Re: [PATCH v4 4/8] bits: introduce fixed-type BIT

Andy Shevchenko
2025-03-06T09:13:03Z

On Wed, Mar 05, 2025 at 09:50:27PM +0000, David Laight wrote:
> On Wed, 5 Mar 2025 21:56:22 +0200
> Andy Shevchenko <andriy.shevchenko@linux.intel.com> wrote:
> > On Thu, Mar 06, 2025 at 02:17:18AM +0900, Vincent Mailhol wrote:
> > > On 06/03/2025 at 00:48, Andy Shevchenko wrote:
> > > > On Wed, Mar 05, 2025 at 11:48:10PM +0900, Vincent Mailhol wrote:
> > > >> On 05/03/2025 at 23:33, Andy Shevchenko wrote:
> > > >>> On Wed, Mar 05, 2025 at 10:00:16PM +0900, Vincent Mailhol via B4 Relay wrote:

...

> > > >>>> +#define BIT_U8(b) (BIT_INPUT_CHECK(u8, b) + (unsigned int)BIT(b))
> > > >>>> +#define BIT_U16(b) (BIT_INPUT_CHECK(u16, b) + (unsigned int)BIT(b))
> > > >>>
> > > >>> Why not u8 and u16? This inconsistency needs to be well justified.
> > > >>
> > > >> Because of the C integer promotion rules, if casted to u8 or u16, the
> > > >> expression will immediately become a signed integer as soon as it is get
> > > >> used. For example, if casted to u8
> > > >>
> > > >> BIT_U8(0) + BIT_U8(1)
> > > >>
> > > >> would be a signed integer. And that may surprise people.
> > > >
> > > > Yes, but wouldn't be better to put it more explicitly like
> > > >
> > > > #define BIT_U8(b) (BIT_INPUT_CHECK(u8, b) + (u8)BIT(b) + 0 + UL(0)) // + ULL(0) ?
> > >
> > > OK, the final result would be unsigned. But, I do not follow how this is
> > > more explicit.
> > >
> > > Also, why doing:
> > >
> > > (u8)BIT(b) + 0 + UL(0)
> > >
> > > and not just:
> > >
> > > (u8)BIT(b) + UL(0)
> > >
> > > ?
> > >
> > > What is that intermediary '+ 0' for?
> > >
> > > I am sorry, but I am having a hard time understanding how casting to u8
> > > and then doing an addition with an unsigned long is more explicit than
> > > directly doing a cast to the desired type.
> >
> > Reading this again, I think we don't need it at all. u8, aka unsigned char,
> > will be promoted to int, but it will be int with a value < 256, can't be signed
> > as far as I understand this correctly.
>
> The value can't be negative, but the type will be a signed one.

Yes, that's what I mentioned above: "int with the value < 256".

> Anything comparing types (and there are a few) will treat it as signed.
> It really is bad practise to even pretend you can have an expression
> (rather that a variable) that has a type smaller than 'int'.
> It wouldn't surprise me if even an 'a = b' assignment promotes 'b' to int.

We have tons of code with u8/u16, what you are proposing here is like
"let's get rid of those types and replace all of them by int/unsigned int".
We have ISAs that are byte-oriented despite being 32- or 64-bit platforms.

> So it is even questionable whether BIT8() and BIT16() should even exist at all.

The point is to check the boundaries and not in the returned value per se.

> There can be reasons to return 'unsigned int' rather than 'unsigned long'.
> But with the type definitions that Linux uses (and can't really be changed)
> you can have BIT32() that is 'unsigned int' and BIT64() that is 'unsigned long
> long'. These are then the same on 32bit and 64bit.

--
With Best Regards,
Andy Shevchenko