When I saw your question, what immediate came to my mind is that .. the function
RSFptr() must be returning pointer to an array and so the statement
(*PGE->RSFPtr())[i] = TempRFState[i]; assigns
TempRFState[i] to the i'th element of that array.
Try to run the following test code for further clarification :
Code: C
#include <stdio.h>
#define SIZE 1000
class Test
{
private:
int Array[SIZE];
public:
int* AccessArray() { return Array; }
};
int main()
{
Test ClassA;
Test* pClassA = &ClassA;
for(int i = 0; i < SIZE; ++i)
(pClassA->AccessArray())[i] = i;
for(int i = 0; i < SIZE; ++i)
printf("%u\n", (pClassA->AccessArray())[i]);
return 0;
}
There
might be some other meaning of (*PGE->RSFPtr())[i] = TempRFState[i];
If I find any other meaning, I'll update this thread