Am Sun, 16 Mar 2025 14:36:11 -0500 schrieb olcott:
On 3/16/2025 2:26 PM, Keith Thompson wrote:
Richard Heathfield <[email protected]> writes:
On 16/03/2025 10:31, Mikko wrote:
False. You didn't say that HHH is a C function. In particular, the
code shown above does not say so.
It scarcely qualifies as C.
For example, it begins by a goto this code:
__asm__("lea eax, DATA1");
__asm__("mov Aborted, eax"); __asm__("lea eax, DATA2");
__asm__("mov execution_trace, eax");
__asm__("mov eax, END_OF_CODE"); __asm__("mov End_Of_Code, eax");
which any C compiler is free to reject.
C99 introduced the asm keyword, but that's spelled asm, not __asm__,
and of course it's not a magic wand, so it can't make an inherently
unportable program work on every platform supported by C compilers.
C/370, for example, would have a fit.
No, C99 didn't introduce the asm keyword. Both C90 and C99 (and all
later editions) document the "asm" keyword as a common extension, but
it's not in the list of keywords. K&R1 (1978) mentions that some
implementations reserve "fortran" and "asm". A conforming C compiler
must accept "asm" as an ordinary identifier.
I don't think Olcott intends HHH to be fully portable C (assuming he
knows what that means). In any case, his claims about "Every
sufficiently competent C programmer" are ludicrous.
*Every sufficiently competent C programmer knows*
typedef void (*ptr)();
int HHH(ptr P);
void Infinite_Loop()
{
HERE: goto HERE;
return;
}
void Infinite_Recursion()
{
Infinite_Recursion(); return;
}
void DDD()
{
HHH(DDD);
return;
}
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
When HHH correctly emulates N steps of the above functions none of them
can possibly reach their own "return" instruction and terminate
normally.
The fun thing is that DDD does terminate, but HHH can't simulate it.
Since HHH does see that same pattern that competent C programmers see it correctly aborts its emulation and rejects these inputs as non
terminating.
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.
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