Re: DD emulated by HHH cannot possibly terminate normally --- x86 code

On 3/1/2025 4:08 PM, Fred. Zwarts wrote:

Op 01.mrt.2025 om 22:12 schreef olcott:

On 3/1/2025 2:22 PM, Fred. Zwarts wrote:

Op 01.mrt.2025 om 20:01 schreef olcott:

On 3/1/2025 10:05 AM, Richard Damon wrote:

On 3/1/25 9:41 AM, olcott wrote:

On 3/1/2025 6:49 AM, Richard Damon wrote:

On 2/28/25 7:47 PM, olcott wrote:

When we hypothesize that the code at machine address 0000213c is
an x86 emulator then we know that DD remains stuck in recursive
emulation and cannot possibly reach its own "ret" instruction
and terminate normally.
When we add the additional complexity that HHH also aborts this
sequence at some point then every level of recursive emulation
immediately stops. This does not enable any DD to ever reach its
"ret" instruction.
>

But then you just negated your first assumption, as a partial
emulator that aborts its emulation, then DD no longer gets stuck.
>

Cannot possibly reach its own "ret" instruction and terminate
normally proves non-termination whether aborted or not.
>

But it DOES terminate

>
DD emulated by HHH never terminates no matter how many times you try
to get away with the straw-man deception of referring to anything at
all besides DD EMULATED BY HHH

HHH emulating DD fails to reach the 'ret' instruction.

>
Because DD calls HHH(DD) in recursive emulation thus non-termination
is entirely the fault of DD.

No, DD calls HHH only once.

 
When DD is correctly emulated by HHH and DD calls HHH(DD)
then DD calls HHH in recursive emulation.
When HHH breaks this otherwise endless chain every level of DD
immediately stops with none of them ever reaching their "ret"
instruction.