1. 1 Arrays Chapter 8 Spring 2006 CS 101 Aaron Bloomfield

2. 2 Introduction to arrays

3. 3 Background  Programmer often need the ability to represent a group of values as a list List may be one-dimensional or multidimensional  Java provides arrays and the collection classes The Vector class is an example of a collection class  Consider arrays first

4. 4 Example  Definitions char[] c; int[] value = new int[10];  Causes Array object variable c is un-initialized Array object variable value references a new ten element list of integers  Each of the integers is default initialized to 0 value 0 0000 -c …

5. 5 An array example int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); 8 is displayed int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt(); Suppose 3 is extracted int[] v = new int[10]; int i = 7; int j = 2; int k = 4; v[0] = 1; v[i] = 5; v[j] = v[i] + 3; v[j+1] = v[i] + v[0]; v[v[j]] = 12; System.out.println(v[2]); v[k] = stdin.nextInt();

6. 6 Array variable definition styles  Without initialization Type of values in list Name of list Brackets indicate array variable being defined ElementType [ ] id; int [] a; int a[];

7. 7 Array variable definition styles  With initialization ElementType [] id = new ElementType [n]; Nonnegative integer expression specifying the number of elements in the array A new array of n elements

8. 8 Where we’ve seen arrays  public static void main (String[] args) Thus, the main() method takes in a String array as the parameter  Note that you can also define it as: public static void main (String args[])  or public static void main (String[] foobar)

9. 9 Basic terminology  List is composed of elements  Elements in a list have a common name Example: a[3] = 5; The common name is ‘a’  The list as a whole is referenced through the common name  List elements are of the same type — the base type  Elements of a list are referenced by subscripting (indexing) the common name

10. 10 Java array features  Subscripts are denoted as expressions within brackets: [ ]  Base (element) type can be any type  Size of array can be specified at run time This is different that pure C! (for the most part, at least)  Index type is integer and the index range must be 0... n-1 Where n is the number of elements Just like Strings indexing!  Automatic bounds checking Ensures any reference to an array element is valid  Data field length specifies the number of elements in the list  Array is an object Has features common to all other objects More on this later…

11. 11 Consider  Segment int[] b = new int[100]; b[-1] = 0; b[100] = 0;  Causes Array variable to reference a new list of 100 integers  Each element is initialized to 0 Two exceptions to be thrown  -1 is not a valid index – too small  100 is not a valid index – too large IndexOutOfBoundsException

12. 12 Consider Point[] p = new Point[3]; p[0] = new Point(0, 0); p[1] = new Point(1, 1); p[2] = new Point(2, 2); p[0].setX(1); p[1].setY(p[2].getY()); Point vertex = new Point(4,4); p[1] = p[0]; p[2] = vertex; p p[0]p[1]p[2] null Point: (0, 0) p p[0]p[1] Point: (1, 1)Point: (2, 2) p[2] Point: (1, 0) p p[0]p[1] Point: (1, 1)Point: (2, 2) p[2] Point: (1, 0) p p[0]p[1] Point: (1, 2)Point: (2, 2) p[2] Point: (1, 0) p p[0]p[1] Point: (1, 2)Point: (2, 2) p[2] vertex Point: (4, 4) Point: (1, 0) p p[0]p[1] Point: (2, 2) p[2] vertex Point: (4, 4) Point: (1, 0) p p[0]p[1]p[2] vertex Point: (4, 4) Point[] p = new Point[3]; p[0] = new Point(0, 0); p[1] = new Point(1, 1); p[2] = new Point(2, 2); p[0].setX(1); p[1].setY(p[2].getY()); Point vertex = new Point(4,4); p[1] = p[0]; p[2] = vertex;

13. 13 New 2005 demotivatiors!

