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On 6/26/2024 2:54 AM, Mikko wrote:When DDD is emulated by H1 then H1 needs to the call to H0 and theOn 2024-06-25 13:19:46 +0000, olcott said:_DDD()
On 6/25/2024 4:48 AM, Mikko wrote:Obviously false. The meaning of H(P,P) is determined by the text of H.On 2024-06-24 14:03:12 +0000, immibis said:Both of them are correct.
On 24/06/24 15:50, olcott wrote:But which emulation? By H or by H1?On 6/24/2024 2:32 AM, Mikko wrote:Verified fact: the emulation is incorrectOn 2024-06-23 13:23:10 +0000, olcott said:int P(ptr2 x)
On 6/23/2024 4:42 AM, Mikko wrote:Sure, but that was not the first thing mentioned in the initial message.When the head line has the words "these verified facts" the message shouldIt is a verified fact that 2 + 3 = 5 according to the semantics
first tell what facts are "these verified facts" and who verified them
before any further discussion.
of arithmetic. Anyone having an opinion that contradicts this is WRONG.
{
int Halt_Status = H(x, x);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
The call from P to H(P,P) when P is correctly emulated
by H cannot possibly return.
The call from P to H(P,P) when P is correctly emulated
by H1 DOES return.
Because P never calls H1(P,P) and P does call H(P,P)
the call from P to H(P,P) returns in the first case
and cannot possibly return in the second case.
The meaning is fully determined by the complier that complies to the
x86 code and the semantics of x86. If one simulator interpretes the
x86 code differently from another simulator then one of them does not
follow the x86 semantics and is therefore incorrect.
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call H0(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
The call from DDD to
the x86 emulator H0(DDD) at machine address 0000217a
when DDD is correctly emulated by H0 cannot possibly return.
When DDD is correctly emulated by H0 then H0 must emulate itself
emulating DDD. This derives recursive emulation.
When DDD is correctly emulated by H1 then H1 NEED NOT emulate itself
emulating DDD. This DOES NOT derive recursive emulation.
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