I was given the task to figure out how to determine odd and even. I could not use %. I used & because I found it out on the internet but I couldn't find a decent for the way it works.
N/A
My sample I created was
`
if ((22 & 1) === 0) {
return true;
} else{
return false;
}`
Returns true
The & bitwise operator works like this:
var isOdd = number & 1;
var isEven = !(number & 1);
(22 & 1) === 0 is true and so that tells you that it is even because number & 1 equals 0 if the number is even and 1 if the number is odd.
In binary notation the right-most bit is the ones place:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
...etc
So as you can see every odd number ends in 1 and every even ends in 0.
When you use & you are making a bit-wise and calculation. When you do X & 1, are you comparing each bit of X against 1 or in binary: 00001 (you can keep extending the zeros to the left) and testing wether both bits are 1.
So for example 22 is 10110 in binary 22 & 1 looks each bit and test if both are true:
1 0 1 1 0
0 0 0 0 1 < no bits are 1 in both number
---------
0 0 0 0 0 < all zeros == 0 so 22 is even
23 is 10111:
1 0 1 1 1
0 0 0 0 1 the last bit is one in both numbers
---------
0 0 0 0 1 < 1 so 23 is odd
Since the last bit is always 1 in odd numbers x & 1 will always be one for odd numbers and zero for evens.
function isEven(number) {
return (number & 1) == 0;
}
const userInput = 4;
console.log(isEven(userInput) == true ? "Even" : "Odd");
