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

On 8/5/2024 2:44 AM, Mikko wrote:

On 2024-08-04 13:11:56 +0000, olcott said:
 

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?

 In another message you have said that when HHH simulates itself
simulating DDD is does not simulate itself simulating itself
simulating DDD. You have not told whether it makes a cup of coffee.
Neither action can be seen in the traces you have shown.
 

HHH and HH and the original H have proved that they simulate
themselves simulating DDD, DD and P for three years now.
They did this by deriving the correct execution trace that
simulating themselves simulating their input would derive.
Maybe all of my reviewers have been saying that I am wrong
about this on the basis of pure bluster in that they are
totally confused by assembly language and don't have the
slightest clue what it means.
--
Copyright 2024 Olcott
"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer