Understand the code to adjust buffer

Question

Could anyone explain the following code to me?

adjustedbuf = (void *) ((uint64_t) buf & ~(pagesize - 1))

I don't get the idea of this statement.

It *is* some sort of "alignment". It might be helpful if you provides some context, the platform, and the value of `pagesize`. — Jonathan Wood, May 26 '14 at 18:46

Mat · Answer 1 · 2014-05-26T20:10:56.293

Assuming pagesize is a power of two, its binary representation will be something like:

  0000 10000000  // assume page size = 2^7 = 128

So pagesize-1 will be:

  0000 01111111  // 127

The negation of that (~) will be a bitmask, with all upper bits set, up to the "page size" bit:

  1111 10000000

If you & that with anything, you end up with a number that is "rounded down" to a multiple of the page size

  1100 10110110
& 1111 10000000
= 1100 10000000

Which is what that statement is doing. It aligns buf to a page size boundary.

_{(If pagesize isn't a power of two, the whole thing doesn't make much sense.)}

score 2 · Answer 2 · edited May 23 '17 at 12:05

2

It's page-aligning the buffer using bit operations. See

for more depth.

edited May 23 '17 at 12:05

Community

answered May 26 '14 at 18:47

bcr

2 Answers2