I am performing bitwise operations, the result of which is apparently being stored as a two's complement number. When I hover over the variable it's stored in I see- num = -2086528968.
The binary of that number that I want is - (10000011101000100001100000111000).
But when I say num.toString(2) I get a completely different binary representation, the raw number's binary instead of the 2s comp(-1111100010111011110011111001000).
How do I get the first string back?
Link to a converter: rapidtables.com/convert/number/decimal-to-binary.html
Put in this number: -2086528968
Follow bellow the result:
var number = -2086528968;
var bin = (number >>> 0).toString(2)
//10000011101000100001100000111000
console.log(bin)
pedro already answered this, but since this is a hack and not entirely intuitive I'll explain it.
I am performing bitwise operations, the result of which is apparently being stored as a two's complement number. When I hover over the variable its stored in I see num = -2086528968
No, the result of most bit-operations is a 32bit signed integer. This means that the bit 0x80000000 is interpreted as a sign followed by 31 bits of value.
The weird bit-sequence is because of how JS stringifies the value, something like sign + Math.abs(value).toString(base);
How to deal with that? We need to tell JS to not interpret that bit as sign, but as part of the value. But how?
An easy to understand solution would be to add 0x100000000 to the negative numbers and therefore get their positive couterparts.
function print(value) {
if (value < 0) {
value += 0x100000000;
}
console.log(value.toString(2).padStart(32, 0));
}
print(-2086528968);
Another way would be to convert the lower and the upper bits seperately
function print(value) {
var signBit = value < 0 ? "1" : "0";
var valueBits = (value & 0x7FFFFFFF).toString(2);
console.log(signBit + valueBits.padStart(31, 0));
}
print(-2086528968);
//or lower and upper half of the bits:
function print2(value) {
var upperHalf = (value >> 16 & 0xFFFF).toString(2);
var lowerHalf = (value & 0xFFFF).toString(2);
console.log(upperHalf.padStart(16, 0) + lowerHalf.padStart(16, 0));
}
print2(-2086528968);
Another way involves the "hack" that pedro uses. You remember how I said that most bit-operations return an int32? There is one operation that actually returns an unsigned (32bit) interger, the so called Zero-fill right shift.
So number >>> 0 does not change the bits of the number, but the first bit is no longer interpreted as sign.
function uint32(value){
return value>>>0;
}
function print(value){
console.log(uint32(value).toString(2).padStart(32, 0));
}
print(-2086528968);
will I run this shifting code only when the number is negative, or always?
generally speaking, there is no harm in running nr >>> 0 over positive integers, but be careful not to overdo it.
Technically JS only supports Numbers, that are double values (64bit floating point values). Internally the engines also use int32 values; where possible. But no uint32 values. So when you convert your negative int32 into an uint32, the engine converts it to a double. And if you follow up with another bit operation, first thing it does is converting it back.
So it's fine to do this like when you need an actual uint32 value, like to print the bits here, but you should avoid this conversion between operations. Like "just to fix it".
Related
How do you print an unsigned integer when using JavaScript's BigInt?
BigInts can be printed as binary representation using toString(2). However for negative values this function just appends a - sign when printing.
BigInt(42).toString(2)
// output => 101010
BigInt(-42).toString(2)
// output => -101010
How do I print the unsigned representation of BigInt(42)? I that with regular numbers you can do (-42 >>> 0).toString(2), however the unsigned right shift seems not to be implemented for BigInt, resulting in an error
(BigInt(-42) >>> BigInt(0)).toString(2)
// TypeError: BigInts have no unsigned right shift, use >> instead
An easy way to get the two's complement representation for negative BigInts is to use BigInt.asUintN(bit_width, bigint):
> BigInt.asUintN(64, -42n).toString(2)
'1111111111111111111111111111111111111111111111111111111111010110'
Note that:
You have to define the number of bits you want (64 in my example), there is no "natural"/automatic value for that.
