when parseInt/parseFloat will return NaN. It happens when no digits could be read:
alert( parseInt('a123') ); // NaN, the first symbol stops the processThe second argument of parseInt(str, radix)
The parseInt() function has an optional second parameter. It specifies the base of the numeral system, so parseInt can also parse strings of hex numbers, binary numbers and so on:
alert( parseInt('0xff', 16) ); // 255
alert( parseInt('ff', 16) ); // 255, without 0x also works
alert( parseInt('2n9c', 36) ); // 123456JavaScript has a built-in Math object which contains a small library of mathematical functions and constants.
A few examples:
Math.random()
Returns a random number from 0 to 1 (not including 1).
alert( Math.random() ); // 0.1234567894322 alert( Math.random() ); // 0.5435252343232 alert( Math.random() ); // ... (any random numbers)
Math.max(a, b, c...) and Math.min(a, b, c...)
Returns the greatest and smallest from the arbitrary number of arguments.
alert( Math.max(3, 5, -10, 0, 1) ); // 5 alert( Math.min(1, 2) ); // 1
Math.pow(n, power)
Returns n raised to the given power.
alert( Math.pow(2, 10) ); // 2 in power 10 = 1024
There are more functions and constants in Math object, including trigonometry, which you can find in the docs for the Math object.
To write numbers with many zeroes:
Append "e" with the zeroes count to the number. Like: 123e6 is the same as 123 with 6 zeroes 123000000.
A negative number after "e" causes the number to be divided by 1 with given zeroes. E.g. 123e-6 means 0.000123 (123 millionths).
For different numeral systems:
Can write numbers directly in hex (0x), octal (0o) and binary (0b) systems.
parseInt(str, base) parses the string str into an integer in numeral system with given base, 2 ≤ base ≤ 36.
num.toString(base) converts a number to a string in the numeral system with the given base.
For regular number tests:
isNaN(value) converts its argument to a number and then tests it for being NaN
Number.isNaN(value) checks whether its argument belongs to the number type, and if so, tests it for being NaN
isFinite(value) converts its argument to a number and then tests it for not being NaN/Infinity/-Infinity
Number.isFinite(value) checks whether its argument belongs to the number type, and if so, tests it for not being NaN/Infinity/-Infinity
For converting values like 12pt and 100px to a number:
Use parseInt/parseFloat for the “soft” conversion, which reads a number from a string and then returns the value they could read before the error.
For fractions:
Round using Math.floor, Math.ceil, Math.trunc, Math.round or num.toFixed(precision).
Make sure to remember there’s a loss of precision when working with fractions.
More mathematical functions:
See the Math object when you need them. The library is very small but can cover basic needs.
importance: 5
Create a script that prompts the visitor to enter two numbers and then shows their sum.
Run the demo
P.S. There is a gotcha with types.
let a = +prompt("The first number?", "");
let b = +prompt("The second number?", "");
alert( a + b );Note the unary plus + before prompt. It immediately converts the value to a number.
Otherwise, a and b would be string their sum would be their concatenation, that is: "1" + "2" = "12".
importance: 4
According to the documentation Math.round and toFixed both round to the nearest number: 0..4 lead down while 5..9 lead up.
For instance:
alert( 1.35.toFixed(1) ); // 1.4
In the similar example below, why is 6.35 rounded to 6.3, not 6.4?
alert( 6.35.toFixed(1) ); // 6.3
How to round 6.35 the right way?
Internally the decimal fraction 6.35 is an endless binary. As always in such cases, it is stored with a precision loss.
Let’s see:
alert( 6.35.toFixed(20) ); // 6.34999999999999964473
The precision loss can cause both increase and decrease of a number. In this particular case the number becomes a tiny bit less, that’s why it rounded down.
And what’s for 1.35?
alert( 1.35.toFixed(20) ); // 1.35000000000000008882
Here the precision loss made the number a little bit greater, so it rounded up.
How can we fix the problem with 6.35 if we want it to be rounded the right way?
