Fall 2011 Final Test

Transcription

1 Fall 2011 Final Test Answer all questions You are allowed to bring in one unmarked textbook on Java and the lecture slides (unmarked). No other material will be allowed. This test has 11 pages. If you decide to show your changes to question 2 using this question paper itself, please write your name and student number below and make sure we staple this test to your exam booklet. Name : Student # : 1

2 Question 1) (Full Marks 20) Consider the following Java application. Indicate what will be the output generated by the application. public class Q4 { public static int process(string s){ StringTokenizer listtokens; listtokens = new StringTokenizer(s, "(/)"); int x1, x2; x1 = Integer.parseInt(listTokens.nextToken()); x2 = Integer.parseInt(listTokens.nextToken()); return x1/x2; public static double[] extract(string s){ StringTokenizer mytokens; String atoken; double result[]; int index = 0; mytokens = new StringTokenizer(s, "; "); System.out.println("We have " + mytokens.counttokens() + " tokens"); result = new double[mytokens.counttokens()]; while (mytokens.hasmoreelements()){ atoken = mytokens.nexttoken(); result[index] = process(atoken); index++; return result; public static void main(string a[]){ double result[]; String input = "(25/6); 3/6; 12/4"; result = extract(input); for (int i = 0; i < result.length; i++){ System.out.println("result[" + i + "]= " + result[i]); Answer : We have 3 tokens result[0]= 4.0 result[1]= 0.0 result[2]= 3.0 Question 2) (Full Marks = 30) Three applications are given below. Each application has one or more compile-time and/or run-time problem (or problems). The error message from the Eclipse system is 2

3 also given below. Select any two of the applications. For each application you selected, a) fix all errors, so that the application works as intended. b) explain what was the problem and how your changes solve the problem. (Hint: There may be other problems which will come up when the problem reported by Eclipse has been fixed.) Application i) This application involves the classes MyClass and MyExtendedClass given below. There are some compile-time errors as indicated by the comments. In fixing the errors, you are not allowed to delete any line or change any statement to a comment. You are only responsible for fixing the compile-time errors by adding one or more lines of code or code fragments. public class MyClass { String a = ""; void m1(int m, int n) { a = a + m; System.out.println("Inside m1 in MyClass " + n); void m2(char ch) { // Method m2 is defined twice with // same signature change one of // the signatures System.out.println("Inside m2 in MyClass " + ch); void m2(char ch) { // Method m2 is defined twice with // same signature change one of // the signatures System.out.println("Inside method m2 in MyClass " + ch); m2(ch); public class MyExtendedClass extends MyClass { String b = ""; void m1(int n) { b = b + n; m1(n, n); System.out.println("Inside m1 in MyExtendedClass " + n); void m2(char ch) { System.out.println("Inside m2 in MyExtendedClass " + ch); public static void main(string args[]){ MyClass obj2 = new MyExtendedClass(); obj2.m1(2); // There is no method with a // matching signature in MyClass // cast to MyExtendedClass obj2.b = "John";// MyClass does not have variable b. // cast to MyExtendedClass 3

4 One fix: public class MyClass { String a = ""; void m1(int m, int n) { a = a + m; System.out.println("Inside m1 in MyClass " + n); void m2(char ch) { System.out.println("Inside m2 in MyClass " + ch); void m2a(char ch) { System.out.println("Inside method3 in MyClass " + ch); m2(ch); public class MyExtendedClass extends MyClass { String b = ""; void m1(int n) { b = b + n; m1(n, n); System.out.println("Inside m1 in MyExtendedClass " + n); void m2(char ch) { System.out.println("Inside m2 in MyExtendedClass " + ch); public static void main(string args[]){ MyClass obj2 = new MyExtendedClass(); ((MyExtendedClass) obj2).m1(2); ((MyExtendedClass) obj2).b = "John"; Application ii) The following application extends class ListGeneral ( available in the additional document you received). The program has no compile-time error but aborted with a run-time error as noted below. Your task is to fix the problem. You must retain the recursive algorithm used below in your corrected version. For your convenience, the algorithm used is given below: 1) (Non-recursive case) if a list is empty, its length is 0. 2) (Recursive case) Otherwise the length of the list is 1 + the length of the list pointed at by variable next in the first element in the list. public class Q3b extends ListGeneral{ public int numelements(){// The method returns how many elements // we have in the linked list we created. restart(); if (endoflist()){ 4

