How to convert a BigInt, to two's complement binary in JavaScript? - javascript

With "normal" numbers(32bit range), i'm using zero fill right shift operator to convert to binary, which works both with positive and negative numbers(results in the two's complement binary):
const numberToConvert = -100
(numberToConvert >>> 0).toString(2);
//Result is correct, in two's complement: '11111111111111111111111110011100'
But how can this be done with a negative BigInt?
If i do:
(-1000000000000000000n >>> 0).toString(2)
I get an error "Uncaught TypeError: Cannot mix BigInt and other types, use explicit conversions"
So then i try to use 0 as a bigint:
(-1000000000000000000n >>> 0n).toString(2)
I get the following error: Uncaught TypeError: BigInts have no unsigned right shift, use >> instead
Doing so, results in the non two's complement binary, with "-" appended to it:
(-1000000000000000000n >> 0n).toString(2)
//Result is:'-110111100000101101101011001110100111011001000000000000000000'
How can I get the two's complement binary, of a negative bigint?


The bitwise operators are for 32-bit integers anyway, and why it doesn't work with BigInt, as quoted in JavaScript Definitive Guide, 7th Ed, David Flanagan, O'Reilly, p. 78:
Shift right with zero fill (>>>): This is the only one of the JavaScript bitwise operators that cannot be used with BigInt values. BigInt does not represent negative numbers by setting the high bit the way that 32-bit integers do, and this operator only makes sense for that particular two’s complement representation.
Also note that it looks like it is giving you two's complement, but in fact, the negative number is converted to 32-bit unsigned integer, and then printed as binary, giving you the impression that it is two's complement:
console.log(-100 >>> 0); // => 4294967196
The two's complement has this property:
You have a number, say 123, which is 01111011 in 8 bit binary, and you want the negative number of that, which is -123.
Two complement says: the answer you want, just treat it as a positive number, and add it with the original number 123, and you will just get all 0's with the overflow of the 8 bit number.
As an example, treating everything as positive, 123 + theAnswerYouWant is 01111011 + 10000101, which is exactly 00000000 with an overflow, which is 100000000 (note the extra 1 in front). In other words, you want 256 - 123, which is 133 and if you render 133 as 8 bit, that's the answer you want.
As a result, you can use 28 to subtract the orignal number, and treat it as a positive number and display it, using .toString(2), which you already have.
The following is for 64 bits:
function getBinary(a, nBits) {
[a, nBits] = [BigInt(a), BigInt(nBits)];
if ((a > 0 && a >= 2n ** (nBits - 1n)) || (a < 0 && -a > 2n ** (nBits - 1n))) {
throw new RangeError("overflow error");
}
return a >= 0
? a.toString(2).padStart(Number(nBits), "0")
: (2n ** nBits + a).toString(2);
}
console.log(getBinary(1000000000000000000n, 64));
console.log(getBinary(-1000000000000000000n, 64));
console.log(getBinary(-1, 64));
console.log(getBinary(-2, 64));
console.log(getBinary(-3, 64));
console.log(getBinary(-4, 64n)); // trying the nBits as a BigInt as a test
console.log(getBinary(2n ** 63n - 1n, 64));
console.log(getBinary(-(2n ** 63n), 64));
// console.log(getBinary(2n ** 63n, 64)); // throw Error
// console.log(getBinary(-(2n ** 63n) - 1n, 64)); // throw Error
Note that you don't have to pad it when a is negative, because for example, if it is 8 bit, the number being displayed is any where from 11111111 to 10000000 and it is always 8 bits.
Some more details:
You may already know ones' complement is just simply flipping the bits (from 0 to 1, and 1 to 0). Another way to think of it is, you add the two numbers together and it will becomes all 1s.
The usual way two's complement is described, is to flip the bits, and add 1 to it. You see, if you start with 11111111 and subtract 01111011 (which is 123 decimal), you get 10000100 and it is exactly the same as flipping the bit. (actually this follows from above: adding them get all 1s, so using all 1s to subtract one of them get the other one.
Well, so if you start with 11111111 and subtract that number, and then add 1, isn't it the same as using 11111111, add 1, and subtract that number? Well, 11111111 plus 1 is 100000000 (note the extra 1 in front) -- that's exactly starting with 2n where n is the n-bit integer, and then subtract that number. So you see why the property at the beginning of this post is true.
In fact, two's complement is designed with such purpose: if we want to find out 2 - 1, to make the computer calculate that, we only need to consider this "two's complement" as positive numbers and add them together using the processor's "add circuitry": 00000010 plus 11111111. We get 00000001 but have a carry (the overflow). If we handle the overflow correctly by discarding it, we get the answer: 1. If we use ones' complement instead, we can't use the same addition circuitry to carry out 00000010 + 11111110 to get a 1 because the result is 00000000 which is 0
Another way to think about (4) is, if you have a car's odometer, and it says 000002 miles so far, how do you subtract 1 from it? Well, if you represent -1 as 9999999, then you just add 999999 to the 2, and get 1000001 but the leftmost 1 does not show on the odometer, and now the odometer will become 000001. In decimal, representing -1 as 999999 is 10's complement. In binary, representing -1 as 11111111 is called two's complement.


