Re: Because I have repeated this same point 500 times in the last three years...

On 2024-07-26 01:49:46 +0000, olcott said:

If you understand the x86 language and can't tell how DDD
emulated by HHH differs from DDD emulated by HHH1 by the
following then you are probably lying about understanding
the x86 language.
 *I did annotate it a little better this time*
 typedef void (*ptr)();
int HHH(ptr P);
int HHH1(ptr P);
 void DDD()
{
   HHH(DDD);
}
 int main()
{
   HHH1(DDD);
}
 *You really don't need to know one damn thing else besides this*
*You really don't need to know one damn thing else besides this*
*You really don't need to know one damn thing else besides this*
 All that you have to know is that HHH and HHH1 are x86 emulators
and that HHH sees that same repeated state (first four lines of DDD)
that anyone knowing the x86 language can see.

The first state of DDD is not repeated in the same execution. Another
exection may see the same states and at least sees the same initial
state but that is not called a repeated state. A repeated state is a
state that has alredy occurred in the same execution.
--
Mikko