Re: [c++std-parallel-1614] Re: Compilers and RCU readers: Once more unto the breach!

["Paul E. McKenney" <paulmck@linux.vnet.ibm.com>]

On Wed, May 20, 2015 at 11:03:00AM +0200, Richard Biener wrote:
> On Wed, May 20, 2015 at 9:34 AM, Jens Maurer <Jens.Maurer@gmx.net> wrote:
> > On 05/20/2015 04:34 AM, Paul E. McKenney wrote:
> >> On Tue, May 19, 2015 at 06:57:02PM -0700, Linus Torvalds wrote:
> >
> >>>  - the "you can add/subtract integral values" still opens you up to
> >>> language lawyers claiming "(char *)ptr - (intptr_t)ptr" preserving the
> >>> dependency, which it clearly doesn't. But language-lawyering it does,
> >>> since all those operations (cast to pointer, cast to integer,
> >>> subtracting an integer) claim to be dependency-preserving operations.
> >
> > [...]
> >
> >> There are some stranger examples, such as "(char *)ptr - ((intptr_t)ptr)/7",
> >> but in that case, if the resulting pointer happens by chance to reference
> >> valid memory, I believe a dependency would still be carried.
> > [...]
> >
> > From a language lawyer standpoint, pointer arithmetic is only valid
> > within an array.  These examples seem to go beyond the bounds of the
> > array and therefore have undefined behavior.
> >
> > C++ standard section 5.7 paragraph 4
> > "If both the pointer operand and the result point to elements of the
> > same array object, or one past the last element of the array object,
> > the evaluation shall not produce an overflow; otherwise, the behavior
> > is undefined."
> >
> > C99 and C11
> > identical phrasing in 6.5.6 paragraph 8
> 
> Of course you can try to circumvent that by doing
> (char*)((intptr_t)ptr - (intptr_t)ptr + (intptr_t)ptr)
> (see https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752 for extra fun).
> 
> Which (IMHO) gets you into the standard language that only makes conversion of
> the exact same integer back to a pointer well-defined(?)

I am feeling good about leaving the restriction and calling out
the two paragraphs in a footnote, then.  ;-)

							Thanx, Paul