Proctors are unable to respond to queries about the interpretation of exam questions. Do your best to answer exam questions as written.

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1 QUEEN'S UNIVERSIY SCHOOL O COMPUING CISC124, WINER ERM, 2011 INAL EXAMINAION 7pm to 10pm, 26 APRIL 2011, Ross Gym HAND IN Answers Are Recorded on Question Paper SOLUION Instructor: Alan McLeod If the instructor is unavailable in the examination room and if doubt exists as to the interpretation of any problem, the candidate is urged to submit with the answer paper a clear statement of any assumptions made. Proctors are unable to respond to queries about the interpretation of exam questions. Do your best to answer exam questions as written. Please write your answers in the boxes provided. Extra space is available on the last page of the exam. he back of any page can be used for rough work. his exam is three hours long and refers exclusively to the use of the Java language. Comments are not required in the code you write. or full marks, code must be efficient as well as correct. his is a closed book exam. No computers or calculators are allowed. Student Number: Problem 1: / 10 Problem 4: / 10 Problem 2: / 35 Problem 5: / 15 Problem 3: / 10 Problem 6: / 10 OAL: / 90

2 Student Number: Page 2 of 17 Problem 1) [10 marks] undamental Java he following complete program runs without errors. Indicate the output of each println() statement in the box beside the statement: public class Problem1 { public static void main (String[] args) { // end main // end Problem1 System.out.println( * 7 ); System.out.println( (1 + 4) * 7 ); System.out.println( 2-3 * 4-5 ); System.out.println( 18 % 5 ); System.out.println( 2 * Math.pow(2, 3) ); System.out.println( 60 / 3 / ); System.out.println( 60 / 3 / (2 + 3) ); System.out.println( 99 / 100 ); System.out.println( 100 / 99 ); System.out.println( 99 / 100 ); System.out.println( 99 / ); System.out.println( (float)(5 / 20) ); System.out.println( (float)5 / 20 ); System.out.println( 5.5E3 / 1000 ); System.out.println( " toes" ); System.out.println( 5 + " toes " ); System.out.println( 5 > 3 && 3!= 2 ); System.out.println(!(5 > 3)!(3!= 2) ); System.out.println( 'a' - 'c' ); System.out.println( "No more\nideas!" ); toes 5 toes 23 true false -2 No more ideas!

3 Student Number: Page 3 of 17 Problem 2) [35 marks] undamental Java, ext ile Input, Encapsulation ext ile Input Syntax: You must import java.io.* and java.util.*. Instantiate a ilereader object with a String filename. he constructor can throw a ilenotoundexception if the file is not found or cannot be opened. You will need the close() method of the ilereader object, which can throw an IOException in the unlikely event that the file cannot be closed. Supply the ilereader object to the constructor of a Scanner object. he Scanner object owns several useful methods including hasnextint() and nextint(). he former returns a boolean and the latter an int. he Scanner class and its methods can be used without having to worry about catching exceptions. Problem Summary: A soda bottling plant has been having problems with the exact volume of soda in their 2L bottles. hey have installed an optical device which is supposed to measure the exact volume in each bottle. he volume should be in-between 1950 and 2050 ml, inclusive, but sometimes the device spits out numbers that are far outside this range. he device saves its measurements of volume in a text file, with one number per line in the file. Here are examples of the contents of two of these data files: Data1.txt Data2.txt Data files can be of any length, but only contain integer numbers, one per line in the file. You must create a class called Volumes that stores and summarizes a set of data from a single file. o be a legal object an instance of your Volume class cannot contain any data values that do not lie between 1950 and 2050 inclusive. A legal object must also contain a minimum of 10 valid volume measurements, but there is no upper limit to the number of measurements. So, your Volume class must have a single public constructor that accepts a single parameter, which is the data filename as a String. he constructor must throw a NoDataException object if the file contains less than 10 valid volume measurements. Accessors in the Volume class will return: he volumes array, consisting of only legal measurements. he size, or number of legal measurements. he minimum legal volume. he maximum legal volume. he average of all legal volumes as a double.

