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Am Wed, 31 Jul 2024 05:52:54 -0500 schrieb olcott:On 7/31/2024 3:54 AM, joes wrote:Am Tue, 30 Jul 2024 16:13:55 -0500 schrieb olcott:>On 7/30/2024 4:07 PM, joes wrote:>Am Tue, 30 Jul 2024 15:05:54 -0500 schrieb olcott:On 7/30/2024 1:48 PM, Fred. Zwarts wrote:Op 30.jul.2024 om 17:14 schreef olcott:On 7/30/2024 9:51 AM, Fred. Zwarts wrote:Op 30.jul.2024 om 16:21 schreef olcott:On 7/30/2024 1:52 AM, Mikko wrote:On 2024-07-29 14:07:53 +0000, olcott said:I proved otherwise. When the abort code is commented out then itBut the abort is not commented out in the running code!
keeps repeating again and again, thus conclusively proving that is
must be aborted or HHH never halts.>I modified the original code by commenting out the abort and it does
endlessly repeat just like HHH correctly predicted.Yes, and that modification makes HHH not call itselfNot at all. It makes HHH stop aborting DDD.
So that HHH and DDD endlessly repeat.
Commenting out a section changes the program.This conclusively proving that this section was required.
You changed only the innerNot at all. I simply disabled the abort and this resulted
HHH's, not the outermost one, thus breaking the recursive simulation.
To prove that a section of code is required we remove thatA program's identity changes with its code. It doesn't matter what I labelbut a different program. You'd need to also comment out the outermostIf your name is Charlie and your leg gets amputated you are still
abort; then it wouldn't halt, but if you change HHH to abort, you
change all copies of it at the same time (to keep the recursive call
structure).
yourself, you don't get renamed to Bill.
it in the source. I can define different functions with the same name.
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