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typedef void (*ptr)();Right.
int HHH(ptr P);
void Infinite_Loop()
{
HERE: goto HERE;
}
void Infinite_Recursion()
{
Infinite_Recursion();
}
void DDD()
{
HHH(DDD);
}
int main()
{
HHH(Infinite_Loop);
HHH(Infinite_Recursion);
HHH(DDD);
}
Every C programmer that knows what an x86 emulator is knows
that when HHH emulates the machine language of Infinite_Loop,
Infinite_Recursion, and DDD that it must abort these emulations
so that itself can terminate normally.
When this is construed as non-halting criteria then simulatingBut only when you can actually PROVE that the full emulation of the program would be non-halting.
termination analyzer HHH is correct to reject these inputs as
non-halting by returning 0 to its caller.
Simulating termination analyzers must report on the behaviorBut if DDD WAS stuck in infinte recursion, so must be HHH, so it isn't a decider.
that their finite string input specifies thus HHH must report
that DDD correctly emulated by HHH remains stuck in recursive
simulation.
Everyone else seems to be flat out dishonest or totally ignorant.Nope, just one person, YOU. Your claims are FULL OF LIES and inconsistant definitions, and you re just to ignorant and stupid to understand that.
At least one of my reviewers does not seem to understand that
infinite recursion does not halt.
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