4 Student Number: Page 4 of 17 Problem 2, Cont.) Your class cannot have any mutators. You must also write a tostring(), an equals(), a compareo() and a clone() method for your class, using the conventions you have been taught. Equality is defined as both Volumes objects being exactly equal. heir arrays should have the same size and the exact same measurements in the same order. Comparison, for compareo(), is based only on the average volume. If compareo() is supplied with a Volumes object which has a higher average volume, it would return -1. If supplied an object with a smaller average volume, then it would return +1 and if the average volumes are exactly equal, it would return 0. You can see the output of the tostring() method in the sample testing program below. As usual, clone() should return a deep copy of the current object. If you code your Volumes class properly, it will produce the same output as that shown appended to the following testing program, which was run using the two data files shown on the previous page. import java.io.ilenotoundexception; public class estvolumes { public static void main(string[] args) { Volumes vols1 = null; try { vols1 = new Volumes("Data1.txt"); catch (ilenotoundexception e) { System.out.println(e.getMessage()); catch (NoDataException e) { System.out.println(e.getMessage()); System.out.println("Data set 1:"); int[] cleandata = vols1.getvolumes(); for (int i = 0; i < cleandata.length; i++) System.out.print(cleanData[i] + " "); System.out.println("\nSize = " + vols1.getsize()); System.out.println("Min = " + vols1.getminvol()); System.out.println("Max = " + vols1.getmaxvol()); System.out.printf("Average = %.1f", vols1.getavgvol()); Volumes vols2 = null; try { vols2 = new Volumes("Data2.txt"); catch (ilenotoundexception e) { System.out.println(e.getMessage()); catch (NoDataException e) { System.out.println(e.getMessage()); System.out.println("\n\nData set 2:"); cleandata = vols2.getvolumes(); for (int i = 0; i < cleandata.length; i++) System.out.print(cleanData[i] + " "); System.out.println("\nSize = " + vols2.getsize());

5 Student Number: Page 5 of 17 Problem 2, Cont.) System.out.println("Min = " + vols2.getminvol()); System.out.println("Max = " + vols2.getmaxvol()); System.out.printf("Average = %.1f", vols2.getavgvol()); System.out.println("\n\nesting tostring:"); System.out.println(vols1); System.out.println(vols2); System.out.println("\nesting compareo:"); System.out.println(vols1.compareo(vols2)); System.out.println(vols2.compareo(vols1)); System.out.println("\nesting equals, clone & compareo:"); System.out.println(vols1.equals(vols2)); Volumes vols1clone = vols1.clone(); System.out.println(vols1.equals(vols1Clone)); System.out.println(vols1.compareo(vols1Clone)); // end main // end estvolumes /* OUPU: Data set 1: Size = 13 Min = 1956 Max = 2015 Average = Data set 2: Size = 15 Min = 1985 Max = 2034 Average = esting tostring: 13 volumes, average = volumes, average = esting compareo: -1 1 esting equals, clone & compareo: false true 0 */ Starting on the next page, write the NoDataException class and then write the Volumes class on the following 3 pages.

6 Student Number: Page 6 of 17 Problem 2, Cont.) he exception class: public class NoDataException extends Exception { public NoDataException(String message) { super(message); he Volumes class: import java.io.*; import java.util.*; public class Volumes { private int[] vols; private int size; private double average; private int minvol; private int maxvol; private void setattr() { double sum = 0; minvol = 2050; maxvol = 0; int vol; for (int i = 0; i < size; i++) { vol = vols[i]; sum += vol; if (vol < minvol) minvol = vol; if (vol > maxvol) maxvol = vol; average = sum / size; // end setattr

7 Student Number: Page 7 of 17 Problem 2, Cont.) private Volumes() { // Only used by clone() method public Volumes(String filename) throws ilenotoundexception, NoDataException { int vol, count; ilereader input = null; input = new ilereader(filename); Scanner readvols = new Scanner(input); size = 0; while (readvols.hasnextint()) { vol = readvols.nextint(); if (vol >= 1950 && vol <= 2050) size++; readvols.close(); try { input.close(); catch (IOException e) { if (size < 10) throw new NoDataException("Not enough legal data in file."); vols = new int[size]; input = new ilereader(filename); readvols = new Scanner(input); count = 0; while (readvols.hasnextint()) { vol = readvols.nextint(); if (vol >= 1950 && vol <= 2050) { vols[count] = vol; count++; readvols.close(); try { input.close(); catch (IOException e) { setattr(); // end constructor