Given only that string of binary digits, there is no way to tell whether this is meant to be a positive BigInt (with a value close to 2n**64n) or a two's complement representation of -42n. So if you want to reverse the conversion later, you'll have to provide this information somehow (e.g. by writing your code such that it implicitly assumes one or the other option).
Relatedly, this is not how -42n is stored internally in current browsers. (But that doesn't need to worry you, since you can create this output whenever you want/need to.)
You could achieve the same result with a subtraction: ((2n ** 64n) - 42n).toString(2) -- again, you can specify how many bits you'd like to see.
Is there something like bitAtIndex for BigInt?
No, because there is no specification for how BigInts are represented. Engines can choose to use bits in any way they want, as long as the resulting BigInts behave as the specification demands.
#Kyroath:
negative BigInts are represented as infinite-length two's complement
No, they are not: the implementations in current browsers represent BigInts as "sign + magnitude", not as two's complement. However, this is an unobservable implementation detail: implementations could change how they store BigInts internally, and BigInts would behave just the same.
What you probably meant to say is that the two's complement representation of any negative integer (big or not) is conceptually an infinite stream of 1-bits, so printing or storing that in finite space always requires defining a number of characters/bits after which the stream is simply cut off. When you have a fixed-width type, that obviously defines this cutoff point; for conceptually-unlimited BigInts, you have to define it yourself.
Here's a way to convert 64-bit BigInts into binary strings:
// take two's complement of a binary string
const twosComplement = (binaryString) => {
let complement = BigInt('0b' + binaryString.split('').map(e => e === "0" ? "1" : "0").join(''));
return decToBinary(complement + BigInt(1));
}
const decToBinary = (num) => {
let result = ""
const isNegative = num < 0;
if (isNegative) num = -num;
while (num > 0) {
result = (num % BigInt(2)) + result;
num /= BigInt(2);
}
if (result.length > 64) result = result.substring(result.length - 64);
result = result.padStart(64, "0");
if (isNegative) result = twosComplement(result);
return result;
}
console.log(decToBinary(BigInt(5))); // 0000000000000000000000000000000000000000000000000000000000000101
console.log(decToBinary(BigInt(-5))); // 1111111111111111111111111111111111111111111111111111111111111011
This code doesn't do any validation, however.
When we don't know the numerical value at advance, can we convert it to BigInt or String without value corruption?
const targetNumber: number = 90071992547409946; // suppose we don't know it in advance
function splitEach3DigitsGroupWithComma(targetNumber: number | bigint | string): string {
if(targetNumber > Number.MAX_SAFE_INTEGER) {
// targetNumber = BigInt(targetNumber); // 90071992547409952n
// targetNumber = String(targetNumber); // 90071992547409950
// We need to do something before proceed!
}
return String(targetNumber).replace(/\B(?=(?:\d{3})+(?!\d))/gu, ",");
}
If the targetNumber is neither already a BigInt nor a string at the point where you first can work with it, then it's a plain number - and in that case, it may only be as precise as a number can be, per the IEEE 754 standard. If you only have a number to work with to begin with, possible more precise values have already been lost earlier.
To maintain precision, ensure that the value stays as a BigInt or string from beginning to end. If it gets converted to a number at any time in between, and then that number gets used, it may have lost precision, depending on the number.
In other words, to do something like this, you need to start with something like
const targetNumber = '90071992547409946';
or
const targetNumber = 90071992547409946n;
It can't be done in the middle of the process, once you already only have a number, while maintaining precision.
If your number is bigger than Number._MAX_SAFE_INTEGER then you can not safely convert into BigInt as is.
Say your big integer is 134640597783270600 which is > Number._MAX_SAFE_INTEGER. If you do like
> n = 134640597783270600;
> BigInt(n);
<- 134640597783270592n // Wrong..!