We should bring it closer to an integer prior to rounding:
alert( (6.35 * 10).toFixed(20) ); // 63.50000000000000000000
Note that 63.5 has no precision loss at all. That’s because the decimal part 0.5 is actually 1/2. Fractions divided by powers of 2 are exactly represented in the binary system, now we can round it:
alert( Math.round(6.35 * 10) / 10 ); // 6.35 -> 63.5 -> 64(rounded) -> 6.4
importance: 5
Create a function readNumber which prompts for a number until the visitor enters a valid numeric value.
The resulting value must be returned as a number.
The visitor can also stop the process by entering an empty line or pressing “CANCEL”. In that case, the function should return null.
Run the demo
Open a sandbox with tests.
function readNumber() {
let num;
do {
num = prompt("Enter a number please?", 0);
} while ( !isFinite(num) );
if (num === null || num === '') return null;
return +num;
}
alert(`Read: ${readNumber()}`);The solution is a little bit more intricate that it could be because we need to handle null/empty lines.
So we actually accept the input until it is a “regular number”. Both null (cancel) and empty line also fit that condition, because in numeric form they are 0.
After we stopped, we need to treat null and empty line specially (return null), because converting them to a number would return 0.
Open the solution with tests in a sandbox.
importance: 4
This loop is infinite. It never ends. Why?
let i = 0;
while (i != 10) {
i += 0.2;
}That’s because i would never equal 10.
Run it to see the real values of i:
let i = 0;
while (i < 11) {
i += 0.2;
if (i > 9.8 && i < 10.2) alert( i );
}None of them is exactly 10.
Such things happen because of the precision losses when adding fractions like 0.2.
Conclusion: evade equality checks when working with decimal fractions.
importance: 2
The built-in function Math.random() creates a random value from 0 to 1 (not including 1).
Write the function random(min, max) to generate a random floating-point number from min to max (not including max).
Examples of its work:
alert( random(1, 5) ); // 1.2345623452 alert( random(1, 5) ); // 3.7894332423 alert( random(1, 5) ); // 4.3435234525
We need to “map” all values from the interval 0…1 into values from min to max.
That can be done in two stages:
If we multiply a random number from 0…1 by max-min, then the interval of possible values increases 0..1 to 0..max-min.
Now if we add min, the possible interval becomes from min to max.
The function:
function random(min, max) {
return min + Math.random() * (max - min);
}
alert( random(1, 5) );
alert( random(1, 5) );
alert( random(1, 5) );importance: 2
Create a function randomInteger(min, max) that generates a random integer number from min to max including both min and max as possible values.
Any number from the interval min..max must appear with the same probability.
Examples of its work:
alert( randomInteger(1, 5) ); // 1 alert( randomInteger(1, 5) ); // 3 alert( randomInteger(1, 5) ); // 5
You can use the solution of the previous task as the base.
The simplest, but wrong solution would be to generate a value from min to max and round it:
function randomInteger(min, max) {
let rand = min + Math.random() * (max - min);
return Math.round(rand);
}
alert( randomInteger(1, 3) );The function works, but it is incorrect. The probability to get edge values min and max is two times less than any other.
If you run the example above many times, you would easily see that 2 appears the most often.
That happens because Math.round() gets random numbers from the interval 1..3 and rounds them as follows:
values from 1 ... to 1.4999999999 become 1 values from 1.5 ... to 2.4999999999 become 2 values from 2.5 ... to 2.9999999999 become 3
Now we can clearly see that 1 gets twice less values than 2. And the same with 3.
There are many correct solutions to the task. One of them is to adjust interval borders. To ensure the same intervals, we can generate values from 0.5 to 3.5, thus adding the required probabilities to the edges:
function randomInteger(min, max) {
// now rand is from (min-0.5) to (max+0.5)
let rand = min - 0.5 + Math.random() * (max - min + 1);
return Math.round(rand);
}
alert( randomInteger(1, 3) );An alternative way could be to use Math.floor for a random number from min to max+1:
function randomInteger(min, max) {
// here rand is from min to (max+1)
let rand = min + Math.random() * (max + 1 - min);
return Math.floor(rand);
}
alert( randomInteger(1, 3) );Now all intervals are mapped this way:
values from 1 ... to 1.9999999999 become 1 values from 2 ... to 2.9999999999 become 2 values from 3 ... to 3.9999999999 become 3
All intervals have the same length, making the final distribution uniform.