More on SUBREG

[Etienne Lorrain <etienne_lorrain at yahoo dot fr>]

> > > The mismatch is this:  It's documented that stores into SUBREG's with
> > > a smaller mode, leave the other bits in undefined state:
> > >   "Storing in a non-paradoxical `subreg' has undefined results for
> > >      bits belonging to the same word as the `subreg'."  (from docu of
> >                       ^^^^^^^^^^^^^^^^
> >
> > It's no mismatch.  Bits belonging to other words remain unchanged.
>
> Ahh OK, thanks for clarifying.  That makes things clean and easy ;)

  Seems to be the reason why, on i386, the change endian function:

extern inline unsigned bswap(unsigned val)
{
    union {
       unsigned l;
       unsigned short s;
       } tmp = { val };
    tmp.s = (tmp.s >> 8) | (tmp.s << 8);
    tmp.l = (tmp.l >> 16) | (tmp.l << 16);
    tmp.s = (tmp.s >> 8) | (tmp.s << 8);
    return tmp.l;
}

  Produces this:
    movw    %w0,%w1
    rolw    $8,%w1
    movw    %w1,%w0
    roll    $16,%0
    movw    %w0,%w1
    rolw    $8,%w1
    movw    %w1,%w0

  (replace %w0 by %ax or 15(%esp), %0 by %eax or 15(%esp), and %w1 by %bx
   if you do not like this notation)

  Instead of the simplest:
    rolw    $8,%w0
    roll    $16,%0
    rolw    $8,%w0

  So it is better to do it in assembly:

extern inline unsigned bswap(unsigned val)
{
    unsigned val;
#if defined(__i386__)
    __asm__ __volatile__ ("
        rolw    $8,%w0  # the 'w' is ignored if '+g' is a reference
        roll    $16,%0
        rolw    $8,%w0
        "
        : "+g" (val)
        );
#else
    __asm__ __volatile__ (" bswap   %0 "
             : "+r" (val));
#endif
    return val;
}

  But sometimes, even on 80486+, the "rol" version would be better
 because bswap assembly instruction only accept a register as
 parameter (when there is no more register available).
 Only the compiler could choose correctly.

  Not really a question, just my €0.02.
  Etienne.

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