Re: [PATCH 8/9] iio: imu: st_lsm6dsx: add event configurability on a per axis basis

[Andy Shevchenko <andriy.shevchenko@intel.com>]

On Thu, Nov 20, 2025 at 12:43:09PM +0100, Francesco Lavra wrote:
> On Thu, 2025-11-20 at 10:05 +0100, Andy Shevchenko wrote:
> > On Tue, Nov 18, 2025 at 12:01:57PM +0100, Francesco Lavra wrote:
> > > On Tue, 2025-11-18 at 11:44 +0100, Andy Shevchenko wrote:
> > > > On Mon, Nov 17, 2025 at 08:23:35PM +0100, Francesco Lavra wrote:
> > > > > On Thu, 2025-10-30 at 15:56 +0200, Andy Shevchenko wrote:
> > > > > > On Thu, Oct 30, 2025 at 12:23:19PM +0100, Francesco Lavra wrote:
> > > > > > > On Thu, 2025-10-30 at 10:24 +0200, Andy Shevchenko wrote:
> > > > > > > > On Thu, Oct 30, 2025 at 08:27:51AM +0100, Francesco Lavra
> > > > > > > > wrote:

...

> > > > > > > > > +       old_enable = hw->enable_event[event];
> > > > > > > > > +       new_enable = state ? (old_enable | BIT(axis)) :
> > > > > > > > > (old_enable
> > > > > > > > > &
> > > > > > > > > ~BIT(axis));
> > > > > > > > > +       if (!!old_enable == !!new_enable)
> > > > > > > > 
> > > > > > > > This is an interesting check. So, old_enable and new_enable
> > > > > > > > are
> > > > > > > > _not_
> > > > > > > > booleans, right?
> > > > > > > > So, this means the check test if _any_ of the bit was set and
> > > > > > > > kept
> > > > > > > > set or
> > > > > > > > none were set
> > > > > > > > and non is going to be set. Correct? I think a short comment
> > > > > > > > would be
> > > > > > > > good to have.
> > > > > > > 
> > > > > > > old_enable and new_enable are bit masks, but we are only
> > > > > > > interested
> > > > > > > in
> > > > > > > whether any bit is set, to catch the cases where the bit mask
> > > > > > > goes
> > > > > > > from
> > > > > > > zero to non-zero and vice versa. Will add a comment.
> > > > > > 
> > > > > > If it's a true bitmask (assuming unsigned long type) then all
> > > > > > this
> > > > > > can be
> > > > > > done
> > > > > > via bitmap API calls. Otherwise you can also compare a Hamming
> > > > > > weights of
> > > > > > them
> > > > > > (probably that gives even the same size of the object file, but
> > > > > > !!
> > > > > > instructions
> > > > > >  will be changed to hweight() calls (still a single assembly
> > > > > > instr on
> > > > > > modern
> > > > > >  architectures).
> > > > > 
> > > > > These are u8 variables, so we can't use the bitmap API.
> > > > 
> > > > OK. But hweight8() can still be used.
> > > > 
> > > > > And I don't
> > > > > understand the reason for using hweight(), given that the !!
> > > > > operators
> > > > > would still be needed.
> > > > 
> > > > No, you won't need !! with that.
> > > 
> > > I still don't understand. Are you proposing to replace `if
> > > (!!old_enable ==
> > > !!new_enable)` with `if (hweight8(old_enable) ==
> > > hweight8(new_enable))`?
> > > That won't work, because we only need to check whether the Hamming
> > > weight
> > > goes from zero to non-zero and vice versa.
> > 
> >        old_enable = hw->enable_event[event];
> >        new_enable = state ? (old_enable | BIT(axis)) :
> >                             (old_enable & ~BIT(axis));
> >        if (!!old_enable == !!new_enable)
> >                return 0;
> > 
> > If I am not mistaken this will do exactly the same in a simpler way.
> > 
> >         old_enable = hw->enable_event[event];
> >         if (state)
> >                 new_enable = old_enable | BIT(axis);
> >         else
> >                 new_enable = old_enable & ~BIT(axis);
> >         if ((new_enable ^ old_enable) != BIT(axis))
> >                 return 0;
> 
> This doesn't look right to me, if new_enable differs from old_enable by
> just one bit (which it does), then the XOR result will always have this bit
> (and no others) set, so (new_enable ^ old_enable) will always equal
> BIT(axis).
> We are not checking if the bit was already set or clear, when this code
> runs we already know that the bit is changing, what we are checking is
> whether all bits are zero before or after this change.

The check I proposed is to have a look for the cases when old_enable was 0 and
the BIT(axis) is set and when the BIT(axis) was _the last_ bit in the mask that
got reset. If it's not the case, the code will return 0. I think my check is
correct.

Should I write a test case?

-- 
With Best Regards,
Andy Shevchenko