Re: Simplified proof that DDD correctly simulated by HHH does not halt

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Sujet : Re: Simplified proof that DDD correctly simulated by HHH does not halt
De : noreply (at) *nospam* example.com (joes)
Groupes : comp.theory
Date : 09. Jun 2024, 21:13:26
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v44usm$3g17f$6@i2pn2.org>
References : 1 2 3 4
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Sun, 09 Jun 2024 13:23:04 -0500 schrieb olcott:
On 6/9/2024 12:59 PM, joes wrote:
Am Sun, 09 Jun 2024 11:07:19 -0500 schrieb olcott:
typedef void (*ptr)(); // pointer to void function 01   void HHH(ptr
P, ptr I)
02   {
03     P(I);
04     return;
05   }
06 07   void DDD(int (*x)())
08   {
09     HHH(x, x);
10     return;
11   }
12 13   int main()
14   {
15     HHH(DDD,DDD);
16   }
17
 
In the above Neither DDD nor HHH ever reach their own return statement
thus never halt.
Most of my reviewers incorrectly believe that when HH(DD,DD) aborts
its simulated input that this simulated input halts.
 
You chopped out the mandatory prerequisite.
Please go back and prove that you understand what infinite recursion is
before proceeding.
Dude, I've got nothing to prove to you. You instead could explain how you
can call a simulation that differs from the direct execution "correct".
Or why H and HH are different.

--
joes

Date Sujet#  Auteur
10 Nov 24 o 

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