On Fri, Apr 2, 2010 at 11:14 AM, Steven Watanabe <watanabesj@gmail.com> wrote:
Can't you just use boost::aligned_storage?
In Christ, Steven Watanabe
That would definitely help (can't recall if it was around back then). Either way, it doesn't solve the other problem of "I have a 12 byte object, what is its worse-case alignment?" I would guess 8, given today's machines, and align it to a double. Actually, no. I always start thinking that way, then realize I'm wrong. The alignment would be 4 (for today's typical machine). ie the largest sizeof(primitive_type) such that it divides into the size. And now that I think about it, maybe there is no reason to not use alignment of 12. Same diff I suppose. So maybe aligned_storage<12, 12> is all I need. I now I can't remember if the original hard problem was getting an answer of 8 given a size of 12 (which is wrong anyhow) or whether I'm forgetting some detail that bit me later. So a question. Given: struct { int a, b, c; }; is there any difference saying it has an alignment of 12 vs an alignment of 4? (assuming sizeof(int) == 4)? And/or is there any difference between boost::aligned_storage<12, 4> and boost_aligned_storage<12, 12>? Tony