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How does bitwise operation work on Booleans?

I came across this challenge on Edabit and couldn't work out this bitwise operation solution.
notNotNot = (a,b) => !!(a%2 >> b)
The challenge:
//Something which is not true is false, but something which is not not true is true!
//Create a function where given n number of "not", evaluate whether it's true or false.
//Examples:
notNotNot(1, true) ➞ false
// Not true
notNotNot(2, false) ➞ false
// Not not false
notNotNot(6, true) ➞ true
// Not not not not not not true
I did some research that that operator:
Shifts right by pushing copies of the leftmost bit in from the left, and let the rightmost bits fall off.
That I reckon I understood (e.g. 5 >> 1 same as 0101 >> 1 which evaluates to 0010), but I can't see how that works with a boolean? I know true evaluates to 1 and false to 0.

The function you gave does not satisfy the challenge. Right shifting will not do what is asked for. For example, your notNotNot(6,true) is false, not true when put through your function.
Your question is about bitwise operation on a boolean though. Since operators like >> and << work on integers, Javascript first converts the boolean value to an integer. So true becomes 1 and false becomes 0. To see this you can shift by zero:
console.log("true is",true >> 0)
console.log("false is", false >> 0)
So bitwise operation on booleans is really just bitwise operation on either 0 or 1.
Using !! is a handy way to convert anything into a boolean. It takes anything that would be considered equivalent to false (such as 0, null, undefined or "") and gives back false. Similarly anything that is truthy (like 14, "hello", [4], {a:1}) and give back true. !! works because the first exclamation mark gives the 'not' of the expression which is always true or false, then the second exclamation mark gives the opposite of that (false or true).
Getting back to the challenge, it wants to apply the not-operator 'a' times and compare to the 'b' value. So something like this would work:
function notNotNot(a, b) { return !!(a%2 - b); }
console.log("notNotNot(1, true)",notNotNot(1, true));
console.log("notNotNot(2, false)",notNotNot(2, false));
console.log("notNotNot(6, true)",notNotNot(6, true));

Bitwise operators always convert their operands to an integer. So, 4 >> true is the same as 4 >> 1 which will do a bit shift right by one position
(decimal) 4 = (binary) 100
(binary) 100 >> 1 = (binary) 010
(binary) 010 = (decimal) 2
console.log(4 >> true);
So, using true or false is a just a roundabout way to use 1 or 0.
The notNotNot function has very simple operation, overall:
a%2 converts the first number into 0 for even or 1 for odd.
>> b shifts right by either 0 positions for false or 1 position for true.
a is odd (1) and b is false = 1
there is zero shifts to the right, so the number remains the same.
a is odd (1) and b is true = 0
the only set bit 1 is shifted right and discarded.
a is even (0) and b is false = 0
there is zero shifts to the right, so the number remains the same.
a is even (0) and b is true = 0
the base number is 0 which doesn't have any bits set, so shifting right any amount does not change it.
!!() converts the result to boolean.
With that said, the solution here is wrong, since notNotNot(2, true) will produce false - a is even and b is true. The expectation is that it will produce true since !!true = true. The same problem is present for any even number and true.
It can be easily fixed by using bitwise XOR instead of right shift:
a is odd (1) and b is false = 1
both match, so they are flipped to 0
a is odd (1) and b is true = 0
they don't match, so we get 1
a is even (0) and b is false = 0
both match, so we get 0
a is even (0) and b is true = 1
they don't match, so we get 1
notNotNot = (a,b) => !!(a%2 ^ b);
console.log("!!true = ", notNotNot(2, true))
console.log("!!!true =", notNotNot(3, true))
console.log("!!false = ", notNotNot(2, false))
console.log("!!!false = ", notNotNot(3, false))
//bonus
console.log("true = ", notNotNot(0, true))
console.log("false = ", notNotNot(0, false))
Just for completeness sake, in case you want a fully bitwise operation:
The modulo operation %2 can be changed to a bitwise AND &1 get the lowest bit. For even numbers, this would yield 0 since you'd be computing
xxx0
&
0001
which is zero. And for odd numbers the same applies but you'd get one as a result:
xxx1
&
0001
So the results of a&1 and a%2 are identical. Furthermore, even though bitwise operations convert the number to a 32-bit signed integer that doesn't matter as the parity would be preserved.
//larger than 31 bits
const largeValue = 2**31 + 1;
//larger than 32 bits
const veryLargeValue = 2**32 + 1
console.log("2**31 + 1 =", largeValue);
console.log("2**32 + 1 =", veryLargeValue);
console.log("2**31 + 1 to 32-bit signed integer =", largeValue | 0);
console.log("2**32 + 1 to 32-bit signed integer = ", veryLargeValue | 0);
const isOddModulo = number =>
console.log(`(${number} % 2) can detect an odd number: ${(number % 2) === 1}`);
const isOddBitwise = number =>
console.log(`(${number} & 1) can detect an odd number: ${(number & 1) === 1}`);
isOddModulo(largeValue);
isOddBitwise(largeValue);
isOddModulo(veryLargeValue);
isOddBitwise(veryLargeValue);

