Dropping Last Bit

Thu, 18 Nov 2021 20:00:20 +0530

Clearing or dropping the rightmost set bit of a number is easy if we use the property of x - 1. There we saw that subtracting one from a number changes the bit pattern so that the trailing zeroes are set to one and the last one is set to 0.

We also know that the remaining numbers to the left of the “one” (referred to as the wall bit) which is set are untouched. Given these two facts, we can see that if we AND the numbers X and X-1,

The bits to the left of the wall bit would not change
The bits to the right of the wall bit, including the bit itself, will be cleared (because for each bit we are ANDing 0 and 1 or 1 and 0)

These two facts combine to give us the final outcome that X & X-1 clears the rightmost set bit in a number:

10010010 1 00 # x
10010010 0 11 # x-1
10010010 0 00 # x & x-1

tags: bitmanip

categories: programming

Understanding X - 1

Thu, 18 Nov 2021 13:28:51 +0530

When subtracting one from a number, its binary representation changes in a predicable and useful manner:

All the trailing consecutive zeroes are flipped to ones
The first bit set to one found starting from the right is cleared to zero

As an example, consider

x = 3 # 11
x - 1 # 10

Here there are no trailing zeroes. So only the second change happens. The first one (which is the least significant bit) is flipped.

We can also see that no bits to the left of the “one” which is flipped are ever affected.

Here the unchanged part, the “wall” bit and the rest are separated by spaces for clarity:

1111100_1_000 -
0000000_0_001
1111100_0_111

The 1 bit in braces is the point where the changes to the bit pattern stop. It is flipped to 0. Everything to its right is flipped to 1. Everything to its left remains untouched.

tags: bitmanip

categories: programming

bitmanip on -abhijat-

Dropping Last Bit

Understanding X - 1