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

Well, the mask would produce an exact multiple of 8 by itself. Adding 7 to asz ensures that you get the next higher multiple.

Answer 2

It's actually adding 7 to the number and rounding down.

This has the desired effect of rounding up to the next multiple of 8. (Adding +8 instead of +7 would bump a value of 8 to 16.)

Answer 3

Well, if you were trying to round down, you wouldn't need the addition. Just doing the masking step would clear out the bottom bits and you'd get rounded to the next lower multiple.

If you want to round up, first you have to add enough to "get past" the next multiple of 8. Then the same masking step takes you back down to the multiple of 8. The reason you choose 7 is that it's the only number guaranteed to be "big enough" to get you from any number up past the next multiple of 8 without going up an extra multiple if your original number were already a multiple of 8.

In general, to round up to a power of two:

unsigned int roundTo(unsigned int value, unsigned int roundTo)
{
    return (value + (roundTo - 1)) & ~(roundTo - 1);
}

Answer 4

The +7 isn't to produce an exact multiple of 8, it's to make sure you get the next highest multiple of eight.

edit: Beaten by 16 seconds and several orders of quality. Oh well, back to lurking.

Answer 5

Uhh, you just answered your own question??? by adding 7, you are guaranteeing the result will be at or above the next multiple of 8. truncating then gives you that multiple.

Answer 6

Without the +7 it will be the biggest multiple of 8 less or equal to your orig number

Answer 7

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.