Two's complement only makes sense with fixed bit lengths. Numbers are converted to 32-bit integers (this is an old convention from back when javascript was messier). BigInt doesn't have that kind of conversion as the length is considered arbitrary. So, in order to use two's complement with BigInt, you'll need to figure out what length you want to use then convert it. Conversion to two's complement is described many places including Wikipedia.
Here, we use the LSB to MSB method since it's pretty easy to implement as string processing in javascript:
const toTwosComplement = (n, len) => {
// `n` must be an integer
// `len` must be a positive integer greater than bit-length of `n`
n = BigInt(n);
len = Number(len);
if(!Number.isInteger(len)) throw '`len` must be an integer';
if(len <= 0) throw '`len` must be greater than zero';
// If non-negative, a straight conversion works
if(n >= 0){
n = n.toString(2)
if(n.length >= len) throw 'out of range';
return n.padStart(len, '0');
}
n = (-n).toString(2); // make positive and convert to bit string
if(!(n.length < len || n === '1'.padEnd(len, '0'))) throw 'out of range';
// Start at the LSB and work up. Copy bits up to and including the
// first 1 bit then invert the remaining
let invert = false;
return n.split('').reverse().map(bit => {
if(invert) return bit === '0' ? '1' : '0';
if(bit === '0') return bit;
invert = true;
return bit;
}).reverse().join('').padStart(len, '1');
};
console.log(toTwosComplement( 1000000000000000000n, 64));
console.log(toTwosComplement(-1000000000000000000n, 64));
console.log(toTwosComplement(-1, 64));
console.log(toTwosComplement(2n**63n-1n, 64));
console.log(toTwosComplement(-(2n**63n), 64));
div.as-console-wrapper{max-height:none;height:100%;}

Related

Different hash results in Java and JavaScript [duplicate]

I'm in a computer systems course and have been struggling, in part, with two's complement. I want to understand it, but everything I've read hasn't brought the picture together for me. I've read the Wikipedia article and various other articles, including my text book.
What is two's complement, how can we use it and how can it affect numbers during operations like casts (from signed to unsigned and vice versa), bit-wise operations and bit-shift operations?

Two's complement is a clever way of storing integers so that common math problems are very simple to implement.
To understand, you have to think of the numbers in binary.
It basically says,
for zero, use all 0's.
for positive integers, start counting up, with a maximum of 2(number of bits - 1)-1.
for negative integers, do exactly the same thing, but switch the role of 0's and 1's and count down (so instead of starting with 0000, start with 1111 - that's the "complement" part).
Let's try it with a mini-byte of 4 bits (we'll call it a nibble - 1/2 a byte).
0000 - zero
0001 - one
0010 - two
0011 - three
0100 to 0111 - four to seven
That's as far as we can go in positives. 23-1 = 7.
For negatives:
1111 - negative one
1110 - negative two
1101 - negative three
1100 to 1000 - negative four to negative eight
Note that you get one extra value for negatives (1000 = -8) that you don't for positives. This is because 0000 is used for zero. This can be considered as Number Line of computers.
Distinguishing between positive and negative numbers
Doing this, the first bit gets the role of the "sign" bit, as it can be used to distinguish between nonnegative and negative decimal values. If the most significant bit is 1, then the binary can be said to be negative, where as if the most significant bit (the leftmost) is 0, you can say the decimal value is nonnegative.
"Sign-magnitude" negative numbers just have the sign bit flipped of their positive counterparts, but this approach has to deal with interpreting 1000 (one 1 followed by all 0s) as "negative zero" which is confusing.
"Ones' complement" negative numbers are just the bit-complement of their positive counterparts, which also leads to a confusing "negative zero" with 1111 (all ones).
You will likely not have to deal with Ones' Complement or Sign-Magnitude integer representations unless you are working very close to the hardware.

