Bit Hack - Round off to multiple of 8

Question

can anyone please explain how this works (asz + 7) & ~7; It rounds off asz to the next higher multiple of 8.

It is easy to see that ~7 produces 11111000 (8bit representation) and hence switches off the last 3 bits ,thus any number which is produced is a multiple of 8.

My question is how does adding asz to 7 before masking [edit] produce the next higher[end edit] multiple of 8 ? I tried writing it down on paper

like :

1 + 7 = 8  = 1|000 (& ~7) -> 1000
2 + 7 = 9  = 1|001 (& ~7) -> 1000
3 + 7 = 10 = 1|010 (& ~7) -> 1000
4 + 7 = 11 = 1|011 (& ~7) -> 1000
5 + 7 = 12 = 1|100 (& ~7) -> 1000
6 + 7 = 13 = 1|101 (& ~7) -> 1000
7 + 7 = 14 = 1|110 (& ~7) -> 1000
8 + 7 = 15 = 1|111 (& ~7) -> 1000

A pattern clearly seems to emerge which has been exploited .Can anyone please help me it out ?

Thank You all for the answers.It helped confirm what I was thinking. I continued the writing the pattern above and when I crossed 10 , i could clearly see that the nos are promoted to the next "block of 8" if I can say so.

Thanks again.

Answer 1

Adding 7 does not produce a multiple of 8. The multiple of 8 is produced by anding with ~7. ~7 is the complement of 7, which is 0xffff fff8 (except using however many bits are in an int). This truncates, or rounds down.

Adding 7 before doing that insures that no value lower than asz is returned. You've already worked out how that works.