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On Wed, 07 May 2025 23:16:37 +0100, Richard Heathfield wrote:Precisely. For once we are in agreement.
On 07/05/2025 22:59, olcott wrote:If invisible() call is unreachable then it isn't an accurate simulation.On 5/7/2025 4:52 PM, Richard Heathfield wrote:>On 07/05/2025 22:46, olcott wrote:I already know the answer.On 5/7/2025 4:30 PM, Richard Heathfield wrote:>
<snip>
>>If the simulation can't reach code that the directly executedIf is was true that it is not a faithful simulation then you would be
program reaches, then it's not a faithful simulation.
>
>
able to show exactly what sequence of instructions would be a
faithful simulation.
If it were false, you'd be able to chop out the unreachable code
without any adverse effects. Can you?
>
<snip>
>
>
Then you already know why your simulation code fails to simulate
correctly... but of course you /don't/ know, so I'll explain.
>
Let us postulate a program that contains a function as follows:
>
void invisible(void)
{
secret();
}
>
When directly executed, the program calls invisible(), but when
simulated, the invisible() call is unreachable.
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