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On 7/14/2024 7:20 PM, joes wrote:WHICH IS JUST BASED ON A FALSEHOOD, PROVING YOU ARE JUST A LIAR.Am Sun, 14 Jul 2024 09:00:55 -0500 schrieb olcott:On 7/14/2024 3:29 AM, joes wrote:>Am Sat, 13 Jul 2024 18:33:53 -0500 schrieb olcott:On 7/13/2024 6:26 PM, joes wrote:Can you elaborate? All runtime instances share the same static code.
I am talking about the inner HHH which is called by the simulated
DDD. That one is, according to you, aborted. Which is wrong, because
by virtue of running the same code, the inner HHH aborts ITS
simulation of DDD calling another HHH.>What are the twins and what is their difference?
Do you disagree with my tracing?
>The directly executed DDD is like the first call of infinite recursion.Not really. Execution does not continue.
The emulated DDD is just like the second call of infinite recursion.
When the second call of infinite recursion is aborted then the first
call halts.void Infinite_Recursion()That would be incorrect.
{
Infinite_Recursion();
}
The above *is* infinite recursion.
A program could emulate the above code and simply skip line 3 causing
Infinite_Recursion() to halt.
>When DDD calls HHH(DDD) HHH returns.Therefore it does not need to be aborted.When DDD correctly emulated by HHH the call never returns as is provenI do not see this below.
below. The executed DDD() has HHH(DDD) skip this call.HHH(DDD) must skip this call itself by terminating the whole DDD>
process.Because this HHH does not know its own machine address HHH only sees>
that DDD calls a function that causes its first four steps to be
repeated. HHH does not know that this is recursive simulation. To HHH it
looks just like infinite recursion.New slave_stack at:1038c4 -- create new process context for 1st DDD>
Begin Local Halt Decider Simulation Execution Trace Stored at:1138cc[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)>
New slave_stack at:14e2ec -- create new process context for 2nd DDD[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation StoppedHow is this detected? Is it also triggered when calling a functionYou never bothered to answer whether or not you have 100%
in a loop?
>
understanding of infinite recursion. If you don't then you
can never understand what I am saying. If you do that I
already proved my point. Here is the proof again:
New slave_stack at:1038c4The followin gis NOT a corrct simulaitohn of the CALL HHH instruciton.
Begin Local Halt Decider Simulation Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55 push ebp ; housekeeping
[00002173][001138bc][001138c0] 8bec mov ebp,esp ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ecSince you define your simulation to be per ths x86 instruction set, the ONLY behavior a "Call HHH" instruction can have is the simulation of the instructions of HHH.
[00002172][0015e2e4][0015e2e8] 55 push ebp ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec mov ebp,esp ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
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