8 Student Number: Page 8 of 17 Problem 2, Cont.) public int getsize() { return size; public int[] getvolumes() { return vols.clone(); public int getminvol() { return minvol; public int getmaxvol() { return maxvol; public double getavgvol() { return average; public boolean equals (Object othervols) { if (othervols instanceof Volumes) { Volumes ov = (Volumes)otherVols; if (ov.size == size) { for (int i = 0; i < size; i++) if (ov.vols[i]!= vols[i]) return false; return true; return false; // end equals public int compareo(volumes ov) { if (average < ov.average) return -1; if (average > ov.average) return 1; return 0; public String tostring() { String out = size + " volumes, average = " + Math.round(average * 10) / 10f; return out; public Volumes clone() { Volumes ov = new Volumes(); ov.vols = vols.clone(); ov.size = size; ov.minvol = minvol; ov.maxvol = maxvol; ov.average = average; return ov; // end Volumes

9 Student Number: Page 9 of 17 Problem 2, Cont.)

10 Student Number: Page 10 of 17 Problem 3) [10 marks] undamental Java, Methods and Parameters he following complete program runs without errors. Indicate the output of each println() statement in the box beside the statement: public class Problem3 { public static int method1(int[] array) { array[0] += 10; return array[0]; // Note that the parameter list for method2 is on two lines: public static int method2(int anum, String astr, int[] array1, float[] array2, int[] array3) { anum = 5000; astr = "Goodbye!"; array1[0] = 100; float[] fnums = {1.5, 2.5; array2 = fnums; array3[0] += 10; return 10 + method1(array3); public static void main(string[] args) { int num = 1000; String astr = "Hello!"; int[] arr1 = {1, 2, 3; float[] arr2 = {0.5, 1.5; int[] arr3 = {5, 6, 7; int retnum = method2(num, astr, arr1, arr2, arr3); System.out.println(num); System.out.println(aStr); System.out.println(arr1[0]); System.out.println(arr2[0]); System.out.println(arr3[0]); System.out.println(retNum); 1000 Hello! // end main // end Problem3

11 Student Number: Page 11 of 17 Problem 4) [10 marks] Generic Methods Here is a version of binary search written to accept int values: public static int binsearch (int[] a, int key) { int lo = 0; int hi = a.length - 1; int mid = (lo + hi) / 2; while (lo <= hi) { if (key < a[mid]) hi = mid - 1; else if (a[mid] < key) lo = mid + 1; else return mid; mid = (lo + hi) / 2; // end while return -1; // end binsearch Write a generic version of the same method designed to operate on arrays of objects of type. Objects of type will implement the Comparable<> interface, which means that objects of type will have a compareo() method that returns a negative number if the supplied object is greater than the object used to invoke compareo(). public static < extends Comparable<>> int binsearch ([] a, key) { int lo = 0; int hi = a.length - 1; int mid = (lo + hi) / 2; while (lo <= hi) { if (key.compareo(a[mid]) < 0) hi = mid - 1; else if (key.compareo(a[mid]) > 0) lo = mid + 1; else return mid; mid = (lo + hi) / 2; // end while return -1; // end binsearch

12 Student Number: Page 12 of 17 Problem 5) [15 marks] Qualitative Java Write a or on the line before each of the following statements to indicate if it is rue or alse: 1. All Boolean comparisons must be carried out before arithmetic operators are evaluated in a Java expression. 2. he && operator always evaluates the expressions on both sides of the operator. 3. Casting operations have the highest precedence in any expression. 4. he ++ post increment operator carries out the increment after the variable is used in an expression. 5. If an expression contains a method call that results in an exception, the assignment operation will still be carried out as the last operation in the expression. 6. Unfortunately the current Java version does not contain the syntax for a for each loop. 7. Switch statements can be used with floating point values to see if they lie in certain, predefined value ranges. 8. Any switch statement construct can be built using a chained if construct. 9. hrown exceptions will be propagated to the main method if they are not caught in a try/catch block. 10. Random file I/O can only be used to read data from very large files. 11. Storing numeric data in ASCII text format will usually produce a larger file than if the data is stored in Binary format. 12. Random file I/O can be slower than any other file I/O technique. 13. Passing an array into a method involves the creation of a new copy of that array in the activation frame for that method. 14. Automatic un-boxing can simplify the use of Wrapper classes with Generic classes. 15. Wrapper classes only contain static methods and attributes.

