Thread: Determine if a number lies on a 4-byte boundary

I've been racking my mathless brain for the last 45 minutes, looked at two books (if Bit Twiddling Hacks counts as a book) all to no avail.

The answer is no doubt really obvious. I'm tired. Very, very tired.

EDIT: Yes, I have googled.

I'm not quite sure what you're asking here. If you want to find whether a number can be stored in 4 bytes, I'd check whether it's less than 2^32.

QuantumPete

I want to know if it's dead on a 4-byte boundary, like is it equal to 32, 64, 96 etc. I guess I could use a lookup table with a search, but I was hoping there's a way of calculating it.

Hmm, you could mod it with for 4...

Or I could shoot myself in the FACE for being so stupid. Thank you.

Originally Posted by ahluka

Or I could shoot myself in the FACE for being so stupid. Thank you.

No problem ;-)

>I want to know if it's dead on a 4-byte boundary, like is it equal to 32, 64, 96 etc.

Code:

    if ((num & 0x1f) == 0)

The normal way to check the alignment is to and with alignment-1 and check that it's zero, e.g.

Code:

   if ((address & 3) == 0) ... // it's aligned
   else ... // unaligned.

If you want to make it generic, you could do:

Code:

   if ((address & sizeof(type)-1) == 0) ...

This works for all types that have sizes that are 2^n, e.g 2, 4, 8, 16, 32, 64, etc.

Note that the parenthesis around (address & x) are necessary, otherwise the compiler will evaluate 3 == 0 [or whatever] first, then perform the and. If it's false, like in this case, it will be false all over, since X & 0 is always zero, so false. Don't ask me how I remember this, just trust me that if you've made the same mistake enough times, you don't forget quite so easily.

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Mats