14. 14 End of lecture on 5 April 2006

15. 15 Explicit initialization  Syntax ElementType [] id = { exp 0, 1,... exp n }; id references an array of n elements. id[0] has value exp 0, id[1] has value exp 1, and so on. Each exp i is an expression that evaluates to type ElementType

16. 16 Explicit initialization  Example String[] puppy = { “pika”, “mila”, “arlo”, “nikki” }; int[] unit = { 1 };  Equivalent to String[] puppy = new String[4]; puppy[0] = “pika"; puppy[1] = “mila"; puppy[2] = “arlo"; puppy[3] = “nikki"; int[] unit = new int[1]; unit[0] = 1;

17. 17 Array members  Member length Size of the array for (int i = 0; i < puppy.length; ++i) { System.out.println(puppy[i]); }

18. 18 Array members  Member clone() Produces a shallow copy Point[] u = { new Point(0, 0), new Point(1, 1)}; Point[] v = u.clone(); v[1] = new Point(4, 30); Point: (0, 0)Point: (1, 1) u u[0]u[1] Point: (0, 0) v v[0]v[1] Point: (1, 1) u u[0]u[1] Point: (0, 0) v v[0]v[1] Point: (1, 1) u u[0]u[1] Point: (4, 30) Point[] u = { new Point(0, 0), new Point(1, 1)}; Point[] v = u.clone(); v[1] = new Point(4, 30);

19. 19  Member clone() Produces a shallow copy Point[] u = { new Point(0, 0), new Point(1, 1)}; Point[] v = u.clone(); v[1].setX(10); Point[] u = { new Point(0, 0), new Point(1, 1)}; Point[] v = u.clone(); v[1].setX(10); Array members Point: (0, 0)Point: (1, 1) u u[0]u[1] Point: (0, 0) v v[0]v[1] Point: (1, 1) u u[0]u[1] Point: (0, 0) v v[0]v[1] Point: (10, 1) u u[0]u[1]

20. 20 Making a deep copy  We want to copy the array and all the objects each element of the array references This is called a deep copy  Example Point[] w = new Point[u.length]; for (int i = 0; i < u.length; ++i) { w[i] = (Point) u[i].clone(); }

21. 21 Making a deep copy

22. 22 Review of arrays  Creating an array: int[] foo = new int[10];  Accessing an array: foo[3] = 7; System.out.print (foo[1]);  Creating an array: String[] bar = new String[10];  Accessing an array: bar[3] = “qux”; System.out.println (bar[1]);

23. 23 How Java represents arrays  Consider int[] a = { 1, 2, 3, 4, 5 }; a 1 2345 + … Array - length = 5 - data = 12345

24. 24 More about how Java represents Arrays  Consider int[] a; int[] b = null; int[] c = new int[5]; int[] d = { 1, 2, 3, 4, 5 }; a = c; d = c; 1 23450 0000 a - b null cd int[] a; int[] b = null; int[] c = new int[5]; int[] d = { 1, 2, 3, 4, 5 }; a = c; d = c;

25. 25 What do these pictures mean? Light beer Light beer Dandy lions Dandy lions Assaulted peanut Assaulted peanut Eggplant Eggplant Dr. Pepper Dr. Pepper Pool table Pool table Tap dancers Tap dancers Card shark Card shark King of pop King of pop I Pod I Pod Gator aide Gator aide Knight mare Knight mare Hole milk Hole milk

26. 26 ArrayTools

27. 27 ArrayTools.java  We want to create a series of general utility methods to be used for arrays  We will put these into an ArrayTools class

28. 28 ArrayTools.java – outline public class ArrayTools { // class constant private static final int MAX_LIST_SIZE = 1000; // sequentialSearch(): examine unsorted list for key public static int sequentialSearch(int[] data, int key) {... // putList (): prints list to screen public static void putList(int[] data) {... // getList(): extract and return up to MAX_LIST_SIZE values public static int[] getList() {... // reverse(): reverses the order of the element values public static void reverse(int[] list) {... // binarySearch(): examine sorted list for a key public static int binarySearch(char[] data, char key) {... }

29. 29 ArrayTools.java method putList()  To print the array: public static void putList(int[] data) { for (int i = 0; i < data.length; ++i) { System.out.println(data[i]); } }  Consider int[] score = { 6, 9, 82, 11, 29, 85, 11, 28, 91 }; putList(score);

