#include <iostream>
using namespace std;

/*-------------------------------------------------
 Contract: main : void -> int

 Purpose: make sure that type casting in C++ works
          like we think it does...

 Examples: not really applicable; it performs the
           given casting and prints the results
           to the screen.

 by: Sharon M. Tuttle
 last modified: 11-3-03
 ----------------------------------------------------*/

int main()
{
    int sumOfRolls = 595, numOfRolls = 150;
    double avgRoll;

    // say, for example, that you had sumOfRolls,
    //    an integer sum of a bunch of dice rolls.
    // And, numOfRolls, an integer number of rolls.
    //
    // Two values that really are int's --- and which
    //    we might want to legitimately non-integer
    //    divide, to get the average roll.

    // if we used integer division:
    avgRoll = sumOfRolls/numOfRolls;
    cout << "avg roll (using int division) is: " << avgRoll 
         << endl;

    // a less-klugy way: cast one of them to a double!

    avgRoll = static_cast<double>(sumOfRolls)/numOfRolls;
    cout << "avg roll is (using casting): " << avgRoll 
        << endl;

    // BUT be careful --- if cast the *result* of the
    //    division, you'll get integer division, and
    //    then simply convert the RESULTING int into
    //    a double...!
    
    avgRoll = static_cast<double>(sumOfRolls/numOfRolls);
    cout << "avg roll is (using wrong casting): " << avgRoll 
        << endl;

    // and, to deliberately convert double to int?
    avgRoll = static_cast<double>(sumOfRolls)/numOfRolls;
    cout << "conversion of double to int: "
         << static_cast<int>(avgRoll) << endl;

    return 0;
}