5 return 0; else{ getnextnode(); return 1 + numelements(); public static void main(string a[]){ Q3b mylist = new Q3b(); mylist.addbeforecurrent("you"); mylist.addbeforecurrent("are"); mylist.addbeforecurrent("how"); System.out.println("Number of elements in mylist is " + mylist.numelements()); Problem encountered: The program aborted with the message: Exception in thread "main" java.lang.stackoverflowerror at Q3b.numElements(Q3b.java:10) Diagnosis- numelements is not making any progress since it starts by calling restart(). One fix : public class Q3b extends ListGeneral{ public int numelements(){ restart(); return countelements(); public int countelements(){ if (endoflist()){ return 0; else{ getnextnode(); return 1 + countelements(); public static void main(string a[]){ Q3b mylist = new Q3b(); mylist.addbeforecurrent("you"); mylist.addbeforecurrent("are"); mylist.addbeforecurrent("how"); System.out.println("Number of elelments in mylist is " + mylist.numelements()); Application iii) The following application converts each object of class String in array mystrings into an integer value and saves the integer value in the array myints. I anticipated two problems: a) the array myints may not be large enough, resulting in an ArrayIndexOutOfBounds Exception. 5

6 b) Some strings in mystrings may have illegal characters, resulting in an exception of the NumberFormatException class. Just to make sure, I threw in another catch block to catch all remaining exceptions and came up with the following code: public class Q3c { public static void main(string a[]){ String mystrings[] = {"312", "341", "3a2", "72"; int myints[] = new int[5]; for (int i = 0; i < mystrings.length; i++){ try{ myints[i] = Integer.parseInt(myStrings[i]); catch(exception e){ System.out.println("Some exception happened"); catch(numberformatexception nfe){//a compilation error here System.out.println("NumberFormatException occured " + "when processing " + mystrings[i]); catch(arrayindexoutofboundsexception abe)//a compilation error // here System.out.println("ArrayIndexOutOfBoundsException " + "occured for i = " + i); Diagnosis: The first catch block catches all exceptions. This is why the other catch blocks are unreachable. One fix : Put the first catch block in the last position. Question 3) This question has two alternatives. Alternative i carries 30 marks, while alternative ii carries 20 marks, so that you automatically lose 10 marks if you answer alternative II. Do not answer alternative ii if you can answer alternative i. Alternative I) (Full Marks 30) This question uses class ListGeneral which was discussed in class. Class ListGeneral is defined in the additional document you have received. Your problem is to define a class called Question3 which includes the following: a) A private variable called mylist of type ListGeneral, 6

7 b) A public method insert(string x) which inserts x at the end of the linked list pointed at by mylist. The method insert must call a recursive method in achieving its objective. c) A public method called reverselist() which returns an object of class ListGeneral corresponding to the reverse of mylist. See example below for an explanation of what is meant by reverse of a list. The method reverselist must call a recursive method in achieving its objective. d) A public method deleteremainingelements(int index) that deletes all elements in mylist, after the element in position index. Here the first element(i.e., the element pointed at by firstnode) has index 0, the element pointed at by the first element has index 1 and so on. You are free to other methods/variables and constructors as needed. Note that I did not ask you to write a tostring() method. Example: Consider the Java application below: public class Test{ public static void main(string a[]){ Question3 alist; alist = new Question3(); alist.insert("year"); alist.insert("new"); alist.insert("happy"); alist.insert("a"); alist.insert("you"); alist.insert("wish"); // Line 10 System.out.println("aList: \n" + alist); System.out.println("aList reversed: \n" + alist.reverselist()); alist.deleteremainingelements(2); // Line 14 System.out.println("aList after deleting all elements " + "after element 2: \n" + alist); After executing line 10, we expect the following linked list that alist is pointing at: Here we assign an index value to each node from the first to the last so that the nodes containing year, new, happy, a, you and wish 7

