Re: Computer architects leaving Intel...

On Tue, 10 Sep 2024 11:21:01 +0300, Michael S
<already5chosen@yahoo.com> wrote:

On Mon, 09 Sep 2024 23:27:24 -0400
George Neuner <gneuner2@comcast.net> wrote:
>

On Sun, 08 Sep 2024 15:36:39 GMT, anton@mips.complang.tuwien.ac.at
(Anton Ertl) wrote:
 

Tim Rentsch <tr.17687@z991.linuxsc.com> writes: 

anton@mips.complang.tuwien.ac.at (Anton Ertl) writes:
 

There was still no easy way to determine whether your software
that calls memcpy() actually works as expected on all hardware, 

>
There may not be a way to tell if memcpy()-calling code will work
on platforms one doesn't have, but there is a relatively simple
and portable way to tell if some memcpy() call crosses over into
the realm of undefined behavior. 

>
1) At first I thought that yes, one could just check whether there is
an overlap of the memory areas.  But then I remembered that you
cannot write such a check in standard C without (in the general case)
exercising undefined behaviour; and then the compiler could eliminate
the check or do something else that's unexpected.  Do you have such a
check in mind that does not exercise undefined behaviour in the
general case? 

 
The result of comparing pointers to two elements of the same array is
defined.  Cast to (char*), both src and dst can be considered to point
to elements of the [address space sized] char array at address zero.
>

>
According to my understanding, your 'can be considered' part is not
codified in the C Standard.
>

Adding size_t to a pointer yields another pointer of the same type.
 
 
All of gcc, clang and MSVC seem happy with this.
>

>
It works. But is it guaranteed to work in the future by some sort of
document? I am pretty sure that no such guarantee exists in gcc and
MSVC docs. I did not look in clang docs. Trying to find anythings in
LLVM/clang docs makes me sad.

I know that it has worked as expected with every version of gcc and
Microsoft I've used since 1988.  [clang I don't use, but I tried it on
godbolt.org with the most recent version]

Will it continue to work ... who knows?


I definitely am NOT an expert on the C standard, but thinking about
it, it occurred to me that if an array is explicitly defined that
*might* cover all memory (or at least all heap), then the compiler
would have to honor any apparent pointers into it.

E.g., char (*all_memory)[] = 0;

None of the compilers at godbolt seem to need this to compare
arbitrary addresses as char*, but all accept it.

Obviously speculation, but it's the best I have.