So you should first check if n is bigger than Number._MAX_SAFE_INTEGER and if so just convert it to string first
> BigInt(n > Number.MAX_SAFE_INTEGER ? n+"" : n);
<- 134640597783270600n // Correct..!
However, if n happens to be a big integer expressed in the exponential form like 7.576507751994453e+29 then this happens;
> BigInt(7.576507751994453e+29+"");
Uncaught SyntaxError: Cannot convert 7.576507751994453e+29 to a BigInt
at BigInt (<anonymous>)
at <anonymous>:1:1
At this time i think to be on the safe side, it's best to convert all those integers greater than Number.MAX_SAFE_INTEGER into String type first to test for exponential representation. So, despite we have a BigInt constructor in JS, apparently we still need a function.
function toBigInt(n){
return n.toString().includes("e") ? BigInt(n)
: BigInt(n+"");
}
I Have a hash function like this.
class Hash {
static rotate (x, b) {
return (x << b) ^ (x >> (32-b));
}
static pcg (a) {
let b = a;
for (let i = 0; i < 3; i++) {
a = Hash.rotate((a^0xcafebabe) + (b^0xfaceb00c), 23);
b = Hash.rotate((a^0xdeadbeef) + (b^0x8badf00d), 5);
}
return a^b;
}
}
// source Adam Smith: https://groups.google.com/forum/#!msg/proceduralcontent/AuvxuA1xqmE/T8t88r2rfUcJ
I use it like this.
console.log(Hash.pcg(116)); // Output: -191955715
As long as I send an integer in, I get an integer out. Now here comes the problem. If I have a floating number as input, rounding will happen. The number Hash.pcg(1.1) and Hash.pcg(1.2) will yield the same. I want different inputs to yield different results. A possible solution could be to multiply the input so the decimal is not rounded down, but is there a more elegant and flexible solution to this?
Is there a way to convert a floating point number to a unique integer? Each floating point number would result in a different integer number.
Performance is important.
This isn't quite an answer, but I was running out of room to make it a comment. :)
You'll hit a problem with integers outside of the 32-bit range as well as with non-integer values.
JavaScript handles all numbers as 64-bit floating point. This gives you exact integers over the range -9007199254740991 to 9007199254740991 (±(2^53 - 1)), but the bit-wise operators used in your hash algorithm (^, <<, >>) only work in a 32-bit range.
Since there are far more non-integer numbers possible than integers, no one-to-one mapping is possible with ordinary numbers. You could work something out with BigInts, but that will likely lead to comparatively much slower performance.
If you're willing to deal with the performance hit, your can use JavaScript buffer functions to get at the actual bits of a floating point number. (I'd say more now about how to do that, but I've got to run!)
Edit... back from dinner...
You can convert JavaScript's standard number type, which is 64-bit floating point, to a BigInt like this:
let dv = new DataView(new ArrayBuffer(8));
dv.setFloat64(0, Math.PI);
console.log(dv.getFloat64(0), dv.getBigInt64(0), dv.getBigInt64(0).toString(16).toUpperCase())
The output from this is:
3.141592653589793 4614256656552045848n "400921FB54442D18"
The first item shows that the number was properly stored as byte array, the second shows the BigInt created from the same bits, and the last is the same BigInt over again, but in hex to better show the floating point data format.
Once you've converted a number like this to a BigInt (which is not the same numeric value, but it is the same string of bits) every possible value of number will be uniquely represented.
The same bit-wise operators you used in your algorithm above will work with BigInts, but without the 32-bit limitation. I'm guessing that for best results you'd want to change the 32 in your code to 64, and use 16-digit (instead of 8-digit) hex constants as hash keys.
Look at those evaluations (actual dump from node 0.10.33)
> parseFloat(2.1e-17) === parseInt(2.1e-17)
false
> parseFloat(2.1e-17 + 2) === parseInt(2.1e-17 + 2)
true
> parseFloat(2.000000000000000000000000000000000009) === parseInt(2.00000000000000000000000000000000000009)
true
How can I tell integers from decimals very near to integers?
