Why force power of 2 for capacity?

[yonderblue]

Perhaps I am not seeing it in the paper, but why force the capacity up to the next power of 2? When rebuilding as in the scalable filter, this decreases the potential load factor.

[jkomyno]

I didn't read the code accurately, but this is probably to force the compiler to avoid using the slow DIV assembly operation when using modulo operations.
If the capacity is a power of 2, you can compute modulos with bitwise ANDs.

Maybe @seiflotfy can expand on this.

[yonderblue]

Seems plausible, but I would think the size of the filter is often more important.

[wtysos11]

If the capacity is not the power of 2, then xor won't be involution. And as the candidate position is calcualted using h2(x) = h1(x) xor hash(fingerprint), the result may change and cause unexpected behaviour.
I answer a similar question in stackoverflow. There is a simple example. Assume fingerprint=128, h1(x) = 32, hash(fingerprint) = 73. The candidate position can be calculated using h2(x) = (h1(x) xor hash(fingerprint)) % capacity.
If capacity = 128, h2(x) = 32 xor 73 = 105, and 105 xor 73 = 32.
However if capacity = 100, h2(x) = 105%100 = 5, 5 xor 73 = 76. In this situation, once element swaps away from origin position, it can never go back. For solution you can see my answer or this answer