I wonder if it could be explained any better than the Wikipedia article.
The basic problem that you are trying to solve with two's complement representation is the problem of storing negative integers.
First, consider an unsigned integer stored in 4 bits. You can have the following
0000 = 0
0001 = 1
0010 = 2
...
1111 = 15
These are unsigned because there is no indication of whether they are negative or positive.
Sign Magnitude and Excess Notation
To store negative numbers you can try a number of things. First, you can use sign magnitude notation which assigns the first bit as a sign bit to represent +/- and the remaining bits to represent the magnitude. So using 4 bits again and assuming that 1 means - and 0 means + then you have
0000 = +0
0001 = +1
0010 = +2
...
1000 = -0
1001 = -1
1111 = -7
So, you see the problem there? We have positive and negative 0. The bigger problem is adding and subtracting binary numbers. The circuits to add and subtract using sign magnitude will be very complex.
What is
0010
1001 +
----
?
Another system is excess notation. You can store negative numbers, you get rid of the two zeros problem but addition and subtraction remains difficult.
So along comes two's complement. Now you can store positive and negative integers and perform arithmetic with relative ease. There are a number of methods to convert a number into two's complement. Here's one.
Convert Decimal to Two's Complement
Convert the number to binary (ignore the sign for now)
e.g. 5 is 0101 and -5 is 0101
If the number is a positive number then you are done.
e.g. 5 is 0101 in binary using two's complement notation.
If the number is negative then
3.1 find the complement (invert 0's and 1's)
e.g. -5 is 0101 so finding the complement is 1010
3.2 Add 1 to the complement 1010 + 1 = 1011.
Therefore, -5 in two's complement is 1011.
So, what if you wanted to do 2 + (-3) in binary? 2 + (-3) is -1.
What would you have to do if you were using sign magnitude to add these numbers? 0010 + 1101 = ?
Using two's complement consider how easy it would be.
2 = 0010
-3 = 1101 +
-------------
-1 = 1111
Converting Two's Complement to Decimal
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!

Like most explanations I've seen, the ones above are clear about how to work with 2's complement, but don't really explain what they are mathematically. I'll try to do that, for integers at least, and I'll cover some background that's probably familiar first.
Recall how it works for decimal: 2345 is a way of writing 2 × 103 + 3 × 102 + 4 × 101 + 5 × 100.
In the same way, binary is a way of writing numbers using just 0 and 1 following the same general idea, but replacing those 10s above with 2s. Then in binary, 1111is a way of writing 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20and if you work it out, that turns out to equal 15 (base 10). That's because it is 8+4+2+1 = 15.
This is all well and good for positive numbers. It even works for negative numbers if you're willing to just stick a minus sign in front of them, as humans do with decimal numbers. That can even be done in computers, sort of, but I haven't seen such a computer since the early 1970's. I'll leave the reasons for a different discussion.
For computers it turns out to be more efficient to use a complement representation for negative numbers. And here's something that is often overlooked. Complement notations involve some kind of reversal of the digits of the number, even the implied zeroes that come before a normal positive number. That's awkward, because the question arises: all of them? That could be an infinite number of digits to be considered.
Fortunately, computers don't represent infinities. Numbers are constrained to a particular length (or width, if you prefer). So let's return to positive binary numbers, but with a particular size. I'll use 8 digits ("bits") for these examples. So our binary number would really be 00001111or 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
To form the 2's complement negative, we first complement all the (binary) digits to form 11110000and add 1 to form 11110001but how are we to understand that to mean -15?
The answer is that we change the meaning of the high-order bit (the leftmost one). This bit will be a 1 for all negative numbers. The change will be to change the sign of its contribution to the value of the number it appears in. So now our 11110001 is understood to represent -1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20Notice that "-" in front of that expression? It means that the sign bit carries the weight -27, that is -128 (base 10). All the other positions retain the same weight they had in unsigned binary numbers.
Working out our -15, it is -128 + 64 + 32 + 16 + 1 Try it on your calculator. it's -15.
Of the three main ways that I've seen negative numbers represented in computers, 2's complement wins hands down for convenience in general use. It has an oddity, though. Since it's binary, there have to be an even number of possible bit combinations. Each positive number can be paired with its negative, but there's only one zero. Negating a zero gets you zero. So there's one more combination, the number with 1 in the sign bit and 0 everywhere else. The corresponding positive number would not fit in the number of bits being used.
What's even more odd about this number is that if you try to form its positive by complementing and adding one, you get the same negative number back. It seems natural that zero would do this, but this is unexpected and not at all the behavior we're used to because computers aside, we generally think of an unlimited supply of digits, not this fixed-length arithmetic.
This is like the tip of an iceberg of oddities. There's more lying in wait below the surface, but that's enough for this discussion. You could probably find more if you research "overflow" for fixed-point arithmetic. If you really want to get into it, you might also research "modular arithmetic".

