Quick Quiz

  /* Please do not write a class specification or any method specification.
   * Instead, please write only a code segment.  The code segment should 
   * use a for-loop to print the even numbers between 1 and 10, inclusive
   */
   for (int number=1; number <=10; number++)
     if ((number % 2) == 0) // Alternate: if ( (number/2) == ((number+1)/2) )
       System.out.println (number);
   }
  

The Modulus (%-Operator)

Notice the solution given above. What's the funny %-sign? It is the modulus operator. Think about integer division. It "loses" the remainder, right? Well, where does it go? Don't know. But the %-operator can find it.

Whereas the /-divide operator, when applied to an int, returns the integer quotient, the %-operator, only defined over ints, returns the remainder.

So, for example, "Five divided by two is two -- with one left over", so 5 / 2 = 2, 5 % 2 = 1.

A Simple Tangle

Let's write some quick code. Given a user's choice, represented as a char, let's print my 'F'irst, 'M'iddle, or 'L'ast name, or 'Q'uit.

  boolean quit = false;

  while (!quit) {

    // Display the menu
    System.out.println ("");
    System.out.println ("");
    System.out.println ("--- Menu ---");
    System.out.println ("F)irst name");
    System.out.println ("M)iddle name");
    System.out.println ("L)ast name");
    System.out.println ("Q)uit");
    System.out.println ("");
    System.out.println ("Enter your choice: ");

    // Get choice
    String choiceString = br.readLine();
    char choice = choiceString.toUpperCase().charAt(0); 


    // Decide what to do -- and do it
    if (choice == 'F') 
      System.out.println ("Gregory");
    else 
      if (choice == 'M')
        System.out.println ("Michael");
      else 
        if (choice == 'L')
          System.out.println ("Kesden");
        else
          if (choice == 'Q')
            quit = true; // or break
          else
            System.out.println ("Please select F, M, L, or Q.");
  }
  

Notice how nested that code became. Does it have to be that complicated? Well, we can rewrite it and make it a bit better. To do that, we'll bail as soon as we get a match, so we can eliminate the nesting:

  boolean quit = false;

  while (!quit) {

    // Display the menu
    System.out.println ("");
    System.out.println ("");
    System.out.println ("--- Menu ---");
    System.out.println ("F)irst name");
    System.out.println ("M)iddle name");
    System.out.println ("L)ast name");
    System.out.println ("Q)uit");
    System.out.println ("");
    System.out.println ("Enter your choice: ");

    // Get choice
    String choiceString = br.readLine();
    char choice = choiceString.toUpperCase().charAt(0); 


    // Decide what to do -- and do it
    if (choice == 'F') { 
      System.out.println ("Gregory");
      continue;
    }

    if (choice == 'M') {
      System.out.println ("Michael");
      continue;
    }

    if (choice == 'L') {
      System.out.println ("Kesden");
      continue;
    }

    if (choice == 'Q') {
      quite = true; // Or simply, "break;" and skip the next line
      continue;
    }

    System.out.println ("Please select F, M, L, or Q.");
  }
  

The above construction is better. But, it highlights the problem. We're trying to make a one-of-n decision, given a construct, the if-else, that is designed to make a one-of-two decision. So, we basically build up the the right number of choices in powers of two.

The switch-statment

When we come back next class, we'll take a look at another Java construct, the switch statement, designed for exactly these 1-of many (or even several of many) types of decisions.