30. 30 ArrayTools.java method getList() public static int[] getList() { Scanner stdin = new Scanner (System.in); int[] buffer = new int[MAX_LIST_SIZE]; int listSize = 0; for (int i = 0; (i < MAX_LIST_SIZE) && stdin.hasNext(); ++i) { buffer[i] = stdin.nextInt(); ++listSize; } int[] data = new int[listSize]; for (int i = 0; i < listSize; ++i) { data[i] = buffer[i]; } return data; }

31. 31 ArrayTools.java method reverse() public static void reverse(int[] data) { int[] clone = data.clone(); for ( int i = 0; i < clone.length; ++i ) { data[i] = clone[clone.length-1-i]; } }  Consider int[] foo = { 1, 2, 3, 4, 5 }; reverse (foo); putList (foo);

32. 32 Demo.java public class Demo { // main(): application entry point public static void main(String[] args) { System.out.println (""); System.out.println ("Enter list of integers:"); int[] numbers = ArrayTools.getList (); System.out.println (""); System.out.println ("Your list"); ArrayTools.putList (numbers); ArrayTools.reverse (numbers); System.out.println (""); System.out.println ("Your list in reverse"); ArrayTools.putList (numbers); System.out.println (); } }

34. 34 ArrayTools demo… ArrayDemo.java ArrayDemo.java

35. 35 End of lecture on 10 April 2006  Also went over HW J7 in detail

36. 36 … main (String args[])

37. 37 Consider that main() method again  public static void main (String args[])  How does one pass in a parameter to the main method? public class MainParameters { public static void main (String args[]) { System.out.println ("Number of paramters to “ + "main(): " + args.length); if ( args.length > 0 ) { for ( int i = 0; i < args.length; i++ ) System.out.println ("parameter " + i + ": '" + args[i] + "'"); } } }

38. 38 Program Demo MainParameters.java MainParameters.java Via JCreator Via JCreator Via the command line Via the command line

39. 39 Basic array searching

40. 40 System.out.println("Enter search value (number): "); int key = stdin.nextInt(); int i; for (i = 0; i < data.length; ++i) { if (key == data[i]) { break; } } if (i != data.length) { System.out.println(key + " is the " + i + "-th element"); } else { System.out.println(key + " is not in the list"); } ++i System.out.println("Enter search value (number): "); int key = stdin.nextInt(); int i; if (key == data[i]) { break; if (i != data.length) { System.out.println(key + " is the " + i + "-th element"); } i < data.length i = 0 Searching for a value

41. 41 Searching for the minimum value  Segment int minimumSoFar = sample[0]; for (int i = 1; i < sample.length; ++i) { if (sample[i] < minimumSoFar) { minimumSoFar = sample[i]; } }

42. 42 ArrayTools.java method sequentialSearch() public static int sequentialSearch(int[] data, int key) { for (int i = 0; i < data.length; ++i) { if (data[i] == key) { return i; } } return -1; }  Consider int[] score = { 6, 9, 82, 11, 29, 85, 11, 28, 91 }; int i1 = sequentialSearch(score, 11); int i2 = sequentialSearch(score, 30);

43. 43 Today’s demotivators

44. 44 Sorting

45. 45 Sorting  Problem Arranging elements so that they are ordered according to some desired scheme  Standard is non-decreasing order Why don't we say increasing order?  Major tasks Comparisons of elements Updates or element movement

46. 46 Selection sorting  Algorithm basis On iteration i, a selection sorting method:  Finds the element containing the i th smallest value of its list v and exchanges that element with v[i]  Example – iteration 0 Swaps smallest element with v[0] This results in smallest element being in the correct place for a sorted result v‘E'‘Q''W''Y''R''T''I''U''P''O' 2015347698 v‘E'‘Q''W''Y''R''T''I''U''P''O' 2015347698 v'Q''E''W''Y''R''T''I''U''P''O' 2015347698

