Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?

On 8/3/24 9:50 AM, olcott wrote:

On 8/3/2024 3:14 AM, Fred. Zwarts wrote:

Op 02.aug.2024 om 22:57 schreef olcott:

Who here is too stupid to know that DDD correctly simulated
by HHH cannot possibly reach its own return instruction?
>
void DDD()
{
   HHH(DDD);
   return;
}
>

>
Which proves that the simulation is incorrect.

 When are you going to understand that you are not allowed
to disagree with the semantics of the x86 language?
 

We can ask you the same question.
The ultimate arbiter of the semantics of the xx86 language is the running of the program as an x86 program.
When we do this, with HHH defined as you defined it, DDD calls HHH which runs for a while simulating a copy of DDD, then deciding to abort its emulation and returning to DDD which returns.
Thus, THIS is the actual behavior as DEFINED by the semantics of the x86 language.
Where do you claim it is wrong?
Your problem is your HHH doesn't correct emulate HHH by the semantics of the x86 language. The only one that does never gives an answer, and thus isn't the decider you define HHH to be.
This proves you to be nothing but an ignorant liar.