8 have indices 0, 1, 2, 3, 4 and 5 respectively. If blist is the reverse of alist, then blist is also a list having six nodes, with the same strings year, new, happy, a, you and wish shown above, but the nodes containing these strings have indices 5, 4, 3, 2, 1, and 0 respectively, so that blist, the reverse of alist, is as shown below: After line 14 is executed, we expect all elements in alist after the element having index 2 to be deleted, so that we expect that alist will be as follows: The output produced by the application is given below. alist: year new happy a you wish alist reversed: wish you a happy new year alist after deleting all elements after element 2: year new happy One possible solution: public class Final11Q1 { ListGeneral mylist; public Final11Q1(){ mylist = new ListGeneral(); 8

9 private void insertatendoflist(string x){ if (mylist.endoflist()){// If the end of the list is encountered, // insert x mylist.addbeforecurrent(x); else { mylist.getnextnode(); // Otherwise, go to the next node // and call the method recursively insertatendoflist(x); public void insert(string x){ // Method inserts starts the search from // the beginning of the list mylist.restart(); // and calls a recursive method // insertatendoflist. insertatendoflist(x); /* * In deleteremainingelements(int index) we have assumed that if the * value of index is refers to a node beyond the last node in the * linked list, nothing happens. Other students may assume that the * value of index is such that this cannot happen. */ public void deleteremainingelements(int index){ // Method deleteremainingelements mylist.restart(); // starts search from the first node while ((index >= 0) && (!mylist.endoflist())){// as long as the mylist.getnextnode();//end of the list is not encountered // and we have not found the node we index --; // are looking for, keep going forward. while (!mylist.endoflist()){// Once we find the first node we are mylist.removecurrent();//looking for, delete it and keep // deleting subsequent nodes until // end of the list is encountered. public ListGeneral reverselist(){ // ReverseList starts by creating an empty list ListGeneral listreversed; // When the process ends, this list will be the reverse of the list. listreversed = new ListGeneral(); mylist.restart(); // start at the beginning return reverselist(listreversed);// This is the recursive method which actually // creates the reversed list. public String tostring(){ 9

10 return mylist.tostring(); private ListGeneral reverselist(listgeneral alist){ Object valueincurrentnode; if (mylist.endoflist()) { // If the end of list is encountered the argument alist // gives the reversed list. return alist; else {// Otherwise, take the value in the current node in mylist // add the value at the beginning valueincurrentnode = mylist.currentvalue(); alist.restart();// of the list alist alist.addbeforecurrent(valueincurrentnode); mylist.getnextnode();// go to the next node in mylist return reverselist(alist);// Call method recursively public static void main(string a[]){ Final11Q1 alist; alist = new Final11Q1(); alist.insert("year"); alist.insert("new"); alist.insert("happy"); alist.insert("a"); alist.insert("you"); alist.insert("wish"); System.out.println("aList: \n" + alist); System.out.println("aList reversed: \n" + alist.reverselist()); alist.deleteremainingelements(2); System.out.println("aList after deleting all elements after element 2: \n" + alist); Alternative II) (Full Marks 20) Consider the Java application given below. It uses ListGeneral defined in the additional document you received. Draw a diagram indicating the data structure you have after the application has executed the last call to addbeforecurrent. Your data structure must show all the objects of class Node and their contents, the values of firstnode, currentnode and previousnode. public class Q1b { public static void main(string a[]){ ListGeneral mylist; mylist = new ListGeneral(); mylist.addbeforecurrent("john"); mylist.addbeforecurrent("tom"); mylist.addaftercurrent("mary"); mylist.restart(); mylist.getnextnode(); mylist.getnextnode(); mylist.removecurrent(); mylist.addbeforecurrent(new Integer(5)); mylist.addbeforecurrent(new Integer(7)); 10

