Bit Hacks

This assumes you have a decent understanding of bitwise operators. All hacks are from MIT 6.172, Fall 2018.

Set the kth Bit

Create an OR mask with a 1 in the position you want to set.

y = x | (1 << k);

Clear the kth Bit

Create an AND mask with a 0 in the position you want to clear.

y = x & ~(1 << k);

Toggle the kth Bit

Create an XOR mask with a 1 in the position you want to toggle.

y = x ^ (1 << k);

Extract a Bit Field

Create a mask with the bits you want to extract and shift them such that the rightmost bit is at the 0 position.

(x & mask) >> shift;

Set a Bit Field

Invert the mask to clear the bits (in case of junk), shift the bits you want to the correct position and OR them.

x = (x & ~mask) | (y << shift);

No-Temp Swap

If we let $a$ be the initial state of $x$ and $b$ the initial state of $y$, we can expand out the following:

$$\begin{array}{rl}a& =x\oplus y\\ b& =a\oplus y=(x\oplus y)\oplus y=x\\ a& =a\oplus b=(x\oplus y)\oplus x=y\end{array}$$

Converting this to C code gives us:

x = x ^ y;
y = x ^ y;
x = x ^ y;

Effectively, we mix $x$ and $y$ temporarily and then unmix them, which is a key feature enabled by XOR.

Minimum of Two Integers

The standard technique for finding the minimum involves a branch (x < y ? x : y), which can hurt branch prediction, a compiler optimization. We can do something more clever with bit hacks:

r = y ^ ((x ^ y) & -(x < y));

This works due to how boolean values work in C, where if x < y, -(x < y) yields -1, which is just all 1s in 2s complement. Thus, y ^ (x ^ y) would just return x. If x > y, -(x < y) yields 0, which after ANDing becomes 0. Thus, y ^ 0 would just return y.

Least-Significant 1

We can compute the 1 bit with the least-significant position by simply ANDing with the inverse of a number and adding 1.

r = x & (-x);

This works because the negative of a number is the same as a number’s 2s complement form, which provides a mask for all but the least-significant digit.