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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD emulated by HHH --- (does not refer to prior posts)
Date: Fri, 6 Sep 2024 07:31:03 -0400
Organization: i2pn2 (i2pn.org)
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On 9/6/24 7:20 AM, olcott wrote:
> On 9/6/2024 5:22 AM, Mikko wrote:
>> On 2024-09-03 13:58:27 +0000, olcott said:
>>>
>>> _DDD()
>>> [00002172] 55         push ebp      ; housekeeping
>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>> [00002175] 6872210000 push 00002172 ; push DDD
>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>> [0000217f] 83c404     add esp,+04
>>> [00002182] 5d         pop ebp
>>> [00002183] c3         ret
>>> Size in bytes:(0018) [00002183]
>>>
>>> Anyone that is not dumber than a box of rocks can tell
>>> that machine address 0000217f is unreachable for every
>>> DDD emulated by HHH according to the semantics of the
>>> x86 language where HHH emulates itself emulating DDD.
>>
>> Anyone who really knows either x86 assembly or machine langage or
>> C can see that the machine address 217f is unreachachable only if
>> the program at 000015d2, named HHH, does not return.
>>
> 
> That is not exactly true. There is a directly executed HHH
> that always returns and a DDD emulated by HHH that calls
> an emulated HHH that never returns.
> 

No, the DDD that is emulated by HHH will return, as the behavior of a 
program doesn't "stop" just because the emulator looking at its behavior 
gave up.

The EMULATION of DDD by HHH is what doesn't return.

Youa are just proving you are just learning impaired as you don't 
understand the difference between those two things,