> > > I'm not entirely convinced this is true; If we'll not enforce the
> > > alignment of this 64-bit field, it's possible there will be
> > > differences between 32-bit and 64-bit machines versions of this struct.
> > > You have to recall that this is going to be copied via DMA between
> > > PF and VF, so they must have the exact same representation of the
> > > structure.
> >
> > Then use properly sized types to fill in all the space in the
> > structure, that's how you guarantee layout, not aligned_u64. Also, do not
> > use
> the packed attribute.
> >
> > struct foo {
> > u32 x;
> > u32 y;
> > u64 z;
> > };
> >
> > 'z' will always be 64-bit aligned.
>
> Perhaps my bit-numeric is a bit weak - why is it so?
> I.e., what prevents `z' from only being 32-bit aligned on a 32-bit machine?
> Isn't it possible that (&x % 8) == 4, (&y % 8) == 0 and (&z % 8) == 4 on
> such a
> platform?
If struct foo were to have been allocated, would it be guaranteed to be 64-bit
aligned on a 64-bit platform and 32-bit aligned on a 32-bit platform?
Assuming it is the case, is it theoretically possible that you'd have 2
different
32-bit platforms, where on one the `z' field would be packed while on the other
it will not, introducing a 4-byte gap [assuming `foo' itself was only 32-bit
aligned]?
