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

On 8/7/24 9:01 AM, olcott wrote:

On 8/7/2024 3:16 AM, Fred. Zwarts wrote:

Op 04.aug.2024 om 15:11 schreef olcott:

On 8/4/2024 1:26 AM, Fred. Zwarts wrote:

Op 03.aug.2024 om 17:20 schreef olcott:>>

When you try to show how DDD simulated by HHH does
reach its "return" instruction you must necessarily
must fail unless you cheat by disagreeing with the
semantics of C. That you fail to have a sufficient
understanding of the semantics of C is less than no
rebuttal what-so-ever.

>
Fortunately that is not what I try, because I understand that HHH cannot possibly simulate itself correctly.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
In other words when HHH simulates itself simulating DDD it
is supposed to do something other than simulating itself
simulating DDD ???  Do you expect it to make a cup of coffee?
>

>
Is English too difficult for you. I said HHH cannot do it correctly.

 *According to an incorrect criteria of correct*
You keep trying to get away with disagreeing with
the semantics of the x86 language. *That is not allowed*
 

No, YOU youse an incorrect criteria of correct, as you don't correctly emulatin the instruction, after all, you have failed ot EVER show a trace produced by HHH of its emulation of its input that show it emulation the instructions at the target of the Call HHH Instruction
Sorry, you are just proving your stupidity.