Firstly, (a,b) => !!(a%2 >> b) does not match the results of the examples. I will break down exactly what it's doing using notNotNot(6, true) ➞ true.
Fist a%2, simply get a divide by 2 return the remainder. So we will get 0 for an even number and 1 for an odd number. a = 6 a%2 = 0 in this case.
Then 0 >> b shift 1 number off from the right because as you said true evaluates to 1. So we get 0 >> 1 = 0.
Last !!(0), is simple, and can be broken down like so, !0 = true, then !true = false.
So if we think about this as long as b is true, we will always get returned false. Let's say we have a = 5, b = true evaluating to 5%2 = 1, 1 >> 1 = 0. You can see because of the mod (%2) we will only ever have 1 or 0 (only ever have 1 digit) and true will always shift off the 1 when we have it.
A simple way to look at this problem is like an isEvenOrNot function. So a is the number we are checking and b is a boolean to check if it's even (true) or not even (false). This works because every second not added will be true.
So a solution using bitwise could be something like: (a,b) => !!(a&1 ^ b).
I will let you have the fun of breaking down why it works! :)
A little bit more into explaining how shift works with a boolean. So true as you said will be 1 and false will be 0. So as shown in your example, 0101 >> true is the same as 0101 >> 1.
I hope this helps.
I used the following as a reference for bitwise: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators

(a%2) //ignore all but the least significant bit (LSB)
(a%2 >> b ) //if TRUE, shifts right, resolves to 0
//if FALSE, no shift, resolves to LSB
// 0 and LSB are both integers so convert to boolean by using logical/boolean NOT
!(a%2 >> b ) //resolves to the boolean which it is NOT
!!(a%2 >> b ) //resolves to the boolean which it is NOT NOT
NB For either boolean,
an even number of NOTs results in the original boolean
an odd number of NOTs results in the opposite boolean
The LSB of any number dictates whether the number is odd or even.(0 even, 1 odd)

I see that your task is:
/* Create a function where given n number of "not",
evaluate whether it's true or false.*/
I don't know why you are writing notnotnot function
for me that is not what the task asks.
So according to the task I made that function not
that accepts a number of "nots" and evaluates them.
The first way
function not(n) {
return Boolean(n - n - 1);
}
The second way using XOr(^)
function not(n) {
return Boolean(bool ^ (bool - 1));
}
The third way using Mod(%) pointed by #VLAZ
function not(n) {
return Boolean(n % 2);
}
The fourth way using bitwise And(&)
function not(n) {
return Boolean(n & 1);
}
Test
not(0)
//> false
not(1)
//> true
not(2)
//> false
not(515)
//> true

