gistfile1.txt

Writing 16 bit value 0xbbaa (ByteCount = 2) at various offsets

+-----------------------------------+---------------------------------------------------------+
|                 DW                |                                                         |
+--------+--------+--------+--------+-------------+-------------------------------------------+
+ byte 3 | byte 2 | byte 1 | byte 0 | Address & 3 | ((LowerAddress & 3) + ByteCount + 3) >> 2 |
+--------+--------+--------+--------+-------------+-------------------------------------------+
|        |        |   bb   |   aa   |     0       | ((0 & 3) + (2 + 3)) >> 2 = 5 / 4 = 1      |
|        |        |        |        |             |                                           |
|        |   bb   |   aa   |        |     1       | ((1 & 3) + (2 + 3)) >> 2 = 6 / 4 = 1      |
|        |        |        |        |             |                                           |
|   bb   |   aa   |        |        |     2       | ((2 & 3) + (2 + 3)) >> 2 = 7 / 4 = 1      |
|        |        |        |        |             |                                           |
|   aa   |        |        |        |     3       | ((3 & 3) + (2 + 3)) >> 2 = 8 / 4 = 2      |
+--------+--------+--------+--------+-------------+-------------------------------------------+

[12:58] promach_, regarding last gist comment. pow2 must be a power of two, otherwise it won't work
[12:58] <promach_> yeah, PCIe is a computer protocol, and it should follow the binary rule of computing

[13:00] 2^n is guaranteed to have n zeros in the least significant bits, and bit n is guaranteed to be 1, and all bits to the left of bit n are guaranteed to be zero
[13:00] == M_D_K [~[email protected]] has quit [Quit: ZNC - http://znc.in]
[13:01] (2^n)-1 is guaranteed to have (n-1) 1's in the least significant bits, and all bits >= bit n are guaranteed to be zero
[13:01] ^ that is why it works

[13:03] promach_, pow2-1 is n 1s, ~(pow2 - 1) has all 1s from position n+1 onwards, (x + (pow2-1)) has a 1 in position n+1 and 0s onwards

why does ROUNDUP(x, pow2) = ((x + (pow2 - 1)) & ~(pow2 - 1)) round up to the next integer in pow (where pow2 must be a power of two ) ?

[21:33] promach_: there are two cases to consider - first when x is already multiple of pow2, then the operation is no-op, check yourself
[21:34] promach_: otherwise, x + pow2 - 1 will be >= next multiple of pow2
[21:34] and & ~(pow2-1) just clears low bits

https://www.reddit.com/r/math/comments/7hxr3z/poweroftwo_integer_roundup_maths_algorithms/