11 System.out.println(myList); Question 4) Answer any one of alternatives I and II given below. (Full Marks 20) Alternative I) Our objective is to develop an information system that allows us to i) store a number of (key, value) pairs, and ii) retrieve information from such a system by specifying the contents of key. You may assume that a given key value appears at most once in the information system. Two examples of such pairs are given below: a) In a student information system, value could be an object of a Student class and the key could be the student number. In this case, you may retrieve an object of class Student by specifying his/her student number as a key value. b) In a bank, value could be an object of a Customer class (for instance containing the name, address and the account balance of the customer) and the key could be the account number. In this case, if you supply an account number as a key value, you may retrieve an object of class Customer corresponding to that account number. Your problem is to develop a Java class called MyInformationSystem with type parameter T which includes the following : a) A private inner class called Pair, with type parameter T, that has the following instance variable value of type T, instance variable key of type String, method keymatches(string keyvalue) which returns true if the supplied keyvalue matches the value of key in the current object, constructor method as needed. b) A private instance variable called mylist of type ArrayList with appropriate type parameter, to store the objects of class Pair<T>. c) Constructor as needed. d) A public method insert(t value, String key) which uses the supplied key and value to create a new object of class Pair, inserts the new object into mylist. (You are free to decide the position in mylist where you wish to insert the new object you created). e) A public method retrieve(string keyvalue) which searches all elements in mylist for an object of class Pair with a key that matches keyvalue. If the search succeeds in locating such an object, it returns the instance variable value in the 11

12 object. If the search fails for any reason, the method should throw an exception of class Exception. A Java main method is given below illustrating the features described above. public static void main(string a[]){ String result; String keylist[] = {"126", "536", "428", "245"; MyInformationSystem<String> namestudentnumberlist; namestudentnumberlist = new MyInformationSystem<String>(); namestudentnumberlist.insert("john", "245");// John has // student number 245 namestudentnumberlist.insert("tom", "126"); namestudentnumberlist.insert("mary", "536"); namestudentnumberlist.insert("mark", "821"); for (int i = 0; i < keylist.length; i++){ try{ result = namestudentnumberlist.retrieve(keylist[i]); System.out.println("For key " + keylist[i] + " the matching value is " + result); catch(exception e){ System.out.println("No Match with " + keylist[i]); The output produced is as follows: For key 126 the matching value is Tom For key 536 the matching value is Mary No Match with 428 For key 245 the matching value is John\ One solution : public class MyInformationSystem<T> { /* * Class Pair is an inner class that simply encapsulates the key and the * value */ private class Pair<T>{ // The type parameter T refers to the type of T value; String key; Pair(T value, String key){ this.value = value; this.key = key; // object that value is pointing at // Constructor just initializes the // instance variables. boolean keymatches(string keyvalue){ // Method keymatches returns return keyvalue.equals(key); // true if the current object // satisfies the condition for retrieval. 12

13 private ArrayList<Pair<T>> mylist; // As stated in the question public MyInformationSystem(){ //Constructor just initializes the ArrayList mylist = new ArrayList<Pair<T>>(); public void insert(t value, String key){ for insert //insert adds the new(key, mylist.add(0, new Pair(value, key));// value) pair as the // first element in mylist public T retrieve(string keyvalue) throws Exception{// 5 marks // retrieve returns Pair<T> keyvaluepair; // the object with a matching key. // If none found, it throws an exception. for (int i = 0; i < mylist.size(); i++){ // Look through all keyvaluepair = mylist.get(i); // elements in mylist //if the ith element has a matching key if ((keyvaluepair.key).equals(keyvalue)){// return the return keyvaluepair.value; // value throw (new Exception("match not found")); // If search fails, // throw an exception Alternative II) You have to develop a Java application involving mouse events and mouse motion events. Details of the application are given below: a) Your application initially displays a square of size 20 X 20 pixels at some convenient location within the frame as shown below. The perimeter of the square should be shown in black. 13

14 Hints: i) use Color.BLACK, ii) drawrect(x0, y0, width, height) displays a rectangle with the bounding box of the rectangle having its top left corner at (x0, y0) and with its dimensions as specified. c) The user can select either the top edge or the left edge of the square by pressing the left mouse button with the cursor very close to the edge (for instance within 5 pixels of the edge) he/she wishes to select. When the user selects an edge, the perimeter of the square should be shown in green (Hint: use Color.GREEN). d) If the user now drags the mouse, the selected edge should move with the mouse, so that the square will become smaller or bigger, depending on whether the user is moving the mouse towards the opposite edge or away from the opposite edge. Remember that a square has the same width and height, so that if the user changes the width, the height should change automatically and if the user is changing the height, the width should change automatically. As long as the left mouse button is pressed, the perimeter of the square should be shown in green. An example of dragging the top edge away from the bottom edge is shown below: 14