Lets analysis solution first
notNotNot(oddNumber, true) ➞ false
notNotNot(evenNumber, true) ➞ true
notNotNot(oddNumber, false) ➞ true
notNotNot(evenNumber, false) ➞ false
Now analysis the for (a,b) => !!(a%2 >> b)
a%2 == 0 ➞ even number
a%2 == 1 ➞ odd number
// For a%2 == 0
a%2 >> b ➞ if b is true ➞ 0 >> 1 ➞ 0 // Not working
a%2 >> b ➞ if b is false ➞ 0 >> 0 ➞ 0
// For a%2 == 1
a%2 >> b ➞ if b is true ➞ 1 >> 1 ➞ 0
a%2 >> b ➞ if b is false ➞ 1 >> 0 ➞ 1
Thats means this is not working for notNotNot(6, true) is true but current solution gives false.
We can you ^(XOR) operator to make it correct Like (a,b) => !!(a%2 ^ b)
Now analysis the for (a,b) => !!(a%2 ^ b)
a%2 == 0 ➞ even number
a%2 == 1 ➞ odd number
// For a%2 == 0
a%2 ^ b ➞ if b is true ➞ 0 ^ 1 ➞ 1 // Now working
a%2 ^ b ➞ if b is false ➞ 0 ^ 0 ➞ 0
// For a%2 == 1
a%2 ^ b ➞ if b is true ➞ 1 ^ 1 ➞ 0
a%2 ^ b ➞ if b is false ➞ 1 ^ 0 ➞ 1
!(a%2 ^ b) use `!` to make int as boolean but solution result will reversed then
!!(a%2 ^ b) use `!` again to reversed it again and make it correct.
Example:
notNotNot = (a,b) => !!(a%2 ^ b);
console.log("!!!!true = ", notNotNot(4, true))
console.log("!!!!false = ", notNotNot(4, false))
console.log("!!!true =", notNotNot(3, true))
console.log("!!!false = ", notNotNot(3, false))

Write Shortage of if-else statement with two cases in one line

There is a javascript expression that assigns name variable to renamed variable.
It is always the same with one exclusion:
renamed = name == 'John' ? 'Johnny' : name;
However, I want two exclusions:
rename John to Johny
rename Alex to Alexander
All other names are assigned with no changes.
Is it possible to write this expression in one string?
renamed = (name == 'John' || name == 'Alex') ? <____> : name;
I need it to be done in one string.
Thank you.

(name === 'John' && 'Johny') || (name === 'Alex' && 'Alexander') || name;
If name is John, then it goes to the next part in the && expression and returns Johny.
If name is Alex, then the like in the first case returns Alexander.
If neither of them is true, then return the name as it is.
Demo
This solution works because, in JavaScript, && operator evaluates the expression in the left and if it is falsy then the value will be returned and the right hand side expression will not be evaluated at all.
If the expression in the left evaluates to be Truthy, then the expression on the right side will be evaluated and the result will be returned as it is. For example
console.log(1 && 2);
# 2
console.log(0 && 2);
# 0
It first evaluates 1, it is Truthy so it 2 is evaluated and the value is returned. That is why it prints 2.
In the second case, 0 is evaluated to be Falsy. So, it is returned immediately. That is why it prints 0.
The same way
console.log("John" && "Johny");
# Johny
John will be evaluated to be Truthy and so Johny will also be evaluated and returned. That is why we get Johny.
As per ECMA 5.1 Standard, Truthiness of an object will be decided, as per the following table
+-----------------------------------------------------------------------+
| Argument Type | Result |
|:--------------|------------------------------------------------------:|
| Undefined | false |
|---------------|-------------------------------------------------------|
| Null | false |
|---------------|-------------------------------------------------------|
| Boolean | The result equals the input argument (no conversion). |
|---------------|-------------------------------------------------------|
| Number | The result is false if the argument is +0, −0, or NaN;|
| | otherwise the result is true. |
|---------------|-------------------------------------------------------|
| String | The result is false if the argument is the empty |
| | String (its length is zero); otherwise the result is |
| | true. |
|---------------|-------------------------------------------------------|
| Object | true |
+-----------------------------------------------------------------------+
Note: The last line, Object, which includes both objects and Arrays.

