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On 5/9/25 10:20 PM, olcott wrote:DDD are HHH are on the same memory space I told you this 50 timesOn 5/9/2025 9:15 PM, Richard Heathfield wrote:Nope, because by your definitions HHH can not correctly emulate that DDD, as it doesn't include the code for the HHH that it calls,On 10/05/2025 02:48, olcott wrote:>On 5/9/2025 8:32 PM, Richard Heathfield wrote:>On 10/05/2025 02:29, olcott wrote:On 5/9/2025 8:15 PM, Richard Heathfield wrote:>On 10/05/2025 01:51, olcott wrote:>On 5/9/2025 7:29 PM, Richard Heathfield wrote:>On 10/05/2025 00:02, olcott wrote:>Correctly emulating one or more instructions <is>>
the correct emulation of 1 or more instructions
of DD. This is a truism.
No, it's not. Correct emulation would entail accurately simulating the whole of DDD's behaviour.
It is stupidly wrong to require the complete
emulation of a non-terminating input.
It is touchingly naive to think you can persuade people to accept incomplete emulation as 'correct'.
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If one instruction is emulated correctly
then is is dishonest to say that zero
instructions were emulated correctly.
Which instruction do you think is emulated correctly?_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)
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Assumes facts not in evidence. Your HHH function is in a translation unit that contains a syntax error.
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It is a truism that when DDD is emulated by HHH
according to the rules of the x86 language that
the first four instructions of DDD would be emulated
and then HHH would also be required to emulate
itself emulating DDD.
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