Re: The actual code of HHH

On 2025-02-27 15:29:31 +0000, olcott said:

On 2/27/2025 3:26 AM, Fred. Zwarts wrote:

Op 27.feb.2025 om 00:09 schreef olcott:

On 2/26/2025 3:52 PM, joes wrote:

Since there is so much talk around, but not really about it,
let's take a look:
https://github.com/plolcott/x86utm/blob/
48b4cbfeb3f486507276a5fc4e9b10875ab24dbf/Halt7.c#L1081
In line 1137, we compute a flag:
u32 Root = Init_Halts_HH(&Aborted, &execution_trace, &decoded, &code_end,
(u32)P, &master_state, &slave_state, &slave_stack);
In line 918, we find it basically checks for the magic number
**execution_trace==0x90909090. What is this unexplained value?
 We then pass the saved flag in line 1143:
if (Decide_Halting_HH(&Aborted, &execution_trace, &decoded,
code_end, End_Of_Code, &master_state, &slave_state, &slave_stack, Root)),
defined in line 1030.
Then we get a switch:
1059    if (Root)  // Master UTM halt decider
Line 1070 is then conditionally skipped:
Needs_To_Be_Aborted_HH((Decoded_Line_Of_Code*)**execution_trace);
defined in line 1012, which (on a jmp or call instruction) calls
u32 Needs_To_Be_Aborted_Trace_HH(Decoded_Line_Of_Code* execution_trace,
                                  Decoded_Line_Of_Code *current)
in line 964, where the abort logic lives. (It basically triggers
on a call or jump to itself.)
 So we only abort depending on the address of the execution trace.
This makes no sense. Why is that?
 

 DD emulated by HHH according to the behavior that the x86
machine code of DD cannot possibly terminate normally thus
HHH is infallibly correct to report that this DD emulated
by HHH (not any other DD in the whole freaking universe)
is not-terminating.

No, it is correct to report that HHH is unable to correctly simulate this halting program up to its end.

 In other words you are totally clueless that infinite
recursion HAS NO END.

Do you mean that having no end enables the simulation to the end?
--
Mikko