47. 47 Selection sorting  Algorithm basis On iteration i, a selection sorting method:  Finds the element containing the i th smallest value of its list v and exchanges that element with v[i]  Example – iteration 1 Swaps second smallest element with v[1] This results in second smallest element being in the correct place for a sorted result v'Q''E''W''Y''R''T''I''U''P''O' 2015347698 v'Q''E''I''Y''R''T''W''U''P''O' 2015347698

48. 48 v'Q''E''I''Y''R''T''W''U''P''O' 2015347698 Selection sorting  Algorithm basis On iteration i, a selection sorting method:  Finds the element containing the i th smallest value of its list v and exchanges that element with v[i]  Example – iteration 2 Swaps third smallest element with v[2] This results in third smallest element being in the correct place for a sorted result v‘O''E''I''Y''R''T''W''U''P'‘Q' 2015347698

49. 49 Selection sorting  Algorithm basis On iteration i, a selection sorting method:  Finds the element containing the i th smallest value of its list v and exchanges that element with v[i]  Example – iteration 3 Swaps fourth smallest element with v[3] This results in fourth smallest element being in the correct place for a sorted result v‘O''E''I''Y''R''T''W''U''P'‘Q' 2015347698 v‘O''E''I''Y'‘P''T''W''U'‘R'‘Q' 2015347698

50. 50 Selection sorting  Algorithm basis On iteration i, a selection sorting method:  Finds the element containing the i th smallest value of its list v and exchanges that element with v[i]  Example – iteration 4 Swaps fifth smallest element with v[4] This results in fifth smallest element being in the correct place for a sorted result v‘O''E''I''Y'‘P''T''W''U'‘R'‘Q' 2015347698 v‘O''E''I''Y'‘P'‘Q''W''U'‘R'‘T' 2015347698

51. 51 ArrayTools.java selection sorting public static void selectionSort(int[] v) { for (int i = 0; i < v.length-1; ++i) { // find the location of the ith smallest element int spot = i; for (int j = i+1; j < v.length; ++j) { if (v[j] < v[spot]) { // is current location ok? // update spot to index of smaller element spot = j; } } // spot is now correct, so swap elements int rmbr = v[i]; v[i] = v[spot]; v[spot] = rmbr; } }

52. 52 Iteration i // find the location of the ith smallest element int spot = i; for (int j = i+1; j < v.length; ++j) { if (v[j] < v[spot]) // is spot ok? // update spot with index of smaller element spot = j; } // spot is now correct, swap elements v[spot] and v[i]

53. 53 From Dubai

54. 54 Binary search

55. 55 Binary search  Given a list, find a specific element in the list List MUST be sorted!  Each time it iterates through, it cuts the search space in half  A binary search is MUCH faster than a sequential search

56. 56 Binary search use  The ‘BS’ in BSDemo is for Binary Search, mind you public class BSDemo { public static void main(String[] args) { int[] numbers = { 9, 3, 1, 8, 4, 6, 10, 2 }; System.out.println ("The original list of numbers:"); ArrayTools.putList(numbers); System.out.println(); ArrayTools.selectionSort(numbers); System.out.println ("The sorted list of numbers:"); ArrayTools.putList(numbers); System.out.println(); System.out.println ("Searching for 0: " + ArrayTools.binarySearch(numbers, 0)); System.out.println ("Searching for 1: " + ArrayTools.binarySearch(numbers, 1)); System.out.println ("Searching for 4: " + ArrayTools.binarySearch(numbers, 4)); System.out.println ("Searching for 5: " + ArrayTools.binarySearch(numbers, 5)); System.out.println ("Searching for 6: " + ArrayTools.binarySearch(numbers, 6)); System.out.println ("Searching for 10: " + ArrayTools.binarySearch(numbers, 10)); System.out.println ("Searching for 11: " + ArrayTools.binarySearch(numbers, 11)); }

