Re: [RFC/PATCH v2] ftrace: Reduce size of function graph entries

[Namhyung Kim <namhyung@kernel.org>]

On Tue, Jun 28, 2016 at 09:57:38PM -0400, Steven Rostedt wrote:
> On Wed, 29 Jun 2016 10:32:34 +0900
> Namhyung Kim <namhyung@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