C / C++ / MFC

// the countof macro for calculating array size
#define countof(X)    (sizeof(X) / sizeof((X)[0]))

#define COUNT_OF(X)  (sizeof(X) / sizeof((X)[0]))
 
void DisplayNumbers(int[6] numbers);
void DisplayNumbers2(int* pNumbers);
void DisplayNumbers3(int* pNumbers, int nSize);
 
void main()
{
   int numbers[] = {1, 2, 3, 5, 8, 13};
 
   // Here it works fine
   int nSize = COUNT_OF(numbers);
 
   DisplayNumbers(numbers);
 
   DisplayNumbers2(numbers);
 
   DisplayNumbers3(numbers, nSize);
}
 
void DisplayNumbers(int[6] numbers)
{
   printf("DisplayNumbers shows:\n   ");
 
   // Here it works fine too, although it's
   // not very useful because you are already
   // stating the size is 6
   int nSize = COUNT_OF(numbers);
   for (int i = 0; i < nSize; ++i)
      printf("%d ", numbers[i]);
 
   printf("\n");
}
 
void DisplayNumbers2(int* pNumbers)
{
   printf("DisplayNumbers2 shows:\n   ");
 
   // Here it doesn't work at all, because for
   // all the compiler knows, pNumbers is just
   // a pointer to an int. Thus, sizeof(X) in
   // the macro will return 4 (i.e., the size
   // of any pointer), sizeof((X)[0]) will
   // return 4 too, and the result for COUNT_OF(X)
   // will then be 1.
   int nSize = COUNT_OF(pNumbers); // will return 1
   for (int i = 0; i < nSize; ++i)
      printf("%d ", numbers[i]);
 
   printf("\n");
}
 
void DisplayNumbers3(int* pNumbers, int nSize)
{
   printf("DisplayNumbers3 shows:\n   ");
 
   for (int i = 0; i < nSize; ++i)
      printf("%d ", numbers[i]);
 
   printf("\n");
}

DisplayNumbers shows:
   1 2 3 5 8 13 
DisplayNumbers2 shows:
   1
DisplayNumbers3 shows:
   1 2 3 5 8 13