Dave Angel wrote:
--- On Sat, 7/16/11, Albert-Jan Roskam<fo...@yahoo.com> wrote:
(in my case 8 bytes) is aligned in the computer's memory such that the
boundary's address is a power of two.
Not quite. Doubleword alignment is alignment on an 8byte boundary. The
address of such a boundary will be a multiple of 8, not a power of two.
Oops, when I said Albert was "right" I meant in terms of aligning with
computer memory, I didn't register the "power of two" bit, which is, as
you say, not correct. Apologies for any confusion caused!
Alan G.
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