Midterm #1CMSC 433Programming Language Technologies and ParadigmsFall 2006October 26, 2006GuidelinesThis exam has 8 pages (including this one); make sure you have them all. Put your name on each pagebefore starting the exam. Write your answers directly on the exam sheets, using the back of the page asnecessary. Bring your exam to the front when you are finished. Please be as quiet as possible.If you have a question, raise your hand. If you feel an exam question assumes something that is notwritten, write it down on your exam sheet. Barring some unforeseen error on the exam, however, youshouldn’t need to do this at all, so be careful when making assumptions.You may avail yourself of the punt rule. If you write down punt for any part of a question, you will earn1/5 of the points for that question (rounded to nearest integer).Use good test-taking strategy: read through the whole exam first, and first answer the questions that areeasiest for you and are worth the most points.Question Points Score1 242 193 194 195 19Total 1001. (24 pt, Short answer)(a) What is the information hiding principle?(b) Why use information hiding?(c) Say you have a class whose data members are all private. Is this enough to ensure that the designdecision hidden in this module remains hidden? If it is, explain why; if not, write a few lines ofJava showing how information hiding can still be violated.(d) Say you have some base c lasse s that are likely to change and want to avoid the fragile base classproblem. What are two alternatives to inheritance that you can use? (One word each)2(e) What is the principle of low coupling and how does it facilitate ease of change?(f) Good or bad pair programming practice? Discuss. While the driver is impleme nting a class, thenavigator is writing down test cases for the class.32. (19 pts) List the line numbers of all violations of the Law of Demeter in the following code.1 public class MainFrame extends JFrame2 {3 private static final String appName = "MainFrame" ;4 private static final String appVer = "0.6" ;5 private Color bgColor = Color.black ;6 private boolean fullScreen = false ;7 private Dimension screenSize ;89 /* Make a new Frame with an ImageCollage as the content pane */10 MainFrame(ImageCollage collage)11 {12 // Set the initial window size, title, background color, close on exit13 setTitle(appName) ;14 setDefaultCloseOperation(EXIT_ON_CLOSE) ;15 setBackground(bgColor) ;16 Toolkit toolkit = Toolkit.getDefaultToolkit() ;17 screenSize = toolkit.getScreenSize() ; // Get (primary) screen resolution18 setSize(screenSize) ;19 toolkit.setDynamicLayout(true) ; // Tell Swing to update the window as it is being resized20 Image icon = toolkit.getImage("icon.gif") ;21 setIconImage(icon) ;22 collage.setCanvasSize(screenSize) ;23 collage.setBackgroundColor(bgColor) ;24 collage.clearCanvas() ;25 setContentPane(collage) ;26 boolean fullScreen = GraphicsEnvironment.getLocalGraphicsEnvironment()27 .getDefaultScreenDevice().isFullScreenSupported() ;28 Debug.d("Screen size " + (int) screenSize.getWidth() + "x" + (int) screenSize.getHeight()29 + ", full screen " + (fullScreen ? "" : "not ") + "supported.") ;30 }31 }43. (19 pt) Design pattern(a) You are writing a diagnostic utility to test the various components of a computer system to ensurethat they will work with your big, expensive enterprise application. You want to write one class foreach computer component you diagnose: CPUTest, MemoryTest, and DiskTest. However, yourutility needs to be c ross- platform, and each class contains some platform-specific code. So foreach type of test, you will have several implementations, one for each platform: PCMemoryTest,MacMemoryTest, LinuxMemoryTest, etc. You’d like your diagnostic utility to create and use theright test classes for the platform its running on, without having to change a lot of code each timeyou add a new platform. You recall that there is a design pattern to solve this problem. Whatdesign pattern would you use?(b) Draw a UML c lass diagram showing the relationships among participants in your design pattern.Include the three types of tests above, for two platforms: PC and Mac.54. (19 pt) Visitor / Double dispatch(a) You have an interface A, implemented by classes B and C and an interface X, implemented byclasses Y and Z.Define a method Test.doubleDispatchclass Test {static void doubleDispatch(A a, X x) { ... }}Use the visitor or double dispatch des ign pattern to use dynamic dispatch so that a method chosento be executed based on the runtime type of a and x. In other words, there should be one methodthat is selected if a is a B and x is a Y, another method selected if a is a B and x is a Z, and soon. You may not use instanceof or any conditional expressions; instead, use the visitor patternor double dispatch to achieve this. You may add any methods you need to add to the interfacesand classes. Label the methods that select for each combination of runtime types (e.g., label onemethod as the BY method). Use space on the following page if needed.6(b) Given the implementation you chose, what would be easier: adding an additional class thatimplemented A, or an additional class that implemented X (in either case, the code would nowneed to select one of six different methods, depending upon the runtime type of a and x).75. (19 pts) Iterator patternThe 3n+1 problem is a simple but open problem in computer science.1Given the following method,a sample call to threeNPlusOne(22) will print the sequence 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1.void threeNPlusOne(int n) {while (true) {System.out.println(n);if (n == 1) return;if (n%2 == 0) n = n/2;else n = 3*n+1;}}Your job is to define a method:Iterator<Integer> threeNPlusOneIterator(int n) {...}that returns an iterator over integer values. Thus,for(int i : threeNPlusOneIterator(22)) System.out.println(i)will print the same sequence as the code above.Note that due to autob oxing, you can pretty much ignore the difference between int and Integer forthis project. Also, the Iterator interface is declared as:interface Iterator<T> {boolean hasNext();T next();void remove();}1It is believed that this sequence will terminate for any positive integer n, but that conjecture is unproven (fame if notfortune to the first person who proves it). It is known to terminate for all positive integer values that can be represented witha int. To win the fame, you need to prove that it terminates for all positive integer
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