2's complement is very useful for finding the value of a binary, however I thought of a much more concise way of solving such a problem(never seen anyone else publish it):
take a binary, for example: 1101 which is [assuming that space "1" is the sign] equal to -3.
using 2's complement we would do this...flip 1101 to 0010...add 0001 + 0010 ===> gives us 0011. 0011 in positive binary = 3. therefore 1101 = -3!
What I realized:
instead of all the flipping and adding, you can just do the basic method for solving for a positive binary(lets say 0101) is (23 * 0) + (22 * 1) + (21 * 0) + (20 * 1) = 5.
Do exactly the same concept with a negative!(with a small twist)
take 1101, for example:
for the first number instead of 23 * 1 = 8 , do -(23 * 1) = -8.
then continue as usual, doing -8 + (22 * 1) + (21 * 0) + (20 * 1) = -3

Imagine that you have a finite number of bits/trits/digits/whatever. You define 0 as all digits being 0, and count upwards naturally:
00
01
02
..
Eventually you will overflow.
98
99
00
We have two digits and can represent all numbers from 0 to 100. All those numbers are positive! Suppose we want to represent negative numbers too?
What we really have is a cycle. The number before 2 is 1. The number before 1 is 0. The number before 0 is... 99.
So, for simplicity, let's say that any number over 50 is negative. "0" through "49" represent 0 through 49. "99" is -1, "98" is -2, ... "50" is -50.
This representation is ten's complement. Computers typically use two's complement, which is the same except using bits instead of digits.
The nice thing about ten's complement is that addition just works. You do not need to do anything special to add positive and negative numbers!

I read a fantastic explanation on Reddit by jng, using the odometer as an analogy.
It is a useful convention. The same circuits and logic operations that
add / subtract positive numbers in binary still work on both positive
and negative numbers if using the convention, that's why it's so
useful and omnipresent.
Imagine the odometer of a car, it rolls around at (say) 99999. If you
increment 00000 you get 00001. If you decrement 00000, you get 99999
(due to the roll-around). If you add one back to 99999 it goes back to
00000. So it's useful to decide that 99999 represents -1. Likewise, it is very useful to decide that 99998 represents -2, and so on. You have
to stop somewhere, and also by convention, the top half of the numbers
are deemed to be negative (50000-99999), and the bottom half positive
just stand for themselves (00000-49999). As a result, the top digit
being 5-9 means the represented number is negative, and it being 0-4
means the represented is positive - exactly the same as the top bit
representing sign in a two's complement binary number.
Understanding this was hard for me too. Once I got it and went back to
re-read the books articles and explanations (there was no internet
back then), it turned out a lot of those describing it didn't really
understand it. I did write a book teaching assembly language after
that (which did sell quite well for 10 years).

Two complement is found out by adding one to 1'st complement of the given number.
Lets say we have to find out twos complement of 10101 then find its ones complement, that is, 01010 add 1 to this result, that is, 01010+1=01011, which is the final answer.

Lets get the answer 10 – 12 in binary form using 8 bits:
What we will really do is 10 + (-12)
We need to get the compliment part of 12 to subtract it from 10.
12 in binary is 00001100.
10 in binary is 00001010.
To get the compliment part of 12 we just reverse all the bits then add 1.
12 in binary reversed is 11110011. This is also the Inverse code (one's complement).
Now we need to add one, which is now 11110100.
So 11110100 is the compliment of 12! Easy when you think of it this way.
Now you can solve the above question of 10 - 12 in binary form.
00001010
11110100
-----------------
11111110

Looking at the two's complement system from a math point of view it really makes sense. In ten's complement, the idea is to essentially 'isolate' the difference.
Example: 63 - 24 = x
We add the complement of 24 which is really just (100 - 24). So really, all we are doing is adding 100 on both sides of the equation.
Now the equation is: 100 + 63 - 24 = x + 100, that is why we remove the 100 (or 10 or 1000 or whatever).
Due to the inconvenient situation of having to subtract one number from a long chain of zeroes, we use a 'diminished radix complement' system, in the decimal system, nine's complement.
When we are presented with a number subtracted from a big chain of nines, we just need to reverse the numbers.
Example: 99999 - 03275 = 96724
That is the reason, after nine's complement, we add 1. As you probably know from childhood math, 9 becomes 10 by 'stealing' 1. So basically it's just ten's complement that takes 1 from the difference.
In Binary, two's complement is equatable to ten's complement, while one's complement to nine's complement. The primary difference is that instead of trying to isolate the difference with powers of ten (adding 10, 100, etc. into the equation) we are trying to isolate the difference with powers of two.
It is for this reason that we invert the bits. Just like how our minuend is a chain of nines in decimal, our minuend is a chain of ones in binary.
Example: 111111 - 101001 = 010110
Because chains of ones are 1 below a nice power of two, they 'steal' 1 from the difference like nine's do in decimal.
When we are using negative binary number's, we are really just saying:
0000 - 0101 = x
1111 - 0101 = 1010
1111 + 0000 - 0101 = x + 1111
In order to 'isolate' x, we need to add 1 because 1111 is one away from 10000 and we remove the leading 1 because we just added it to the original difference.
1111 + 1 + 0000 - 0101 = x + 1111 + 1
10000 + 0000 - 0101 = x + 10000
Just remove 10000 from both sides to get x, it's basic algebra.

