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On 8/19/24 9:38 PM, olcott wrote:The data is provided, but I did not say that the dataOn 8/19/2024 8:24 PM, Richard Damon wrote:No, it is just a basic fact that NO emulator can properly emulate the about input past the 4th instruction, as the data just isn't provided.On 8/19/24 8:50 PM, olcott wrote:>
>
void DDD()
{
HHH(DDD);
return;
}
>
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
>
*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)
PERIOD.--
Note, above you say HHH, below you say HHHn, thus they are not related problems,
Sorry, you are just proving yourself to be an ignorant liar.
>Nope, and HHHn creates a DDDn that will call HHHn that will return to it after emulating n steps, and then DDDn will halt
X = DDD emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD
Z = DDD never stops running
>
The above claim boils down to this: (X ∧ Y) ↔ Z
>
void EEE()
{
HERE: goto HERE;
}
>
HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.
>
You are just proving you don't understand the difference between the behavior of the actuall DDDn to the partial simulation of it by HHHn.
DDDn's behavior doesn't stop just because the emulator looking at it stopped, but continues to the final state,
The Emulation of DDDn by HHHn DOES stop when HHHn stops emulating, and thus doesn't tell you what happens afterwards unless you can form an actual valid inductive argument to carry the emulation forward.
Sorry, you are just proving how stupid you are.
And, that you are nothing but a cheat that trys to twist the words.
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