On Tue, Jun 28, 2016 at 09:57:38PM -0400, Steven Rostedt wrote: > On Wed, 29 Jun 2016 10:32:34 +0900 > Namhyung Kim <namhy...@kernel.org> wrote: > > > > is at 32-37 > > > > > > For a total of 38 bytes. I'm betting that without the packed, the 4 > > > extra bytes will always be at the end. > > > > Woundn't it be 36 or 40 bytes? :) > > Ug, don't know what I was counting then. I added the 64 bit version as > a after thought. > > > > > > > > > If the compiler places it incorrectly without any attribute, it will > > > fail to read the long long if the arch requires 64 bits to be 8 bytes > > > aligned. The alignment is meaningless here. All we need is "packed" and > > > be done with it. It's only going to truncate the 4 bytes at the end of > > > the structure if that. > > > > I agree that in-struct alignment preserved without the 'aligned' > > attribute but I'm not sure whether it's guaranteed that the *start* > > address of the struct is still in proper alignment boundary. > > > > IOW the struct ftrace_graph_ret should be placed at 8-byte boundary in order > > to keep alignment of struct members. Is it guaranteed after applying > > 'packed'? > > > > My point is, a structure doesn't change size depending on where it is > located. Thus, if a structure contains a 8 byte field, that must be 8 > bytes aligned due to architecture constraints, then it had better be > aligned that way everywhere. If it is not, then accessing the 8 byte > fields will cause issues. > > The only thing that the "packed" changes, is it removes the last 4 > padded bytes of the structure. It doesn't change anything else. Hence, > the alignment is just extra and unneeded.
Ok then, I'll change it to use 'packed' only in v3. Thanks, Namhyung