The word complement derives from completeness. In the decimal world the numerals 0 through 9 provide a complement (complete set) of numerals or numeric symbols to express all decimal numbers. In the binary world the numerals 0 and 1 provide a complement of numerals to express all binary numbers. In fact The symbols 0 and 1 must be used to represent everything (text, images, etc) as well as positive (0) and negative (1).
In our world the blank space to the left of number is considered as zero:
35=035=000000035.
In a computer storage location there is no blank space. All bits (binary digits) must be either 0 or 1. To efficiently use memory numbers may be stored as 8 bit, 16 bit, 32 bit, 64 bit, 128 bit representations. When a number that is stored as an 8 bit number is transferred to a 16 bit location the sign and magnitude (absolute value) must remain the same. Both 1's complement and 2's complement representations facilitate this.
As a noun:
Both 1's complement and 2's complement are binary representations of signed quantities where the most significant bit (the one on the left) is the sign bit. 0 is for positive and 1 is for negative.
2s complement does not mean negative. It means a signed quantity. As in decimal the magnitude is represented as the positive quantity. The structure uses sign extension to preserve the quantity when promoting to a register [] with more bits:
[0101]=[00101]=[00000000000101]=5 (base 10)
[1011]=[11011]=[11111111111011]=-5(base 10)
As a verb:
2's complement means to negate. It does not mean make negative. It means if negative make positive; if positive make negative. The magnitude is the absolute value:
if a >= 0 then |a| = a
if a < 0 then |a| = -a = 2scomplement of a
This ability allows efficient binary subtraction using negate then add.
a - b = a + (-b)
The official way to take the 1's complement is for each digit subtract its value from 1.
1'scomp(0101) = 1010.
This is the same as flipping or inverting each bit individually. This results in a negative zero which is not well loved so adding one to te 1's complement gets rid of the problem.
To negate or take the 2s complement first take the 1s complement then add 1.
Example 1 Example 2
0101 --original number 1101
1's comp 1010 0010
add 1 0001 0001
2's comp 1011 --negated number 0011
In the examples the negation works as well with sign extended numbers.
Adding:
1110 Carry 111110 Carry
0110 is the same as 000110
1111 111111
sum 0101 sum 000101
SUbtracting:
1110 Carry 00000 Carry
0110 is the same as 00110
-0111 +11001
---------- ----------
sum 0101 sum 11111
Notice that when working with 2's complement, blank space to the left of the number is filled with zeros for positive numbers butis filled with ones for negative numbers. The carry is always added and must be either a 1 or 0.
Cheers

2's complement is essentially a way of coming up with the additive inverse of a binary number. Ask yourself this: Given a number in binary form (present at a fixed length memory location), what bit pattern, when added to the original number (at the fixed length memory location), would make the result all zeros ? (at the same fixed length memory location). If we could come up with this bit pattern then that bit pattern would be the -ve representation (additive inverse) of the original number; as by definition adding a number to its additive inverse always results in zero. Example: take 5 which is 101 present inside a single 8 bit byte. Now the task is to come up with a bit pattern which when added to the given bit pattern (00000101) would result in all zeros at the memory location which is used to hold this 5 i.e. all 8 bits of the byte should be zero. To do that, start from the right most bit of 101 and for each individual bit, again ask the same question: What bit should I add to the current bit to make the result zero ? continue doing that taking in account the usual carry over. After we are done with the 3 right most places (the digits that define the original number without regard to the leading zeros) the last carry goes in the bit pattern of the additive inverse. Furthermore, since we are holding in the original number in a single 8 bit byte, all other leading bits in the additive inverse should also be 1's so that (and this is important) when the computer adds "the number" (represented using the 8 bit pattern) and its additive inverse using "that" storage type (a byte) the result in that byte would be all zeros.
1 1 1
----------
1 0 1
1 0 1 1 ---> additive inverse
---------
0 0 0

