Re: [PATCH v1 1/2] overflow: Allow to sum a few arguments at once

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

On Wed, Jun 17, 2026 at 10:30:56PM +0100, David Laight wrote:
> On Wed, 17 Jun 2026 14:56:09 +0200
> Johannes Berg <johannes@sipsolutions.net> wrote:
> > On Wed, 2026-06-17 at 13:12 +0200, Andy Shevchenko wrote:
> > > Convert size_add() to take variadic argument, so we can simplify users
> > > with using a macro only once.  
> > 
> > > +#define __size_add3(addend1, addend2, addend3, addend4, ...)			\
> > > +	__size_add(__size_add2(addend1,  addend2, addend3), addend4)
> > > +#define __size_add4(addend1, addend2, addend3, addend4, addend5, ...)		\
> > > +	__size_add(__size_add3(addend1,  addend2, addend3, addend4), addend5)  
> > 
> > I guess it's not going to really matter, but it would generate fewer
> > calls to have something more like
> > 
> > #define __size_add3(a1, a2, a3, a4) \
> > 	size_add(size_add(a1, a2), size_add(a3, a4))
> > #define __size_add4(a1, a2, a3, a4, a5) \
> > 	size_add(size_add(a1, a2), size_add(a3, a4, a5))
> > 
> > as a binary tree, rather than only cutting one off every time. Not sure
> > that results in hugely different code though - maybe fewer overflow
> > checks?

Good question. I'm also thinking that one-by-one may expand in too much of
preprocessor code (haven't checked myself).

> The binary tree stands a chance of executing less slowly because the leaf
> adds can be executed in parallel.
> Excluding the saturation checks (wtf is it called size_add() not
> saturating_add() ?) (a + b) + (c + d) will usually execute faster than
> ((a + b) + c) + d because the (a + b) and (c + d) can execute at the
> same time; unfortunately gcc will always generate the latter.

I'm confused. "unfortunately... the latter"? You meant "the former"?

> > Although your version make it really completely equivalent to the
> > nl80211.c code, clearly it doesn't matter if all the values are "good",
> > and I believe the overflow behaviour means it doesn't matter for the
> > overflow case either?

Indeed. Whenever the value is saturated, the rest is just matter of sequential
unlikely branches taken.

-- 
With Best Regards,
Andy Shevchenko