It seems that JS (or at least V8) doesn't care about digits smaller than 10^-16 when doing calculations, even if the 64bit representation used by the language (reference) should handle it.
Your examples are pretty much straight forward to explain. First thing to note is, that parseInt() and parseFloat() take a string as an input. So you inputs first get converted to string, before actually getting parsed.
The first is easy to see:
> parseFloat(2.1e-17) === parseInt(2.1e-17)
false
// look at the result of each side
parseFloat(2.1e-17) == 2.1e-17
parseInt(2.1e-17) == 2
When parsing the string "2.1e-17" as integer, the parse will stop at the dot as that is no valid digit and return everything it found until then, which is just 2.
> parseFloat(2.1e-17 + 2) === parseInt(2.1e-17 + 2)
true
// look at the result of each side
parseFloat(2.1e-17 + 2) == 2
parseInt(2.1e-17 + 2) == 2
Here the formula in the parameter will be evaluated first. Due to the limitations of floating point math (we just have 52bit for the mantissa and can't represent something like 2.000000000000000021), this will result in just 2. So both parseX() function get the same integer parameter, which will result in the same parsed number.
> parseFloat(2.000000000000000000000000000000000009) === parseInt(2.00000000000000000000000000000000000009)
true
Same argument as for the second case. The only difference is, that instead of a formula, that gets evaluated, this time it is the JavaScript parser, which converts your numbers just to 2.
So to sum up: From JavaScript's point of view, your numbers are just the same. If you need more precision, you will have to use some library for arbitrary precision.
This is something I learned from ReSharper
instead of using expressions like
if (2.00001 == 2) {}
try
if (Math.abs(2.00001 - 2) < tolerance) {}
where tolerance should be an aceptable value for you for example .001
so all values wich difference is less than .001 will be equals
Do you really need 10^-16 precision I mean that is why 1000 meter = 1 kilometer, just change the unit of the output so you dont have to work with all those decimals
I am trying to perform some bitshift operations and dealing with binary numbers in JavaScript.
Here's what I'm trying to do. A user inputs a value and I do the following with it:
// Square Input and mod with 65536 to keep it below that value
var squaredInput = (inputVal * inputVal) % 65536;
// Figure out how many bits is the squared input number
var bits = Math.floor(Math.log(squaredInput) / Math.log(2)) + 1;
// Convert that number to a 16-bit number using bitshift.
var squaredShifted = squaredInput >>> (16 - bits);
As long as the number is larger than 46, it works. Once it is less than 46, it does not work.
I know the problem is the in bitshift. Now coming from a C background, I know this would be done differently, since all numbers will be stored in 32-bit format (given it is an int). Does JavaScript do the same (since it vars are not typed)?
If so, is it possible to store a 16-bit number? If not, can I treat it as 32-bits and do the required calculations to assume it is 16-bits?
Note: I am trying to extract the middle 4-bits of the 16-bit value in squaredInput.
Another note: When printing out the var, it just prints out the value without the padding so I couldn't figure it out. Tried using parseInt and toString.
Thanks
Are you looking for this?
function get16bitnumber( inputVal ){
return ("0000000000000000"+(inputVal * inputVal).toString(2)).substr(-16);
}
This function returns last 16 bits of (inputVal*inputVal) value.By having binary string you could work with any range of bits.
Don't use bitshifting in JS if you don't absolutely have to. The specs mention at least four number formats
IEEE 754
Int32
UInt32
UInt16
It's really confusing to know which is used when.
For example, ~ applies a bitwise inversion while converting to Int32. UInt16 seems to be used only in String.fromCharCode. Using bitshift operators converts the operands to either UInt32 or to Int32.
In your case, the right shift operator >>> forces conversion to UInt32.
When you type
a >>> b
this is what you get:
ToUInt32(a) >>> (ToUInt32(b) & 0x1f)