I am student studying programming.
As far as I know, Javascript saves number as float.
However, bitwise operator in Javascript run as a type of number is integer.
For instance,
> 1 | 2 // -> 3
> 2 << 4 // -> 32
How is it possible?
I find official documentation(mdn web docs), but I can not find the reason.
The bitwise operators in JavaScript convert numbers to 32-bit integers before performing the operation. This means that even though JavaScript saves numbers as floats, the bitwise operators treat them as 32-bit integers. This is why you are able to use the bitwise operators on numbers and get integer results.
As #rviretural_001 mentioned, JavaScript does some automatic conversions by spec. For example, in addition to IEEE 754 doubles to 32 signed integers:
0 == '0' // true
String to Integer (Shift Right)
'4' >> 1 // 2
Converting to Integer (Bitwise OR)
'1.3' | 0 // 1
This last one was used in asm.js for 32 signed integers
just to contribute to the answers above note that the conversion of a floating-point number to an integer is done by removing the fractional part of the number.
For example, if you have the number 4.5 and you use a bitwise operator on it, JavaScript will convert it to the integer 4 before performing the operation.
This tutorial might help you to find out more https://www.programiz.com/javascript/bitwise-operators
Also note that performing bitwise operations on floating-point numbers can lead to unexpected results. This is due to the way floating-point numbers are represented in memory. So, it's recommended to avoid using bitwise operators on floating-point numbers in JavaScript.
Related
This question already has answers here:
What is JavaScript's highest integer value that a number can go to without losing precision?
(21 answers)
Closed 4 months ago.
How to increase this number(you can try it on the browser console):
36893488147419103000 + 1
The result of this is:
36893488147419103000
The number stays the same no changes to it why is that? and how can I increase it by 1?
For big integers you should use the BigInt (Big Integer) type.
Note 1: you almost always cannot mix BigInt numbers with Numbers (eg for math operations) without first performing an explicit conversion.
Note 2: JSON does not currently natively support BigInt values. As a workaround you can use strings (eg. '1n' for the values and then use a reviver function when calling JSON.parse.
JavaScript currently only has two numeric types: double-precision IEEE 754 floating point Numbers, and Big Integers which can be used to represent arbitrarily large integers. You can declare a BigInt number literal using the suffix n, eg. 1n.
IEEE 754 Numbers can "only" accurately represent integers up to and including Number.MAX_SAFE_INTEGER, which has a value of 2^53 - 1 or 9,007,199,254,740,991 or ~9 quadrillion.
From MDN:
Double precision floating point format only has 52 bits to represent the mantissa, so it can only safely represent integers between -(253 – 1) and 253 – 1. "Safe" in this context refers to the ability to represent integers exactly and to compare them correctly. For example, Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2 will evaluate to true, which is mathematically incorrect. See Number.isSafeInteger() for more information.
A "Decimal" number type, that will be able to represent arbitrarily precise decimal numbers, is under development.
Obviously the number is internally represented as a floating point number.When the value you want do add to this number is less then the value of the least significant bit, it will not change the the value.
The only way would be to use floating point numbers with a higher resolution i.e. with a higher number of significant bits.
Double precision floating point format only has 52 bits to represent the mantissa, so it can only safely represent integers between -(253 – 1) and 253 – 1. See Number.MAX_SAFE_INTEGER. Larger numbers may not be able to be represented exactly.
I used to think that all numbers are stored in computers in 2’s complement format. And there is no -0 in 2’s complement. But in IEEE754, there are both +0 and -0. And I read that in JavaScript, all numbers are IEEE754. So now I am confused. What’s more, in JavaScript, ~a = -a -1. I used to think it was caused by the change of 2’s complement of number a. But if every number in JavaScript is stored as IEEE754, how can we convert it as 2’s complement?
JavaScript cheats: It represents numbers using the IEEE-754 basic 64-bit binary floating-point format but converts them to 32-bit integers for bitwise operations and then converts them back.
JavaScript is an implementation of ECMAScript. The ECMAScript 2018 Language Specification (9th edition, June 2018) specifies the unary ~ operator in clause 12.5.8 (shown as 12.5.10 in the table of contents but 12.5.8 in the text!). The relevant part is:
UnaryExpression : ~ UnaryExpression
1. Let expr be the result of evaluating UnaryExpression.
2. Let oldValue be ? ToInt32(? GetValue(expr)).
3. Return the result of applying bitwise complement to oldValue. The result is a signed 32-bit integer.
The ToInt32 operation truncates the number to an integer (removing any fraction part, rounding toward zero) and then maps the number to the interval [−231, +231) modulo 232.
I am not an expert in bitwise operators, but i often see a pattern which is used by programmers of 256k demos at competitions. Instead of using Math.floor() function, double bitwise NOT operator is used ~~ ( maybe faster ? ).
Like this:
Math.floor(2.1); // 2
~~2.1 // 2
Search revealed that there are more patterns that used the same way:
2.1 | 0 // 2
2.1 >> 0 // 2
When playing with this in the dev console, i have noticed a behavior that i'm not sure i understand fully.
Math.floor(2e+21); // 2e+21
~~2e+21; // -1119879168
2e+21 | 0; // -1119879168
What is happening under the hood ?
As Felix King pointed out, the numbers are being converted to 32 bit signed integers. 2e9 is less than the maximum positive value of a signed int, so this works:
~~(2e9) //2000000000
But when you go to 2e10, it can't use all the bits, so it just takes the lowest 32 bits and converts that to an int:
~~(2e10) //-1474836480
You can verify this by using another bitwise operator and confirming that it's grabbing the lowest 32 bits:
2e10 & 0xFFFFFFFF // also -1474836480
~~(2e10 & 0xFFFFFFFF) // also -1474836480
Math.floor is built to account for large numbers, so if accuracy over a big range is important then you should use it.
Also of note: The ~~ is doing truncation, which is the same as flooring for positive numbers only. It won't work for negatives:
Math.floor(-2.1) // -3
~~(-2.1) // -2
As stated in the MDN Docs, and here I quote,
The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.
This means tha when you apply a bitwise operator, for instance ~, to 2.1 it is first converted to an integer, and only then is the operator applied. This effectively achieves the rounding down (floor) effect for positive numbers.
As to why these operators are used, instead of the much nicer to grasp Math.floor, there are two main reasons. For one, these operators may be considerably faster to achieve the same result. Besides performance, some people just want the shortest code possible. All three operators you mentioned achieve the same effect, but ~~ just happens to be the shortest, and arguably the easiest to remember.
Given that the float to integer conversion happens before the bitwise operators are applied, let's see what happens with ~~. I'll represent our target number (2, after the convertion from 2.1) using 8 bits, instead of 32, for shortness.
2: 0000 0010
~2: 1111 1101 (-3)
~~2: 0000 0010
So, you see, we apply an operator to retrieve only the integer part, but we can't apply only one bitwise not, because it would mess up the result. We revert it to the desired value applying the second operator.
Regarding your last example, take into account that the number you're testing with, 2e+21, is a relatively big number. It's a 2 followed by twenty-one zeroes. It simply doesn't fit as a 32-bit integer (the data-type it is being converted to, when you apply the bitwise operators). Just look at the difference between your number and what a 32-bit signed integer can represent.
Max. Integer: 2147483647
2e+21: 2000000000000000000000
How about binary?
Max. Integer: 01111111111111111111111111111111
2e+21: 11011000110101110010011010110111000101110111101010000000000000000000000
Quite big, huh?
What really happens under the hood is that Javascript is truncating your big number to what it can represent in 32 bits.
110110001101011100100110101101110001011 10111101010000000000000000000000
^---- Junk ----^
When we convert our truncated number to decimal, we get back what you're seeing.
Bin: 10111101010000000000000000000000
Dec: -1119879168
Conversely, Math.floor accounts for the big numbers and avoids truncating them, which is one of the possible reasons for it being slower, although accurate.
As I see in examples, the functionality if ~~ and Math.floor is the same. Both of them round a number downward (Am I think correct?)
Also I should mention that according to this test ~~ is faster than Math.floor: jsperf.com/math-round-vs
So I want to know, is there any difference between ~~ and Math.floor?
Yes, bitwise operators generally don’t play well with negative numbers. f.ex:
~~-6.8 == -6 // doesn’t round down, simply removes the decimals
Math.floor(-6.8) == -7
And you also get 0 instead of NaN, f.ex:
~~'a' == 0
Math.floor('a') == NaN
In addition to David answer:
One of the things that I have noticed about bitwise operations in JavaScript is that it can be convenient for smaller values, but doesn’t always work for larger values. The reason this is the case is that bitwise operators will only work fully for operands which can be fully expressed in a 32-bit signed format. In other words, using bitwise operations will only produce numbers that are in the range of -2147483648 (-231) to 2147483647 (231 – 1). In addition, if one of the operands used is outside of that range, the last 32 bits of the number will be used instead of the specified number.
http://cwestblog.com/2011/07/27/limits-on-bitwise-operators-in-javascript/
This limitation can easily be found when working with Date, assume you are rounding a milliseconds value:
Math.floor(1559125440000.6) // 1559125440000
~~1559125440000.6 // 52311552
I've recently discovered some other ways to remove the fractional part of numeric values in JavaScript other than Math.floor(n), specifically the double bitwise NOT operator ~~n and performing a bitwise or with 0 n|0.
I'd like to know what are the difference between these approaches and what the different scenarios are where one method is recommended over another.
The operands of all bitwise operators are converted to signed 32-bit integers:
Math.floor(2147483648) // 2147483648
2147483648 | 0 // 2147483648
~~2147483648 // 2147483648
Math.floor(2147483649) // 2147483649
2147483649 | 0 // -2147483647
~~2147483649 // -2147483647
So use Math.floor();
Be clear to the next person looking at your code and use Math.floor().
The performance gain of 1%-40% isn't really worth it, so don't make your code confusing and hard to maintain.
(I entirely agree with josh's answer: favor clear maintainable code.)
Here is an explanation on the other bit-wise approaches:
The bit-wise operators work because they only operator on 32-bit (signed) integers but numbers in JavaScript are all IEEE-754 values. Thus, there is an internal conversion (truncation, not floor!) that happens to operands for bit-wise operators.
The applied bit-wise operation (e.g. n<<0, ~~n or n|0) then acts as an identity function which "does nothing" to the converted values: that is, all of these approaches rely on the same conversion applied to bit-wise operands.
Try n as a negative number or a value outside of [-231, 231-1]:
(-1.23|0) // -1
Math.floor(-1.23) // -2
var x = Math.pow(2, 40) + .5
x|0 // 0
Math.floor(x) // 1099511627776
Happy coding.