Many of the answers so far nicely explain why two's complement is used to represent negative numbers, but do not tell us what two's complement number is, particularly not why a '1' is added, and in fact often added in a wrong way.
The confusion comes from a poor understanding of the definition of a complement number. A complement is the missing part that would make something complete.
The radix complement of an n digit number x in radix b is, by definition, b^n-x.
In binary 4 is represented by 100, which has 3 digits (n=3) and a radix of 2 (b=2). So its radix complement is b^n-x = 2^3-4=8-4=4 (or 100 in binary).
However, in binary obtaining a radix's complement is not as easy as getting its diminished radix complement, which is defined as (b^n-1)-y, just 1 less than that of radix complement. To get a diminished radix complement, you simply flip all the digits.
100 -> 011 (diminished (one's) radix complement)
to obtain the radix (two's) complement, we simply add 1, as the definition defined.
011 +1 ->100 (two's complement).
Now with this new understanding, let's take a look of the example given by Vincent Ramdhanie (see above second response):
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!
Should be understood as:
The number starts with 1, so it's negative. So we know it is a two's complement of some value x. To find the x represented by its two's complement, we first need find its 1's complement.
two's complement of x: 1111
one's complement of x: 1111-1 ->1110;
x = 0001, (flip all digits)
Apply the sign -, and the answer =-x =-1.

I liked lavinio's answer, but shifting bits adds some complexity. Often there's a choice of moving bits while respecting the sign bit or while not respecting the sign bit. This is the choice between treating the numbers as signed (-8 to 7 for a nibble, -128 to 127 for bytes) or full-range unsigned numbers (0 to 15 for nibbles, 0 to 255 for bytes).

It is a clever means of encoding negative integers in such a way that approximately half of the combination of bits of a data type are reserved for negative integers, and the addition of most of the negative integers with their corresponding positive integers results in a carry overflow that leaves the result to be binary zero.
So, in 2's complement if one is 0x0001 then -1 is 0x1111, because that will result in a combined sum of 0x0000 (with an overflow of 1).

2’s Complements: When we add an extra one with the 1’s complements of a number we will get the 2’s complements. For example: 100101 it’s 1’s complement is 011010 and 2’s complement is 011010+1 = 011011 (By adding one with 1's complement) For more information
this article explain it graphically.

Two's complement is mainly used for the following reasons:
To avoid multiple representations of 0
To avoid keeping track of carry bit (as in one's complement) in case of an overflow.
Carrying out simple operations like addition and subtraction becomes easy.

Two's complement is one of the ways of expressing a negative number and most of the controllers and processors store a negative number in two's complement form.

In simple terms, two's complement is a way to store negative numbers in computer memory. Whereas positive numbers are stored as a normal binary number.
Let's consider this example,
The computer uses the binary number system to represent any number.
x = 5;
This is represented as 0101.
x = -5;
When the computer encounters the - sign, it computes its two's complement and stores it.
That is, 5 = 0101 and its two's complement is 1011.
The important rules the computer uses to process numbers are,
If the first bit is 1 then it must be a negative number.
If all the bits except first bit are 0 then it is a positive number, because there is no -0 in number system (1000 is not -0 instead it is positive 8).
If all the bits are 0 then it is 0.
Else it is a positive number.

To bitwise complement a number is to flip all the bits in it. To two’s complement it, we flip all the bits and add one.
Using 2’s complement representation for signed integers, we apply the 2’s complement operation to convert a positive number to its negative equivalent and vice versa. So using nibbles for an example, 0001 (1) becomes 1111 (-1) and applying the op again, returns to 0001.
The behaviour of the operation at zero is advantageous in giving a single representation for zero without special handling of positive and negative zeroes. 0000 complements to 1111, which when 1 is added. overflows to 0000, giving us one zero, rather than a positive and a negative one.
A key advantage of this representation is that the standard addition circuits for unsigned integers produce correct results when applied to them. For example adding 1 and -1 in nibbles: 0001 + 1111, the bits overflow out of the register, leaving behind 0000.
For a gentle introduction, the wonderful Computerphile have produced a video on the subject.

The question is 'What is “two's complement”?'
The simple answer for those wanting to understand it theoretically (and me seeking to complement the other more practical answers): 2's complement is the representation for negative integers in the dual system that does not require additional characters, such as + and -.

Two's complement of a given number is the number got by adding 1 with the ones' complement of the number.
Suppose, we have a binary number: 10111001101
Its 1's complement is: 01000110010
And its two's complement will be: 01000110011

Reference: Two's Complement (Thomas Finley)
I invert all the bits and add 1. Programmatically:
// In C++11
int _powers[] = {
1,
2,
4,
8,
16,
32,
64,
128
};
int value = 3;
int n_bits = 4;
int twos_complement = (value ^ ( _powers[n_bits]-1)) + 1;

You can also use an online calculator to calculate the two's complement binary representation of a decimal number: http://www.convertforfree.com/twos-complement-calculator/

The simplest answer:
1111 + 1 = (1)0000. So 1111 must be -1. Then -1 + 1 = 0.
It's perfect to understand these all for me.