renamed = (name == 'john') ? 'johney': (name == 'alex'? 'alexander' : name);

you can try this:
renamed = (name == 'John' ? 'Johnny' : (name == 'Alex' ? 'Alexander' : name));

JavaScript if(x) vs if(x==true)

In JavaScript , in which cases the following statements won't be logically equal ?
if(x){}
and
if(x==true){}
Thanks

They are not at all equal.
if (x)
checks if x is Truthy where as the later checks if the Boolean value of x is true.
For example,
var x = {};
if (x) {
console.log("Truthy");
}
if (x == true) {
console.log("Equal to true");
}
Not only an object, any string (except an empty string), any number (except 0 (because 0 is Falsy) and 1) will be considered as Truthy, but they will not be equal to true.
As per ECMA 5.1 Standards, in if (x), Truthiness of x will be decided, as per the following table
+-----------------------------------------------------------------------+
| Argument Type | Result |
|:--------------|------------------------------------------------------:|
| Undefined | false |
|---------------|-------------------------------------------------------|
| Null | false |
|---------------|-------------------------------------------------------|
| Boolean | The result equals the input argument (no conversion). |
|---------------|-------------------------------------------------------|
| Number | The result is false if the argument is +0, −0, or NaN;|
| | otherwise the result is true. |
|---------------|-------------------------------------------------------|
| String | The result is false if the argument is the empty |
| | String (its length is zero); otherwise the result is |
| | true. |
|---------------|-------------------------------------------------------|
| Object | true |
+-----------------------------------------------------------------------+
Note: The last line object, which includes both objects and Arrays.
But in the later case, as per The Abstract Equality Comparison Algorithm,
If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.
If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
value of x will be converted to a number and that number will be checked against true.
Note:
In JavaScript, true is 1 and false is 0.
console.log(1 == true);
# true
console.log(0 == false);
# true

Several cases evaluate to false in the first form, such as empty string, 0, undefined, null.
If you want to be a bit more semantic about it, try the bang bang in front of the expression:
if(!!x){...}
this will convert the expression result to a truthy representing the same semantically. This is closer to an analogue to the expression you describe (x == true)
Also be aware that == is value comparions with type coercion, eg "3" == 3, whereas === asserts equal typing too.
So they are not the same, but often logically represent the same test, thanks to the semantics of the language and the !! you can use

Smart math regular expression

I know there are smart ppl out there. At the moment, I want to use one single and the most elegant regular expression to separate numbers to achieve some really simple math calculation.
It needs support some fuzzy number guessing for example:
1+.2 should equiv to 1 + 0.2
.1-.2 should equiv to 0.1 - 0.2
-.2+-100.2 should equiv to -0.2 - 100.2
Here is a quick demo I made up for you to play around
http://jsfiddle.net/av57A/60/
Is that possible to achieve those logic in one elegant regular expression?
Thanks.

I got it. Nice practice in regex for me. Also thanks for showing me jsfiddle. Never seen anything like that b4.
reg1 : ^[\-\d{1,}|\-*\.*\d*+]{1,}
reg2 (g) : \.{2}
reg3 (g) : ^(([\-]{0,})(\d*\.*\d*)?)|([\+\-\*\/]{1,})|((\d*)+(\.*\d*)?)
---
2-1 : 2 | - | 1
.1+.2 : 0.1 | + | 0.2
.2+1 : 0.2 | + | 1
-1-1.3 : -1 | - | 1.3
2+1 : 2 | + | 1
-2+1 : -2 | + | 1
2+-1 : 2 | + | -1
1+1.1 : 1 | + | 1.1
2.1+1 : 2.1 | + | 1
-1.0-1.3 : -1 | - | 1.3
-1-.3 : -1 | - | 0.3

When should you use parentheses inside an if statement's condition?

