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

[Richard Biener <richard.guenther@gmail.com>]

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(?)

Richard.

> Jens