Re: DDD correctly emulated by HHH --- Correct Emulation Defined 2

On 3/21/25 8:02 PM, olcott wrote:

 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]
 For every HHH at machine address 000015d2 that emulates
a finite number of steps of DDD according to the
semantics of the x86 programming language no DDD
ever reaches its own "ret" instruction halt state.
 

So, you demonstrate your utter stupidity and use of incorrect definitions.
For EVERY HHH at machine address 000015d2 that emulates just a finite number of steps and return, then the PROGRAM DDD WILL reach its own return instruction, as the behavior of a program is DEFINED by its direct execution.
All you have done is proven that you statement is self-contradictory, as any HHH that only emulated a finite number of steps of DDD, doesn't correctly emulate DDD per the x86 programming language unless it reached that return instruction, as no x86 instruction says the execution stopped there, but your emulation did.
So, all you did what establish that you don't understand the meaning of the words you used.
Note, the bigger problem is that the above DDD is not a "Program" per the definitions, as it doesn't include all its code, but needs to inlcude as part of its definition the function you referenced, and thus each HHH gets a DIFFERENT DDD, and thus you can't compare results of different HHHs to talk about the behavior of their inputs.