57. 57 Binary search use demo… BSDemo.java BSDemo.java

58. 58 Chapter 2: Computer bugs

59. 59 Binary search public static int binarySearch (int[] data, int key) { int i = 0; // left endpoint of search interval int j = data.length-1; // right endpoint of search interval while ( i < j ) { int m = (i+j)/2; if ( key > data[m] ) { i = m+1; } else { j = m; } } if ( key == data[i] ) { return i; } else { return -1; } }

60. 60 if ( key == data[i] ) { return i; } else { return -1; } if ( key == data[i] ) { return i; } else { return -1; } int i = 0; int j = data.length-1; int i = 0; int j = data.length-1; while ( i < j ) { int m = (i+j)/2; if ( key > data[m] ) { i = m+1; } else { j = m; } while ( i < j ) { int m = (i+j)/2; if ( key > data[m] ) { i = m+1; } else { j = m; } } Binary search, take 1 2468101214161820 a0a0 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 a7a7 a8a8 a9a9 ijm public static int binarySearch (int[] data, int key) { 0 key 14 945776656 returns: 6 data

61. 61 Binary search  But what if the element is not in the list?

62. 62 if ( key == data[i] ) { return i; } else { return -1; } if ( key == data[i] ) { return i; } else { return -1; } int i = 0; int j = data.length-1; int i = 0; int j = data.length-1; while ( i < j ) { int m = (i+j)/2; if ( key > data[m] ) { i = m+1; } else { j = m; } while ( i < j ) { int m = (i+j)/2; if ( key > data[m] ) { i = m+1; } else { j = m; } } Binary search, take 2 2468101214161820 a0a0 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 a7a7 a8a8 a9a9 ijm public static int binarySearch (int[] data, int key) { 0 key 15 945776 returns: data 7

63. 63 Binary search  A somewhat alternative view of what a binary search does…

64. 64 How long does a binary search take?  Given a array of 64 elements 1 st iteration cuts the array to 32 2 nd iteration cuts the array to 16 3 rd to 8 4 th to 4 5 th to 2 6 th to 1  Given a array of 1024 elements 1 st iteration cuts the array to 512... 10 th iteration cuts the list to 1 element  Thus, the binary search takes log 2 n iterations! Where n is the size of the array

65. 65 Binary search vs. sequential search  Assume the array has n elements  Sequential search takes n iterations to find the element  Binary search takes log 2 n iterations to find the element  Consider a list of 1 million elements Binary search takes about 20 iterations Sequential search takes 1,000,000 iterations  Consider a list of 1 trillion elements Binary search takes about 40 iterations Sequential search takes 1,000,000,000,000 iterations

66. 66 End of lecture on 12 April 2006

67. 67 Multi-dimensional arrays

68. 68 Multidimensional arrays  Many problems require information be organized as a two- dimensional or multidimensional list  Examples Matrices Graphical animation Economic forecast models Map representation Time studies of population change Microprocessor design

69. 69 Example  Segment int[][] m = new int[3][]; m[0] = new int[4]; m[1] = new int[4]; m[2] = new int[4];  Produces When an array is created, each value is initialized! m m[0]m[1]m[2] 00000000 0000 m[2][0]m[2][1]m[2][2]m[2][3] m[0][0]m[0][1]m[0][2]m[0][3]m[1][0]m[1][1]m[1][2]m[1][3] m

70. 70 Example  Alternative int[][] m = new int[3][4];  Produces m m[0]m[1]m[2] 00000000 0000 m[2][0]m[2][1]m[2][2]m[2][3] m[0][0]m[0][1]m[0][2]m[0][3]m[1][0]m[1][1]m[1][2]m[1][3]

71. 71 Multidimensional array visualization  A multi-dimensional array declaration (either one): int[][] m = new int[3][4];  How we visualize it: 000 000 000 000 0 0 0 0 0 0 0 0 0 0 0 0 or

72. 72 Example  Segment for (int c = 0; c < m.length; ++c) { for (int r = 0; r < m[c].length; ++r) { System.out.print("Enter a value: "); m[c][r] = stdin.nextInt(); } 0 0 0 0 0 0 0 0 0 0 0 0

73. 73 Rows by columns or columns by rows?  Consider int[][] m = new int[3][4];  Is that 3 rows by 4 columns or 3 columns by 4 rows?  The answer is that it can be either As long as you are consistent with your column/row placement 000 000 000 000 or 000 000 0000 0 0

74. 74 Rows by columns or columns by rows?  This makes it 3 columns by 4 rows: for (int c = 0; c < m.length; ++c) for (int r = 0; r < m[c].length; ++r) { System.out.print("Enter a value: "); m[c][r] = stdin.nextInt(); }  This makes it 3 rows by 4 columns: for (int c = 0; c < m.length; ++c) for (int r = 0; r < m[c].length; ++r) { System.out.print("Enter a value: "); m[r][c] = stdin.nextInt(); }

75. 75 Example  Segment String[][] s = new String[4][]; s[0] = new String[2]; s[1] = new String[2]; s[2] = new String[4]; s[3] = new String[3];  Produces

76. 76 Multidimensional array visualization  Segment String[][] s = new String[4][]; s[0] = new String[2]; s[1] = new String[2]; s[2] = new String[4]; s[3] = new String[3];  Produces  Called a “ragged” array 000 000 0 0 0 0 0 0 0 0 0 0 0 0 0 or 0 0 0

77. 77 Explicit Initialization  Segment int c[][] = {{1, 2}, {3, 4}, {5, 6}, {7, 8, 9}};  Produces

78. 78 Matrices  A two-dimensional array is sometimes known as a matrix because it resembles that mathematical concept  A matrix a with m rows and n columns is represented mathematically in the following manner

79. 79 Matrix addition  Definition C = A + B c ij = a ij + b ij c ij is sum of the elements in the same row and column of A and B

80. 80 Matrix addition public static double[][] add(double[][] a, double[][] b) { // determine number of rows in solution int m = a.length; // determine number of columns in solution int n = a[0].length; // create the array to hold the sum double[][] c = new double[m][n]; // compute the matrix sum row by row for (int i = 0; i < m; ++i) { // produce the current row for (int j = 0; j < n; ++j) { c[i][j] = a[i][j] + b[i][j]; } } return c; }

81. 81 Homework J8  You will be creating a Universe class The Universe class contains a 2-D array In each spot will be a Location object  (from a previous HW)  Lab 10 is going to be a Map class Will print out the 2-D Universe via text

82. 82 Alien Song AlienSong.mpg AlienSong.mpg

83. 83 Vector class This is also the review for the third midterm

84. 84 Limitations of arrays  You can’t change their size once created This can be a big problem!  So we will create a new class that will operate like an array: We can store and get elements by index number It will automatically increase in size as needed And other fancy features…  Let’s call the class Vector As we are basically writing the java.util.Vector class

85. 85 Properties of our Vector class  It needs to have an array to hold the values  As our internal array will often be bigger than the number of elements in the Vector, we need a size as well More on what this means in a slide or two…  Not much else…

86. 86 Methods in our Vector class  Insert and remove elements into the Vector  Get an element from the Vector  Find the length  Print it out to the screen  What happens when the array field is full, and we want to add an element? We will need to increase the size of the array So we need a method to do that as well

87. 87 Our first take on our Vector class public class Vector { private Object array[]; private int size = 0; Vector() { array = new Object[100]; } Vector(int length) { array = new Object[length]; } }  What does this mean? We’ll see that a bit later… But briefly, it means the array can store any object

88. 88 Adding an element to our Vector public void add (Object o) { array[size++] = o; }  Pretty easy!  But what if the array is full?  We need a way to increase the capacity of the array

89. 89 Increasing the Vector’s array’s capacity private void increaseCapacity() { int oldSize = array.length; Object newArray[] = new Object[2*oldSize]; for ( int i = 0; i < oldSize; i++ ) newArray[i] = array[i]; array = newArray; }  And our new add() method: public void add (Object o) { if ( size == array.length ) increaseCapacity(); array[size++] = o; }

