Re: Simulating termination analyzers for dummies --- The only reply until addressed

On 6/19/2024 12:45 PM, joes wrote:

Am Wed, 19 Jun 2024 08:47:17 -0500 schrieb olcott:

On 6/19/2024 4:30 AM, Mikko wrote:

On 2024-06-18 16:45:42 +0000, olcott said:

On 6/18/2024 11:37 AM, Mikko wrote:

On 2024-06-18 16:30:46 +0000, olcott said:

On 6/18/2024 11:21 AM, Mikko wrote:

On 2024-06-17 03:33:50 +0000, olcott said:

 

The subject line is incorrect. The OP of "Simulating termination
analyzers for dummies" should tell what a "simulating termination
analyzer" is.
The OP of this thread does not.

I state the prerequisites if you don't have them then you cannot
understand. If you have them then what I say is self-evidently true.

What do you call this, argument from obscurity? You should explain better.
 

void DDD()
{
   H0(DDD);
}
_DDD()
[000020a2] 55         push ebp      ; housekeeping
[000020a3] 8bec       mov ebp,esp   ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404     add esp,+04   ; housekeeping
[000020b2] 5d         pop ebp       ; housekeeping
[000020b3] c3         ret           ; never gets here
Size in bytes:(0018) [000020b3]
Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer