Why does (x-1) toggle all the bits from the rightmost set bit of x?

问题

Why does this property hold true?

say the kth bit from right side is the first set bit in number 'x'. (x-1) will toggle every bit upto kth bit from right side.

I can verify this property by writing the bit sequence for the numbers but I don't understand as to why this property is true? Can anyone help me with a simple proof or intuition as to why this works?

回答1:

Let's do a hand subtraction

  xxx100...00
- xxx000...01
─────────────
  xxx011...11

x represents bits that we don't care

Starting from the right, 10 - 1 = 1, borrowing 1 from the next column

Then the next one is 0 - 0, minus the borrow, which also results in 0 - 1 = 1 borrows 1. This sequence continues until the bit in subtrahend is 1, now we have 1 - 0 - borrow = 1 - 1 = 0

As the result, all the bits up to the rightmost set bit will be inverted

回答2:

It's the same reason if you do (x-1) where x is a base 10 integers all the zeros (if any) on the right will become nines.

9324930000000 - 1 = 9324929999999

来源：https://stackoverflow.com/questions/50869508/why-does-x-1-toggle-all-the-bits-from-the-rightmost-set-bit-of-x