Sujet : Re: This function proves that only the outermost HHH examines the execution trace
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 30. Jul 2024, 08:15:12
Autres entêtes
Organisation : -
Message-ID : <v8a3u0$ucb1$1@dont-email.me>
References : 1 2 3 4 5 6 7 8
User-Agent : Unison/2.2
On 2024-07-29 16:55:46 +0000, olcott said:
On 7/28/2024 4:10 AM, Mikko wrote:
On 2024-07-27 18:14:52 +0000, Alan Mackenzie said:
olcott <polcott333@gmail.com> wrote:
Stopping running is not the same as halting.
DDD emulated by HHH stops running when its emulation has been aborted.
This is not the same as reaching its ret instruction and terminating
normally (AKA halting).
I think you're wrong, here. All your C programs are a stand in for
turing machines. A turing machine is either running or halted. There is
no third state "aborted". An aborted C program certainly doesn't
correspond with a running turing machine - so it must be a halted turing
machine.
So aborted programs are halted programs. If you disagree, perhaps you
could point out where in my arguments above I'm wrong.
May I disagree? An "aborted" Turing machine is a runnung Turing machine.
A Turing machine has no notion of being aborted.
That's correct. But you have used the word anyway.
-- Mikko
…