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On 6/26/2024 6:55 PM, Richard Damon wrote:Nope, since you didn't put in the rule, and if you had it would have shown that you lied, as if H0 is a pure function then the call to H0 emulated by H0 needs to have the same behaivor as the direct call to H0 by main.On 6/26/24 7:46 PM, olcott wrote:We have already been over that you know that you cheated.On 6/26/2024 6:41 PM, Richard Damon wrote:>On 6/26/24 9:42 AM, olcott wrote:>On 6/26/2024 6:02 AM, Richard Damon wrote:>On 6/25/24 11:42 PM, olcott wrote:>>>
That is not the way that it actually works.
That the the way that lies are defined.
Source for you claim?
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Where is you finite set of steps from the truthmakers of the system to that claim?
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_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call H0(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
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The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.
Sure it can. I have shown an H0 that does so.
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I already told you that example does not count.
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I can't keep repeating those details or others
that so far have no idea what an x86 emulator is
will be baffled beyond all hope of comprehension.
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WHy not?
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