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
        int Array[SIZE];

        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