I have a if condition like so:
$(document).ready(function(){
var name = 'Stack';
var lastname = 'Overflow';
if( name == 'Stack' && lastname == 'Overflow' )
alert('Hi Stacker!');
});
So my alert is fired...
If I put my condition inside a brackets like this:
$(document).ready(function(){
var name = 'Stack';
var lastname = 'Overflow';
if( (name == 'Stack') && (lastname == 'Overflow') ){
alert('Hi Stacker!');
}
});
My alert is also fired...
My question is: When and why I should use parentheses inside my if condition? Thank you!

There is no much difference in your example.
You can read that for reference
Operator Precedence (JavaScript)

You use it to force associations, in your example this won't change anything. But consider this case:
A and B or C
is a lot different from
A and (B or C)
Here you would be preventing the expression from being understood as (A and B) or C which is the natural way javascript and maths do things.

Because of operator precedence :
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence
you'd better read this if you want more details
But basically == (equality) weights more than && (logical-and) , so A == B is evaluated before C && D given
C <- A == B
and
D <- E == F
so adding parenthesis to C or D dont matter,in that specific situation.

Brackets can be used, if multiple conditions needs to be checked.
For ex: User is also a Stacker if full name is 'StackOverflow' then add another condition with 'Or' operator.
if(((name == 'Stack') && (lastname == 'Overflow')) || FullName =='StackOverflow')
As you can see, name and last name are placed inside one bracket meaning that it gives a result either true or false and then it does OR operation with FullName condition.
And having brackets around name, lastname and fullname fields are optional since it doesn't make any difference to the condition. But if you are checking other condition with FullName then group them into bracket.

Brackets are never required in such situations.
Of course, try to read first solution (without) and second solution (with). Second one it's clear, fast-readable and easy to mantain.
Of course if you have to change precedence (in this case you just have an AND condition, but what if you need and AND and an OR? 1 and 2 or 3 priority changes between "(1 and 2) or 3" - "1 and (2 or 3)"

Brackets () are used to group statements.
Ex - if you have an expression '2 + 3 * 5'.
There are two ways of reading this: (2+3)*5 or 2+(3*5).
To get the correct o/p based on operator precedence, you have to group the correct expression like * has higher precedence over +, so the correct one will be 2+(3*5).

first bracket requires for complex condition to ensure operator precedence correctly. lets say a example
var x=1,y=1,z=0;
if(x==0 && y==1 || z==0)
{
//always true for any value of x
}
what will happen here
(0 && 1 ||1) ----> (0 ||1)----->1
&& has high precedence over ||
if(x==0 && (y==1 || z==0)) alert('1');
{
//correct way
}
if you do not use (y==1 || z==0) bracket the condition always will be true for any value of x.
but if you use (..) the condition return correct result.

Conditional statements can be grouped together using parenthesis. And is not only limited to if statements. You can run the example below in your Chrome Developer Tools.
Example 1:
Console Execution
false && false || true
// true
flow
false && false || true
| | |
|________| |
| |
| |
false or true
| |
|_________________|
|
|
true
Example 2:
Console Execution
false && (false || true)
// false
flow
false && (false || true)
| |
|______________|
|
|
false
Helpful resources for playing around with JSInterpreter and AST's:
https://neil.fraser.name/software/JS-Interpreter/
https://esprima.org/demo/parse.html#

Consider the statement
if( i==10 || j == 11 && k == 12 || l == 13)
what you would want is if either i is 10 or j is 11 And either k is 12 or l is 13 then the result shouldbe true, but say if i is 10, j is 11, k is 10 and l is 13 the condition will fail, because the fate of equation is decided at k as aoon as && comes in picture. Now if you dont want this to happen the put it like this
if( (i==10 || j == 11) && (k == 12 || l == 13))
In this ORs will be executed first and the result will be true.