fun Int.floorEven() = this and 0x01.inv()

I was reading a project and i saw this code but i could not understand why this code flooring the number to even i know 01 inverse equals to -2 because of 2's complement equals to 1's complement - 1 and 2 complent is negative of that number but i could not understand why this code will always work?

it clears the lowest bit and that is equivalent to flooring to even. consider that

this shr 1 shl 1

is equivalent to

this.floorDiv(2) * 2

which also clears the lowest bit.

Firstly correct me if i am wrong 0x used as a prefix for hexadecimal numbers right?And that's why 0x01 means 01 and inv() just makes it 10 anding with any number will clear least significant bit.But i also tried 0x10 without inv method but it did not work.Can you explain why? @ephemient

inv() makes it 0b11111111….1110

But no one wants to type 31 1s, so it makes sense to do a 1.inv(). The 0x just makes it clear we’re doing bit math.

ok i got it now thank you both

When we do a bitwise and with 0b1111111…110, we end up with the most significant 31 bits getting and’ed with 1, and (x and 1) == x. The last gets and’ed with 0, and (x and 0) == 0.

but there is a thing

I’ve done (this - this % 2) before for floor even.

same as writing

1.inv()

as well

yeah both works

but hex numbers are usually written with an even number of digits, because each pair is one byte

Just style. Hexadecimal is mainly used in bitmath, so the 0x makes it clear we’re doing bitmath. Some people, such as myself, like having an even number of digits.

it is kinda naming convention?

there are certainly exceptions, for example the space of Unicode codepoints is 21 bits