Re: [PATCH] introduction: document bitfield notation

["Michael S. Tsirkin" <mst@redhat.com>]

On Mon, Mar 05, 2018 at 05:04:40PM +0100, Cornelia Huck wrote:
> On Mon, 5 Mar 2018 16:26:11 +0200
> "Michael S. Tsirkin" <mst@redhat.com> wrote:
> 
> > On Mon, Mar 05, 2018 at 03:20:35PM +0100, Cornelia Huck wrote:
> > > On Wed, 28 Feb 2018 21:16:32 +0200
> > > "Michael S. Tsirkin" <mst@redhat.com> wrote:
> > >   
> > > > Bitfields are a useful and familiar way to specify sub-byte structure
> > > > layout. The only issue is that bitfield order isn't portable across
> > > > architectures.  Document that we list bitfields from least to
> > > > most significant one, and warn about portability issues.
> > > > 
> > > > Signed-off-by: Michael S. Tsirkin <mst@redhat.com>
> > > > ---
> > > >  introduction.tex | 18 ++++++++++++++++++
> > > >  1 file changed, 18 insertions(+)
> > > > 
> > > > diff --git a/introduction.tex b/introduction.tex
> > > > index 979881e..3cb7a70 100644
> > > > --- a/introduction.tex
> > > > +++ b/introduction.tex
> > > > @@ -157,5 +157,23 @@ in little-endian byte order.
> > > >  in big-endian byte order.
> > > >  \end{description}
> > > >  
> > > > +When documenting sub-byte data fields, C-like bitfield notation
> > > > +is used. Fields within an integer are always listed in order,
> > > > +from the least significant to the most significant bit.
> > > > +
> > > > +For example:
> > > > +\begin{lstlisting}
> > > > +be16 A : 15;
> > > > +be16 B : 1;
> > > > +\end{lstlisting}
> > > > +documents the value A stored in the low 15 bit of a 16 bit
> > > > +integer and the value B stored in the high bit of the 16 bit
> > > > +integer, the integer in turn using the big-endian byte order.
> > > > +
> > > > +Note that this notation typically matches the way bitfields are
> > > > +packed by C compilers on little-endian architectures but not the
> > > > +way bitfields are packed by C compilers on big-endian
> > > > +architectures.
> > > > +
> > > >  \newpage
> > > >    
> > > 
> > > I must admit that this explanation confuses me a bit.  
> > 
> > What it is saying is that this is equivalent to
> > 
> > CPU_TO_BE16(B << 15 | A)
> > 
> > Maybe adding this part will clarify things?
> > 
> > 
> > > Would some kind
> > > of graphic representation be more helpful?  
> > 
> > I'm not good at graphics :)
> 
> Me neither :) But pseudo-graphics might be enough.
> 
> > 
> > > For example, on s390 I would expect the structure to look like the
> > > following:
> > > 
> > > |0  ..  14 | 15 |
> > > |    A     |  B |
> > > 
> > > If you included another example for little-endian byte order, this
> > > would clear up things more, I think.  
> > 
> > 
> > It's BE so I think it's
> > 
> > | 15 |14  ..  0 |
> > | B  |    A     |
> > 
> 
> But that's the same, no?

I just tried to show that B is in byte 0,
assuming bytes are numbered 0,1,2,3 left to right.

> Or it's just IBM bitorder striking again...

That's what I'm saying these graphics do not help at all.

It's an integer, B is the most significant bit. Integer is stored
in BE format, thus B is the MSB in the first byte.

Let me know whether writing 
CPU_TO_BE16(B << 15 | A)

helps clarify things.