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On 5/16/2025 10:33 PM, Richard Heathfield wrote:WHich means you need to pass it the source code of the funciton.On 17/05/2025 03:55, olcott wrote:It is possible to create a C function thatOn 5/16/2025 9:44 PM, Richard Heathfield wrote:>On 17/05/2025 03:24, olcott wrote:
<snip>
>>>>>
When you dishonestly remove the context that you are
replying to fools might think that your rebuttal has merit.
The context you claim was 'dishonestly' removed is:
>
void DDD()
{
HHH(DDD);
return;
}
>
with which we are all too, too familiar.
>
The context merely shows that the only information HHH receives is a pointer to a function.
>
That's not enough for HHH to be able to do what you claim for it *within the rules of C*.
>
Unless there is also an interpreter also written in C.
No, not even then, for reasons I have already explained.
>Any competent C programmer would know that C programs>
can be simulated by C interpreters. If they don't know
this then that are not competent.
A C interpreter (eg CH or CINT, both of which have Wiki pages, in case you're interested) doesn't simulate C code. It interprets C code. You don't pass C code to HHH in the form of a char * - "void DDD()\n{\n\tHHH(DDD);\n\treturn;\n}\n", say - to HHH(). You pass a function pointer. All HHH() can do with that pointer value is:
>
simulates the source-code of other C functions.
The essential idea of this is a C interpreter.
The actual HHH uses x86 emulation that is wayNo, and the fact that you think x86 simulation is way over peoples heads just shows how little you understand about it.
over most peoples heads. When I said that HHH
simulates DDD reviewers are not free to ignore
the word "simulate".
They do this because they only glance at aNo, the rebuttals are not "fake". Your understanding of them is though.
couple of my words to artificially contrive
some fake rebuttal.
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