Need for a JavaScript/ecma262 ToInt32 algorithm explanation

I'm trying to understand how JS engines convert a JS Number (Float64) to a 32-bit signed integer. I read that one can quickly convert a 64 bit float to a 32 bit signed integer with the bitwise OR like:
-8589934590 | 0 // which gives 2
I can't understand where does the 2 come from. According to the spec, the ToInt32 algorithm does this (the bold text is mine, not the spec's):
Let number be ? ToNumber(argument): -8589934590 is already a Number
If number is NaN, +0, -0, +∞, or -∞, return +0.: No
Let int be the Number value that is the same sign as number and whose magnitude is floor(abs(number)): -8589934590 is already an integer
Let int32bit be int modulo 2³² Since 2³² is positive the result should also be positive. In JS the remainder operator uses the sign of the left operand, so to get a modulo in this case (where -8589934590 is negative), we negate it: let int32bit = 8589934590 % 2**32 // 4294967294 which has 32 bit length 0b11111111111111111111111111111110
If int32bit ≥ 2³¹, return int32bit - 2³²; otherwise return int32bit. int32bit is smaller 2³¹ (since it's negative), so I use int32bit which equals -2 (Even if we consider 0b11111111111111111111111111111110 an unsigned integer, then it's greater 2³¹ and int32bit - 2³² still equals -2
Could someone, please, explain, do I correctly understand the ToInt32 algorithm and the bitwise OR operator?

Your step 4 is wrong. Modulo is defined by the spec as:
The notation “x modulo y” (y must be finite and nonzero) computes a value k of the same sign as y (or zero) such that abs(k) < abs(y) and x-k = q × y for some integer q.
So -8589934590 is our x, and 2**32 is our y, from that we also know that k must be positive. If we choose q = -1 we can solve the equation to k = -4294967294. That is however not a valid solution, as k (negative) does not have the same sign as y (positive). If we choose q = -2 instead, we get k = 2.
So for negative numbers x and positive numbers y, q * y will always have to result in a smaller number than x for k to be positive. Thus if we are transforming that to positive numbers (like you did), we are looking for the larger multiple of the number not the smaller one. E.g. if we take 2 % 3, that'll return 2 (2 - 2 = 3 * 0), whereas -2 modulo 3 will return 1 (-2 -1 = 3 * -1).

Unexpected negative value in Int32Array

const x = new Int32Array(1);
x[0] = 699044815921;
console.log(x[0]);
-1034853327
Who can explain why there is a negetive number?

Int32Array allows 32 bits per value, with 32nd bit (from the right) being reserved to specify the sign of the number. The number you're trying to fit is (699044815921).toString(2).length == 40 bits long, so 8 leftmost bits are discarded, 32nd bit is interpreted as a sign bit, and you get what you get as a result.

Negative numbers to binary string in JavaScript

Anyone knows why javascript Number.toString function does not represents negative numbers correctly?
//If you try
(-3).toString(2); //shows "-11"
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
I am really curious why it doesn't work properly or what is the reason it works this way?
I've searched it but didn't find anything that helps.

Short answer:
The toString() function takes the decimal, converts it
to binary and adds a "-" sign.
A zero fill right shift converts it's operands to signed 32-bit
integers in two complements format.
A more detailed answer:
Question 1:
//If you try
(-3).toString(2); //show "-11"
It's in the function .toString(). When you output a number via .toString():
Syntax
numObj.toString([radix])
If the numObj is negative, the sign is preserved. This is the case
even if the radix is 2; the string returned is the positive binary
representation of the numObj preceded by a - sign, not the two's
complement of the numObj.
It takes the decimal, converts it to binary and adds a "-" sign.
Base 10 "3" converted to base 2 is "11"
Add a sign gives us "-11"
Question 2:
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
A zero fill right shift converts it's operands to signed 32-bit integers. The result of that operation is always an unsigned 32-bit integer.
The operands of all bitwise operators are converted to signed 32-bit
integers in two's complement format.

