On 11/11/21 11:00 AM, olcott wrote:
#define ptr uintptr_t
void P(ptr x)
{
H(x, x);
}
int H(ptr x, ptr y)
{
((void(*)(ptr))x)(y);
return 1;
}
int main()
{
H((ptr)P, (ptr)P);
return 0;
}
Yes.
So all you have proven is that if H unconditionally executes its input
you get infinite recursion.
And, H will never return.
If H isn't that function, then the computation of P changes so you have
no proof for what the behavior of that machine is.
On 11/11/21 11:31 AM, olcott wrote:
On 11/11/2021 10:14 AM, Richard Damon wrote:
On 11/11/21 11:00 AM, olcott wrote:
#define ptr uintptr_t
void P(ptr x)
{
H(x, x);
}
int H(ptr x, ptr y)
{
((void(*)(ptr))x)(y);
return 1;
}
int main()
{
H((ptr)P, (ptr)P);
return 0;
}
Yes.
So all you have proven is that if H unconditionally executes its
input you get infinite recursion.
And, H will never return.
If H isn't that function, then the computation of P changes so you
have no proof for what the behavior of that machine is.
_P()
[00001a5e](01) 55 push ebp
[00001a5f](02) 8bec mov ebp,esp
[00001a61](03) 8b4508 mov eax,[ebp+08]
[00001a64](01) 50 push eax // push P
[00001a65](03) 8b4d08 mov ecx,[ebp+08]
[00001a68](01) 51 push ecx // push P
[00001a69](05) e810000000 call 00001a7e // call H
[00001a6e](03) 83c408 add esp,+08
[00001a71](01) 5d pop ebp
[00001a72](01) c3 ret
Size in bytes:(0021) [00001a72]
If H simulates the x86 machine language of its input and sees that its
simulated P is calling H with the same parameters that H was called
with H can abort its simulation of P and correctly report that P would
never reach its final state at 1a72.
No it can't.
If H can abort its simulation, then it needs to take into account that H
can abort its simulation. PERIOD.
H, in making that conclusion, is PRESUMING that the called H will not
abort its simulation, which it is wrong about.
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