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On 11/11/2024 4:54 AM, Mikko wrote:But the behaivor of DDD does.On 2024-11-09 14:36:07 +0000, olcott said:That you pretend to not understand my clear words does
>On 11/9/2024 7:53 AM, Mikko wrote:>On 2024-11-08 14:41:57 +0000, olcott said:>
>On 11/8/2024 3:57 AM, joes wrote:>Am Thu, 07 Nov 2024 15:56:31 -0600 schrieb olcott:>On 11/7/2024 3:24 PM, joes wrote:There is no "do not return" instruction.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:The code itself does say that within the semantics of the x86 languageThe code by itself doesn’t say "do not return". That is a semanticThat is not what the machine code of DDD that calls the
machine code of HHH says.
property.
as I have been saying all long hundreds of times.
>When the instance of HHH that DDD calls aborts simulating, it returnsDDD emulated by HHH does have different behavior than DDD emulated byYes, because DDD calls HHH.Right, so that is part of the input, or it can't beAre you really so ignorant of these things that you think
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.
that the fact that HHH returns to main() causes its emulated
DDD to reach its own final state?
>There is only one program DDD, although it is invoked multiple times.Just repeating your errors, and not even trying to refute theBut the PROGRAM DDD, that it is emulating does. Just its own
PARTIAL emulation of it is aborted before it gets there.
errors pointed out, I guess that means you accept these as
errors.
We don’t care whether HHH actually simulates the return as long as it
actually derives (not guesses) the right result.
HHH1 or directly executed DDD.
DDD emulated by CANNOT POSSIBLY HALT no matter WTF HHH does: abort or
NEVER abort.
to the simulated DDD, which then halts.
>There <is> a key distinguishing difference in the behavior of DDDThat difference is not due to 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.
>
The semantic property of the finite string pair: HHH/DDD
unequivocally entails that DDD never reaches its final halt state.
No, it does not. You might say that the semantic property of the
finite string "Olcott is an idiot" unequvocally entails that Olcott
is an idiot but it does not.
The semantic property of the finite string pair: HHH/DDD
unequivocally entails that DDD never reaches its final halt
state WITHIN THE SEMANTICS OF THE X86 LANGUAGE.
The expression "The semantic property" is incorrect when it is not
clear from context which semantic property is meant. Note that a
string per se does not have semantic properties, they all come
from interpretrations.
>
not mean that my words are not clear.
The relevant semantic property of the finite string pair
HHH/DDD where DDD is emulated by HHH according to the
semantics of the x86 language is that DDD cannot possibly
reach its "return" instruction final halt state.
The fact that DDD defines a pathological relationship withRight, it needs to be taken into account by HHH, to see that if it aborts and returns that the HHH that DDD calls will also return.
HHH cannot be simply ignored and must be accounted for. The
actual computation itself does involve HHH emulating itself
emulating DDD. To simply pretend that this does not occur
seems dishonest.
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