> > 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?
