// simplified step count of sum of n numbers#include <iostream.h>int count = 0;template<class T>T Sum(T a[], int n){// Return sum of numbers a[0:n - 1].   for (int i = 0; i < n; i++)      count += 2;   count += 3;    return 0;}void main(void){   float a[6] = {1, 2, 3, 4, 5, 6};   cout << Sum(a,6) << endl;   cout << count << endl;}