Re: The philosophy of computation reformulates existing ideas on a new basis ---x86 code is a liar?

Am Thu, 07 Nov 2024 15:56:31 -0600 schrieb olcott:

On 11/7/2024 3:24 PM, joes wrote:

Am Thu, 07 Nov 2024 10:31:41 -0600 schrieb olcott:

On 11/7/2024 5:56 AM, Richard Damon wrote:

On 11/6/24 11:39 PM, olcott wrote:

 

That is not what the machine code of DDD that calls the
machine code of HHH says.

The code by itself doesn’t say "do not return". That is a semantic
property.

The code itself does say that within the semantics of the x86 language
as I have been saying all long hundreds of times.

There is no "do not return" instruction.

Right, so that is part of the input, or it can't be
emulated.
The Machine code of HHH says that it will abort its
emulation and return, so that is the only correct result
per the x86 language.

Are you really so ignorant of these things that you think
that the fact that HHH returns to main() causes its emulated
DDD to reach its own final state?

Yes, because DDD calls HHH.
 

But the PROGRAM DDD, that it is emulating does. Just its own
PARTIAL emulation of it is aborted before it gets there.

Just repeating your errors, and not even trying to refute the
errors pointed out, I guess that means you accept these as
errors.

There is only one program DDD, although it is invoked multiple times.
We don’t care whether HHH actually simulates the return as long as it
actually derives (not guesses) the right result.

DDD emulated by HHH does have different behavior than DDD emulated by
HHH1 or directly executed DDD.
DDD emulated by CANNOT POSSIBLY HALT no matter WTF HHH does: abort or
NEVER abort.

When the instance of HHH that DDD calls aborts simulating, it returns
to the simulated DDD, which then halts.

There <is> a key distinguishing difference in the behavior of DDD
emulated by HHH and DDD emulated by HHH1 or directly executed. It is
ridiculously stupid to simply ignore this for three f-cking years.

That difference is not due to DDD.

--
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.