Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

Op 28.feb.2025 om 00:28 schreef olcott:

On 2/27/2025 2:45 PM, joes wrote:

Am Thu, 27 Feb 2025 14:13:13 -0600 schrieb olcott:

On 2/27/2025 3:50 AM, Fred. Zwarts wrote:

Op 27.feb.2025 om 06:00 schreef olcott:

On 2/26/2025 5:50 PM, Richard Damon wrote:

On 2/26/25 9:49 AM, olcott wrote:

On 2/26/2025 3:48 AM, Fred. Zwarts wrote:

Op 26.feb.2025 om 05:50 schreef olcott:

On 2/25/2025 10:21 PM, Richard Damon wrote:

>

You already know that you are stupidly wrong about the emulation
being incorrect or you would have provided the correct emulation
sequence long ago.

>

What are the first 15 lines of DD correctly emulated by HHH?

The error in the simulation occurs already at the 5th instruction,
the 'call 000015c3'. Instead of simulating this instruction,

You dodged the actual question.
What are the correct first 15 lines of DD emulated by HHH.

And you are missing the point that HHH can not correctly emulate 15
line of this DD, as it is missing information and thus the task is
improperly defined.

If we assume that HHH emulates the above machine code then there is
NO MORE relevant information needed and we know that machine address
0000213c of DD is followed by machine address 00002133 of DD.

 

No. Relevant information includes the code at address 0x213c and
everything it points to.
>

 _DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
 THE ONLY FREAKING THING THAT NEED BE KNOWN
ABOUT THE CODE AT THAT ADDRESS
(TO DETERMINE THE SEQUENCE OF MACHINE INSTRUCTIONS OF DD)
IS THAT IT EMULATES THE MACHINE INSTRUCTIONS OF DD.
  From this base assumption we know that the first
five instructions of DD are repeated three times
when the first 15 instructions of DD are emulated.
 

Not necessarily. A good decider would prevent infinite recursion and base its decision on other facts and then return. If we assume that HHH is a good decider, we can skip the call to it and use its return value and continue with 2141, then we will reach 2155.
The only problem is that we then discover that HHH's return value was incorrect. This was expected, because a correct solution does not exist. Olcott want us to draw a square circle.