From Newsgroup: comp.lang.c

On 8/12/2025 3:47 AM, Mikko wrote:

On 2025-08-11 14:34:50 +0000, olcott said:

So you understand that HHH(DD)==0 on the basis of the
behavior of DD correctly simulated by HHH?

No but on the basis of failure to simulate correctly the behavoiur of DD,
in particular the point where the HHH called by DD aborts its simulation
and returns 0 to DD which then halts.

Can DDD emulated by HHH according to the semantics
of the x86 language reach its own emulated "ret"
instruction final halt state?

void DDD()
{
HHH(DDD);
return;
}

_DDD()
[00002192] 55 push ebp
[00002193] 8bec mov ebp,esp
[00002195] 6892210000 push 00002192 // push DDD
[0000219a] e833f4ffff call 000015d2 // call HHH
[0000219f] 83c404 add esp,+04
[000021a2] 5d pop ebp
[000021a3] c3 ret
Size in bytes:(0018) [000021a3]

(a) HHH executes.

(b) HHH creates a separate process context having
its own stack and set of 16 virtual registers
to simulate DDD using cooperative multi-tasking.

(c) HHH emulates the first four instructions of DDD
in this separate process context.

(d) HHH emulates an instance of itself in this same
process context.

(e) This HHH instance creates another process context
for its own emulated DDD instance.

(f) This HHH instance emulates the first three
instructions of DDD in this separate process
context and sees the call 000015d2 instruction.

It is a verified fact HHH did emulate DDD according
to the semantics of the x86 language. That *is* the
ultimate measure of correct emulation. Disagreeing
with the x86 language is necessarily incorrect.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2