90. 90 Methods can be private as well  Notice that the increaseCapacity() method is called only by the add() method when necessary  It’s not ever going to be called by whomever is using our Vector  Thus, we will make it private  That means that only other Vector methods can call it

91. 91 Removing an element from a Vector public Object remove (int which) { Object ret = array[which]; for ( int i = which; i < array.length-1; i++ ) array[i] = array[i+1]; array[array.length-1] = null; size--; return ret; }

92. 92 Miscellaneous other methods public int size() { return size; } public Object get (int which) { return array[which]; }

93. 93 Our toString() method public String toString() { String ret = "["; for ( int i = 0; i < size; i++ ) { ret += array[i]; if ( i != size-1 ) ret += ", "; } ret += "]"; return ret; }

94. 94 Using our Vector  This code is in a separate class called VectorUsage public static void main (String[] args) { Vector v = new Vector(); for ( int i = 12; i < 30; i++ ) { v.add (String.valueOf(i)); } System.out.println (v); System.out.println (v.size()); String s = (String) v.get(5); System.out.println (s); v.remove (5); System.out.println (v); v.remove (5); System.out.println (v); }

95. 95 Program Demo VectorUsage.java VectorUsage.java

96. 96 The “real” Vector class  Java provides a Vector class In java.util  It contains all of the methods shown

97. 97 Program Demo VectorUsage.java VectorUsage.java But using java.util.Vector But using java.util.Vector

98. 98 What about those errors?  When compiled with java.util.Vector, we see: Note: C:\...\VectorUsage.java uses unchecked or unsafe operations. Note: Recompile with -Xlint:unchecked for details.  You can ignore these They deal with generics, which you will see in future courses The program was still compiled

99. 99 More on using the Vector class  To add a Location object l to the end of a Vector v v.add(l);  To get the Location object at the end of the Vector v Location l = (Location) v.get(v.size()-1);  To remove a Location object from the end of a Vector v Location l = (Location) v.remove(v.size()-1); This both removes the object from the Vector and stores the removed value into l

100. 100 Very unofficial demotivators

101. 101 End of lecture on 17 Apr 2006

102. 102 Wrapper classes

103. 103 But what about adding variables?  The add method takes an Object as a parameter public void add (Object o) { Although we haven’t seen it yet, this means you can add any object you want to the vector  Primitive types (i.e. variables) are not objects How can they be added?  The solution: wrapper classes!

104. 104 The Integer wrapper class  This is how you add an int variable to a Vector: int x = 5; Integer i = new Integer(x); vector.add (i); //… Integer j = (Integer) v.get(0); int y = j.intValue();  Pretty annoying syntax – we’ll see how to get around it in a bit…

105. 105 More on wrapper classes  All the primitive types have wrapper classes Usually, the names are just the capitalized version of the type  I.e. Double for double, Byte for byte, etc. Two exceptions: int and char  int has Integer  char has Character

106. 106 More on wrapper classes  Consider this code: int x = 5; vector.add (x); //… int y = vector.get(0);  Does this code work? It shouldn’t  As we are adding a variable (not an object) to a vector But it does work!  Why?

107. 107 Auto-boxing  Java 1.5 will automatically “wrap” a primitive value into it’s wrapper class when needed And automatically “unwrap” a wrapper object into the primitive value  So Java translates the previous code into the following: int x = 5; vector.add (new Integer(x)); //… int y = ((Integer)vector.get(0)).intValue();  This is called autoboxing And auto-unboxing (unauto-boxing?) This does not work in Java 1.4 or before

108. 108 More on auto-boxing  Consider the following code: Double d = 7.5; Double e = 6.5; Double f = d + e; System.println (f);  This is doing a lot of auto-boxing (and auto-unboxing): Double d = new Double(7.5); Double e = new Double(6.5); Double f = newDouble(d.doubleValue() + e.doubleValue()); System.println (f);

109. 109 A solution to commenting your code The commentator: http://www.cenqua.com/commentator/ The commentator: http://www.cenqua.com/commentator/ http://www.cenqua.com/commentator/