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On 10/15/2024 10:17 AM, joes wrote:Am Tue, 15 Oct 2024 08:11:30 -0500 schrieb olcott:On 10/15/2024 6:35 AM, Richard Damon wrote:On 10/14/24 10:13 PM, olcott wrote:On 10/14/2024 6:50 PM, Richard Damon wrote:On 10/14/24 11:18 AM, olcott wrote:On 10/14/2024 7:06 AM, joes wrote:Am Mon, 14 Oct 2024 04:49:22 -0500 schrieb olcott:On 10/14/2024 4:04 AM, Mikko wrote:On 2024-10-13 12:53:12 +0000, olcott said:
Oh, did you take out the check if HHH is the root simulator?There are no static root variables. There never has been any "not a pureYes! It really has different code, by way of the static Root variable.It explains in great detail that another different DDD (same machinehttps://chatgpt.com/share/6709e046-4794-8011-98b7-27066fb49f3e WhenI did that, and it admitted that DDD halts, it just tries to justify
you click on the link and try to explain how HHH must be wrong when
it reports that DDD does not terminate because DDD does terminate it
will explain your mistake to you.
why a wrong answer must be right.
code different process context) seems to terminate only because the
recursive emulation that it specifies has been aborted at its second
recursive call.
No wonder it behaves differently.
function of its inputs" aspect to emulation.
Every termination analyzer that emulates itself emulating its input hasThat point can never come in the complete simulation of a non-
always been a pure function of this input up to the point where
emulation stops.
By "pure" I mean having no side effects. You mean total vs. partial.Non-terminating C functions do not ever return, thus cannot possibly beYou err because you fail to understand how the same C/x86 functionDo explain how a pure function can change.
invoked in a different process context can have different behavior.
pure functions.
HHH is a pure function of its input the whole time that it is emulating.I thought DDD was fixed to only call HHH(DDD)?
DDD has no inputs and is allowed to be any finite string of x86 code.
Inputs to HHH are by no means required to ever return AT ALL.
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