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On 5/26/2025 12:25 PM, Richard Heathfield wrote:HHH fails to correctly simulate the program specified in the input, because it aborts one cycle too soon, which make that it does not see the abort specified in the input. That does not change the specification in the input. One cycle more is needed to see that the input specified a halting program, but HHH is programmed to be blind for that part of the specification in the input.On 26/05/2025 17:24, olcott wrote:HHH(DDD) does correctly report on the behavior that itsOn 5/26/2025 11:10 AM, Richard Heathfield wrote:>On 26/05/2025 16:42, olcott wrote:>no>
C function can see its own caller.
So because DDD calls HHH, HHH can't analyse the halting behaviour of DDD.
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Got it.
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I didn't say that.
Yes, you did.
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On 24/5/2025 in Message-ID <100sr6o$ppn2$3@dont-email.me> you said:
>You are a damned liar when you say that I said>
that HHH must report on the behavior of its caller.
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No HHH can report on the behavior of its caller
for the same reason that no function can report
on the value of the square-root of a dead cat.
Your words.
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Since DDD is HHH's caller, according to you HHH can't report on DDD's behaviour.
input specifies.
_DDD()
[00002192] 55 push ebp
[00002193] 8bec mov ebp,esp
[00002195] 6892210000 push 00002192
[0000219a] e833f4ffff call 000015d2 // call HHH
[0000219f] 83c404 add esp,+04
[000021a2] 5d pop ebp
[000021a3] c3 ret
Size in bytes:(0018) [000021a3]
How many recursive emulations does HHH have to
wait before its emulated DDD magically halts
on its own without ever needing to be aborted?
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