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On 7/22/2025 9:12 PM, Richard Damon wrote:And that same world class simulator shows, when simulating exactly the same input, that it is possible to reach the final halt state of the simulation. Apparently, you introduced some bugs when modifying this simulator into your HHH, so that HHH now fails to reach that final halt state.On 7/22/25 11:21 AM, olcott wrote:We have agreed on that hundreds of times and you keep forgetting.On 7/22/2025 9:19 AM, joes wrote:>Am Mon, 21 Jul 2025 09:19:23 -0500 schrieb olcott:>
>The correct measure of the behavior of the input to HHH(DDD)The measure of HHH(DDD) is HHH(DDD)?
is DDD simulated by HHH according to the semantics of the C programming
language.
>
void DDD()
{
HHH(DDD);
return;
}
>
executed HHH emulates DDD that calls emulated HHH(DDD)
that emulates DDD that calls emulated emulated HHH(DDD)
But that isn't your HHH, so just a lie.
>
Your HHH only PARTIALLY emulates its input and DOES abort
>>>
machine stack stack machine assembly
address address data code language
======== ======== ======== ============== =============
[000021be][00103872][00000000] 55 push ebp
[000021bf][00103872][00000000] 8bec mov ebp,esp
[000021c1][0010386e][0000219e] 689e210000 push 0000219e // push DDD
[000021c6][0010386a][000021cb] e823f4ffff call 000015ee // call HHH
New slave_stack at:103916
And here you show that your idea of "correct simulation" is just a lie.
>
The *ONLY* correct x86 simulation of the call HHH is to go into HHH.
>
Even if you do a funtional simulation of HHH, then it must do what HHH actually does, and none of the following is the simulation of the above DDD, and the function simulation of HHH must show it returning 0, since that *IS* what a call to HHH(DDD) ultimately does.I directly know that HHH directly emulates itself
>
because I used a world class emulator with decades
of development effort and implemented it in the
multi-tasking operating system that I wrote.
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