Re: DDD emulated by HHH diverges from DDD emulated by HHH1

On 6/4/2025 10:36 PM, olcott wrote:

On 6/4/2025 9:10 PM, dbush wrote:

On 6/4/2025 10:06 PM, olcott wrote:

On 6/4/2025 8:48 PM, dbush wrote:

On 6/4/2025 9:39 PM, olcott wrote:

On 6/4/2025 8:28 PM, dbush wrote:

>

>
DDD emulated by HHH1 emulating HHH  DDD emulated by HHH emulating 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
                                     ### HHH aborts ###
>

>
*That is dishonest*

>
No, that is the *fact* of what happens when HHH1 emulates DDD
>

HHH1 emulates DDD once and this matches

The one time that the outer HHH emulates DDD, up to the point that HHH aborts
>

>
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
>
*THE MATCHING IS NOW USED UP*
*THE MATCHING IS NOW USED UP*
*THE MATCHING IS NOW USED UP*

>
False.  HHH1 is still simulating the left side by simulating HHH as shown above.
>

 The is not the way that side-by-side comparisons of
DDD simulated by HHH1 versus DDD simulated by HHH
are made.

False.  That is *exactly* how they are made.  Anything that the directly executed HHH1 emulates, at any level, is part of the emulation by HHH1. And it is being compared against the emulation made by the direct execution HHH, *not* against any simulated HHH which is really part of DDD.

 There is a list of DDD instructions that are directly
simulated by HHH1 and directly simulated by HHH right
here:
 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 never directly emulates any other DDD.
Another instance of HHH does directly emulate DDD.

Irrelevant.  The directly executed instance of HHH1 and the directly executed instance of HHH do exactly the same thing up to the point that HHH aborts, as you have *explicitly* admitted:
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.