15 e) If the user releases the left mouse button, the rectangle will remain at the same position it occupied the last time it was displayed. The only difference is that the perimeter of the square should be shown in black. Hints: For your convenience, the java application for the Tictactoe problem, discussed in class, is given in the appendix that you have received. One solution: public class ChangeSquare extends JFrame{ int referencepointx, referencepointy; // gives the coordinates of the top // left corner of bounding box int lastmousepositionx, lastmousepositiony;// coordinates of last mouse position int heightorwidth; Color squarecolor; int whichedgeisselected;//user can select either the left edge or the top edge. // if left edge is selected whichedgeisselected = 1 // otherwise it is 2. boolean perimeterselected = false;// true if the user presses mouse button on // the left edge or the top edge of the square public ChangeSquare(){ super("showing square"); referencepointx = 100; // initial coordinate of bounding box referencepointy = 100; heightorwidth = 20; squarecolor = Color.BLACK; addmouselistener( new MouseAdapter(){ public void mousepressed(mouseevent e ){ // if mouse is pressed lastmousepositionx = e.getx(); // get the coordinates //where the lastmousepositiony = e.gety(); // mouse was pressed, perimeterselected = false; if (onperimeter(lastmousepositionx, lastmousepositiony)){ // If the user presses mouse button on the top edge or the // left edge of the square perimeterselected is true and // the color becomes green perimeterselected = true; squarecolor = Color.GREEN; repaint(); public void mousereleased(mouseevent e ){ // when the left mouse button is released 15

16 ); // the color becomes black again and the flag is reset. // perimeterselected is false squarecolor = Color.BLACK; perimeterselected = false; repaint(); addmousemotionlistener( new MouseMotionAdapter(){ public void mousedragged( MouseEvent e ){ int xvalue, yvalue; int newmousepositionx, newmousepositiony; if (perimeterselected){ // There is a reaction to mouse dragging only if the // user has selected the perimeter by pressing left // mouse button ); setsize(300,300); setvisible(true); newmousepositionx = e.getx(); newmousepositiony = e.gety(); updatesquare(newmousepositionx, newmousepositiony); // update the size of the square and the // coordinates of the left corner of the // bounding box. repaint(); /* * updatesquare is called when the perimeter is selected and the user has dragged * the mouse. The method determines how the top left corner of the bounding box * the dimension of the square and the position of the last mouse movement * have to be updated. The arguments are the current coordinates of the mouse. */ private void updatesquare(int newmousepositionx, int newmousepositiony){ int changeinx, changeiny; if (whichedgeisselected == 1){ // If left edge is selected only the x-coordinate of the top left // corner will change. The amount of change in the x-coordinate // as well as the dimension of the square is the amount of the // mouse displacement. Note that referencepointx is increased // while the dimension is decreased by the amount of change. changeinx = newmousepositionx - lastmousepositionx; referencepointx += changeinx; heightorwidth -= changeinx; 16

17 lastmousepositionx = newmousepositionx; else { // Same as above except that we consider the change in y- // coordinate. changeiny = newmousepositiony - lastmousepositiony; referencepointy += changeiny;// heightorwidth -= changeiny; lastmousepositiony = newmousepositiony; /* * onperimeter returns true if the user has selected the top edge or the * left edge. Otherwise it returns false. * An edge is selected if the user presses the left mouse button with the * cursor within 5 pixels of an edge. */ private boolean onperimeter(int mousepositionx, int mousepositiony){ if ((Math.abs(mousePositionX - referencepointx)< 5) && (mousepositiony >= referencepointy) && (mousepositiony <= referencepointy + heightorwidth)){ // The user has selected the left edge if the mouse is within // 5 pixels of the x-value of the top left corner of bounding box // and y-value of mouse is within the top and the bottom edge whichedgeisselected = 1; return true; else if ((Math.abs(mousePositionY - referencepointy)< 5) && (mousepositionx >= referencepointx) && (mousepositionx <= referencepointx + heightorwidth)){ // Same as above for the top edge whichedgeisselected = 2; return true; else return false; public void paint(graphics g){ super.paint(g); g.setcolor(squarecolor); g.drawrect(referencepointx, referencepointy, heightorwidth, heightorwidth); public static void main(string args[]){ ChangeSquare myframe; myframe = new ChangeSquare(); 17