Re: The execution trace of HHH1(DDD) shows the divergence

On 6/7/2025 10:32 AM, olcott wrote:

The execution trace of HHH1(DDD) shows the divergence
of DDD emulated by HHH from DDD emulated by HHH1.
 int main()
{
   HHH1(DDD);
}
 Shows that DDD emulated by HHH and DDD emulated by
HHH1 diverges as soon as HHH begins emulating itself
emulating DDD.
 *From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1              DDD emulated by HHH
[00002183] push ebp               [00002183] push ebp
[00002184] mov ebp,esp            [00002184] mov ebp,esp
[00002186] push 00002183 ; DDD    [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH    [0000218b] call 000015c3 ; HHH
*HHH1 emulates DDD once then HHH emulates DDD once, these match*
 The next instruction of DDD that HHH emulates is at
the machine address of 00002183.
 The next instruction of DDD that HHH1 emulates is at
the machine address of 00002190.

False.
The next instruction of DDD that both HHH and HHH1 emulates is at the machine address of 000015c3, as the algorithm DDD consists of the machine code of the function DDD, the machine code of the function HHH, and the machine code of everything that HHH calls down to the OS level.
The traces are exactly the same up to the point that HHH aborts, as shown by the side-by-side trace posted by Mike, and as you have agreed on the record *multiple times*.
On 6/4/2025 12:38 PM, olcott wrote:
 > On 6/4/2025 4:20 AM, Fred. Zwarts wrote:
 >>
 >> That did not answer the question: WHAT INSTRUCTION, correctly simulated did that?
 >
 > When HHH1(DDD) simulates DDD it never simulates itself.
 > When HHH(DDD) simulates DDD then simulates itself simulating
 > DDD the first instruction that this simulated HHH simulates
 > diverges from the simulation that HHH1 did.
 >
 >> You cannot point to any instruction interpreted differently by the two simulators.
 >
 > There are no instructions interpreted differently.
On 6/3/2025 10:54 PM, dbush wrote:
 > On 6/3/2025 10:51 PM, olcott wrote:
 >> On 6/3/2025 9:42 PM, dbush wrote:
 >>> On 6/3/2025 10:29 PM, olcott wrote:
 >>>> On 6/3/2025 8:57 PM, dbush wrote:
 >>>>> So how exactly do HHH and HHH1 emulate the first instruction of HHH
 >>>>> differently?
 >>>>>
 >>>>
 >>>> The question is incorrect.
 >>>> HHH emulates DDD two times and HHH1 emulates DDD one time
 >>>> the whole second time is the divergence.
 >>>
 >>> There is no divergence if the instructions are emulated exactly the
 >>> same in both cases.
 >>
 >> HHH1(DDD) emulates DDD exactly one time.
 >> HHH(DDD) emulates DDD exactly two times.
 >>
 >> The whole second time that HHH emulates DDD is
 >> divergence.
 >>
 >
 > Let the record show that Peter Olcott has failed to identify an
 > instruction that HHH and HHH1 emulated differently.
 >
 >>
 >>> What happened before either emulation started is irrelevant.
 >>>
 >>> The only way for the emulations to diverge is if there is a
 >>> particular instruction such that X happens if HHH emulates it and Y
 >>> happens if HHH1 emulates it.  Again, what happened before either
 >>> emulation started is irrelevant.
 >>>
 >>>
 >>> So I'll ask one more time:  how exactly do HHH and HHH1 emulate the
 >>> first instruction of HHH, or *any* instruction that is part of the
 >>> emulation of DDD, differently?
 >>>
 >>> Failure to provide the above explanation in your next reply or within
 >>> one hour of your next post in this newsgroup will be taken as your
 >>> official on-the-record admission that the emulations of DDD performed
 >>> by HHH and HHH1 do *not* diverge but are in fact the same up to the
 >>> point that HHH aborts.
 >
 > Therefore, as per the above criteria:
 >
 > Let The Record Show
 >
 > That Peter Olcott
 >
 > Has *officially* admitted:
 >
 > That the emulations of DDD by HHH and HHH1 in fact do *not* diverge but
 > are in fact the same up to the point that HHH aborts.