Re: DDD specifies recursive emulation to HHH and halting to HHH1

On 4/2/25 10:45 PM, olcott wrote:

On 4/2/2025 9:10 AM, joes wrote:

Am Mon, 31 Mar 2025 18:34:17 -0500 schrieb olcott:

On 3/31/2025 5:54 PM, dbush wrote:

No, it is YOUR misconception.  The algorithm DDD consists of the
function DDD, the function HHH, and everything that HHH calls down to
the OS level.
>

We have already been over this.
HHH(DDD) and HHH1(DDD) have the same inputs all the way down to the OS
level. The ONLY difference is that DDD does not call HHH1(DDD) in
recursive emulation.

 

That is actually not a difference but the same that DDD calls HHH.
>

 That DDD calls HHH(DDD) instead of HHH1(DDD)
causes the behavior of DDD emulated by HHH and
DDD emulated by HHH1 to differ.
 

How?
The fact that both are emulating the EXACT SAME instruction sequence says they should see the exact same behavior if they are actually emulating at the instruction level.
A call instruction doesn't care if the routine being called has been called before, it just doesn't what it does.
Your problem is you can't keep a straight definition of what you are doing in your mind, perhaps because you don't have one.
You have effectively admitted, by failing to respond for years, that your claim here is invalid, as you can't show how they actually differ without just admitting that you are not looking at the emulation as per the x86 language.
It seems you don't understand what that means, or maybe it is just the concept of needing to tell the truth that you don't understand.