-3 >>> 0 (right logical shift) coerces its arguments to unsigned integers, which is why you get the 32-bit two's complement representation of -3.
http://en.wikipedia.org/wiki/Two%27s_complement
http://en.wikipedia.org/wiki/Logical_shift

var binary = (-3 >>> 0).toString(2); // coerced to uint32
console.log(binary);
console.log(parseInt(binary, 2) >> 0); // to int32
on jsfiddle
output is
11111111111111111111111111111101
-3

.toString() is designed to return the sign of the number in the string representation. See EcmaScript 2015, section 7.1.12.1:
If m is less than zero, return the String concatenation of the String "-" and ToString(−m).
This rule is no different for when a radix is passed as argument, as can be concluded from section 20.1.3.6:
Return the String representation of this Number value using the radix specified by radixNumber. [...] the algorithm should be a generalization of that specified in 7.1.12.1.
Once that is understood, the surprising thing is more as to why it does not do the same with -3 >>> 0.
But that behaviour has actually nothing to do with .toString(2), as the value is already different before calling it:
console.log (-3 >>> 0); // 4294967293
It is the consequence of how the >>> operator behaves.
It does not help either that (at the time of writing) the information on mdn is not entirely correct. It says:
The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.
But this is not true for all bitwise operators. The >>> operator is an exception to the rule. This is clear from the evaluation process specified in EcmaScript 2015, section 12.5.8.1:
Let lnum be ToUint32(lval).
The ToUint32 operation has a step where the operand is mapped into the unsigned 32 bit range:
Let int32bit be int modulo 232.
When you apply the above mentioned modulo operation (not to be confused with JavaScript's % operator) to the example value of -3, you get indeed 4294967293.
As -3 and 4294967293 are evidently not the same number, it is no surprise that (-3).toString(2) is not the same as (4294967293).toString(2).

Just to summarize a few points here, if the other answers are a little confusing:
what we want to obtain is the string representation of a negative number in binary representation; this means the string should show a signed binary number (using 2's complement)
the expression (-3 >>> 0).toString(2), let's call it A, does the job; but we want to know why and how it works
had we used var num = -3; num.toString(-3) we would have gotten -11, which is simply the unsigned binary representation of the number 3 with a negative sign in front, which is not what we want
expression A works like this:
1) (-3 >>> 0)
The >>> operation takes the left operand (-3), which is a signed integer, and simply shifts the bits 0 positions to the left (so the bits are unchanged), and the unsigned number corresponding to these unchanged bits.
The bit sequence of the signed number -3 is the same bit sequence as the unsigned number 4294967293, which is what node gives us if we simply type -3 >>> 0 into the REPL.
2) (-3 >>> 0).toString
Now, if we call toString on this unsigned number, we will just get the string representation of the bits of the number, which is the same sequence of bits as -3.
What we effectively did was say "hey toString, you have normal behavior when I tell you to print out the bits of an unsigned integer, so since I want to print out a signed integer, I'll just convert it to an unsigned integer, and you print the bits out for me."

Daan's answer explains it well.
toString(2) does not really convert the number to two's complement, instead it just do simple translation of the number to its positive binary form, while preserve the sign of it.
Example
Assume the given input is -15,
1. negative sign will be preserved
2. `15` in binary is 1111, therefore (-15).toString(2) gives output
-1111 (this is not in 2's complement!)
We know that in 2's complement of -15 in 32 bits is
11111111 11111111 11111111 11110001
Therefore in order to get the binary form of (-15), we can actually convert it to unsigned 32 bits integer using the unsigned right shift >>>, before passing it to toString(2) to print out the binary form. This is the reason we do (-15 >>> 0).toString(2) which will give us 11111111111111111111111111110001, the correct binary representation of -15 in 2's complement.

Math.random and arithmetic shift

If I have the following code in JavaScript:
var index1 = (Math.random() * 6) >> 0;
var index2 = Math.floor(Math.random() * 6);
The results for index1 or index2 are anywhere between 0 and 6.
I must be confused with my understanding of the >> operator. I thought that by using arithmetic shift that the results for index1 would be anywhere between 1 and 6.
I am noticing, however that I don't need to use Math.floor() or Math.round() for index1 if I use the >> operator.
I know I can achieve this by adding 1 to both indexes, but I was hoping there was a better way of ensuring results are from 1 to 6 instead of adding 1.
I'm aware that bitwise operators treat their operands as a sequence of 32 bits (zeros and ones), rather than as decimal, hexadecimal, or octal numbers. For example, the decimal number nine has a binary representation of 1001. Bitwise operators perform their operations on such binary representations, but they return standard JavaScript numerical values.
UPDATE:
I saw the original usage in this CAAT tutorial on line 26 and was wondering whether that would actually return a random number between 1 and 6 and it seems it would only ever return a random number between 0 and 6. So you would never actually see the anim1.png fish image!
Thank you in advance for any enlightenment.

Wikipedia says '(Arithmetic right shifts for negative numbers would be equivalent to division using rounding towards 0 in one's complement representation of signed numbers as was used by some historic computers.)'
Not exactly an answer, but the idea is that >> 0 is really specific and shouldn't be used in general for getting a range between 1 and 6.
Most people would tell you to do
Math.floor((Math.random()*10)+1);

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