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On 11/13/2024 2:37 AM, joes wrote:Whch isn't the required semantic property (or even a semantic property at all) thus your agruement is just based on a strawman.Am Tue, 12 Nov 2024 16:45:10 -0600 schrieb olcott:You can keep dishonestly removing the code that IOn 11/12/2024 5:22 AM, Mikko wrote:>On 2024-11-11 15:15:09 +0000, olcott said:On 11/11/2024 5:06 AM, Mikko wrote:On 2024-11-09 14:56:14 +0000, olcott said:On 11/9/2024 3:53 AM, Mikko wrote:Only if the called HHH doesn’t halt. Why doesn’t it?THere is similar ambiguity in x86 semantics as there are operationAnyone with sufficient understanding of the x86 language fully well
codes that are defined on some x86 processor models but undefined on
others, and it is also undefined what happens on a jump to a address in
a non-exstent or uninitialised memory.
knows that no DDD emulated by any HHH can possibly reach past its own
[0000217a] machine address.
>
am referring to.
_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]
DDD emulated by HHH keeps cycling through the first
four instructions from [00002172] to [0000217a] and
thus cannot possibly reach past its own machine address
[0000217a] whether HHH halts or not.
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