13 Student Number: Page 13 of 17 Problem 5, Cont.) 16. he process of functional decomposition helps to reduce the number of methods contained in a program. 17. Proper JUnit tests can be carried out in any order. 18. he rules of DD describe how you should write your unit tests to ensure test coverage of methods you have already written. 19. A object that extends Jrame inherits a frame with a title bar and the normal window title bar controls. 20. Components laid down in a JPanel using lowlayout will always be touching each other. 21. he text attribute of a JLabel in a running GUI program can only be changed in code. 22. here is more than one listener class that can respond to mouse initiated events. 23. System-generated component re-draws invoke the paint(graphics g) method for that component. 24. Code-generated component re-draws can also be carried out by invoking the paint(graphics g) method directly. 25. o invoke a component s repaint() method you must supply the Graphics object associated with that component to the repaint() method. 26. Java is not capable of associating a transparency level with a Color object. 27. You should not attach a window s animation code to a imer thread since it is a daemon thread. 28. Many listener objects can be attached to a imer thread. 29. Once an instance of a hread object is started, it cannot be stopped until the application is finished. 30. It is impossible to convert a Java application into an applet. Applets must be written from scratch.

14 Student Number: Page 14 of 17 Problem 6) [10 marks] GUI Programming Starting on this page and continuing on the next is a complete GUI class that displays a window without error, when run: import javax.swing.*; import java.awt.*; public class Problem6 extends Jrame { public Problem6() { super(); setsize(540, 600); setdefaultcloseoperation(jrame.exi_on_close); setitle("problem 6"); setlocation(20, 20); int rows = 2; int cols = 4; JPanel jpanel1 = new JPanel(new GridLayout(rows, cols)); JButton[] buttons = new JButton[rows * cols]; for (int i = 0; i < rows * cols; i++) { buttons[i] = new JButton("Button " + i); buttons[i].setont(new ont("arial", ont.bold, 18)); jpanel1.add(buttons[i]); JPanel jpanel2 = new JPanel(new lowlayout(lowlayout.cener)); JLabel[] labels = new JLabel[rows * cols]; for (int i = 0; i < rows * cols; i++) { labels[i] = new JLabel("Label " + i); labels[i].setont(new ont("arial", ont.bold, 18)); if (i % 2 == 0) labels[i].setvisible(false); jpanel2.add(labels[i]); add(jpanel1, BorderLayout.NORH); APanel jpanel3 = new APanel(new BorderLayout()); jpanel3.add(jpanel2, BorderLayout.SOUH); add(jpanel3, BorderLayout.CENER); // end constructor

15 Student Number: Page 15 of 17 Problem 6, Cont.) private class APanel extends JPanel { public APanel(LayoutManager lm) { super(lm); public void paint(graphics g) { super.paint(g); int strokesize = 5; int width = getwidth(); int height = getheight(); int gap = 50; BasicStroke stroke = new BasicStroke(strokeSize); Graphics2D g2d = (Graphics2D)g; g2d.setstroke(stroke); g2d.setcolor(color.black); g2d.drawoval(0 + gap, 0 + gap, width - 2 * gap, height - 2 * gap); // end paint(g) // end APanel class public static void main(string[] args) { Problem6 prob6 = new Problem6(); prob6.setvisible(true); // end main // end Problem6 On the next page, draw the appearance of this window when it is first displayed. Other than for the font size, all numeric dimensions are in pixels. Don t worry about what the font looks like. Use the following key to draw your components, changing the text where necessary: A Label A Button

16 Student Number: Page 16 of 17 Problem 6, Cont.)

17 Student Number: Page 17 of 17 (Extra page)