Which code snippet finds the largest value in an array that is only partially full?

Chapter 7 – Arrays and Array Lists

Chapter Goals

• To collect elements using arrays and array lists
• To use the enhanced for loop for traversing arrays and array lists
• To learn common algorithms for processing arrays and array lists
• To work with two-dimensional arrays
• To understand the concept of regression testing

Arrays

• An array collects a sequence of values of the same type.
• Create an array that can hold ten values of type double:

  new double[10]

  • The number of elements is the length of the array
  • The new operator constructs the array
• The type of an array variable is the type of the element to be stored, followed by [].
• To declare an array variable of type double[]

  double[] values; 
   

• To initialize the array variable with the array:

  double[] values = new double[10]; 
  

• By default, each number in the array is 0
• You can specify the initial values when you create the array

  double[] moreValues = { 32, 54, 67.5, 29, 35, 80, 115, 44.5, 100, 65 };

Arrays

• To access a value in an array, specify which “slot” you want to use
  • use the [] operator

  values[4] = 35; 
  

• The “slot number” is called an index.
• Each slot contains an element.
• Individual elements are accessed by an integer index i, using the notation array[i].
• An array element can be used like any variable.

  System.out.println(values[4]);

Arrays

Figure 1 An Array of Size 10

Syntax 7.1 Arrays

Arrays

• The elements of arrays are numbered starting at 0.
• The following declaration creates an array of 10 elements:

  double[] values = new double[10];

• An index can be any integer ranging from 0 to 9.
• The first element is values[0]
• The last element is values[9]
• An array index must be at least zero and less than the size of the array.
• Like a mailbox that is identified by a box number, an array element is identified by an index.
  
  

Arrays - Bounds Error

• A bounds error occurs if you supply an invalid array index.
• Causes your program to terminate with a run-time error.
• Example:

  double[] values = new double[10];
  values[10] = value; // Error

• values.length yields the length of the values array.
• There are no parentheses following length.

Declaring Arrays

Array References

• An array reference specifies the location of an array.
• Copying the reference yields a second reference to the same array.
• When you copy an array variable into another, both variables refer to the same array

  int[] scores = { 10, 9, 7, 4, 5 };
  int[] values = scores; // Copying array reference

• You can modify the array through either of the variables:

  scores[3] = 10;
  System.out.println(values[3]); // Prints 10



Figure 2 Two Array Variables Referencing the Same Array

Using Arrays with Methods

• Arrays can occur as method arguments and return values.
• An array as a method argument

  public void addScores(int[] values)
  {
     for (int i = 0; i < values.length; i++)
     {
        totalScore = totalScore + values[i];
     }
  }

• To call this method

  int[] scores = { 10, 9, 7, 10 };
  fred.addScores(scores);

• A method with an array return value

  public int[] getScores()
      

Partially Filled Arrays

• Array length = maximum number of elements in array.
• Usually, array is partially filled
• Define an array larger than you will need
  

  final int LENGTH = 100;
  double[] values = new double[LENGTH];

• Use companion variable to keep track of current size: call it currentSize

Partially Filled Arrays

• A loop to fill the array
  

  int currentSize = 0;
  Scanner in = new Scanner(System.in);
  while (in.hasNextDouble())
  {
     if (currentSize < values.length)
     {
        values[currentSize] = in.nextDouble();
        currentSize++;
     }
  }

• At the end of the loop, currentSize contains the actual number of elements in the array.
• Note: Stop accepting inputs when currentSize reaches the array length.

Partially Filled Arrays

Partially Filled Arrays

• To process the gathered array elements, use the companion variable, not the array length:

  for (int i = 0; i < currentSize; i++) 
  { 
     System.out.println(values[i]); 
  }

• With a partially filled array, you need to remember how many elements are filled.
  
  
  
  

Self Check 7.1

Declare an array of integers containing the first five prime numbers

• Answer:

  int[] primes = { 2, 3, 5, 7, 11 };

Self Check 7.2

Assume the array

has been initialized as described in Self Check 1. What does it contain after executing the following loop?

for (int i = 0; i < 2; i++)
{
    primes[4 - i] = primes[i];
}

• Answer: 2, 3, 5, 3, 2

Self Check 7.3

Assume the array

has been initialized as described in Self Check 1. What does it contain after executing the following loop?

for (int i = 0; i < 5; i++)
{
   primes[i]++;
}

• Answer: 3, 4, 6, 8, 12

Self Check 7.4

Given the declaration

int[] values = new int[10];

write statements to put the integer 10 into the elements of the array values with the lowest and the highest valid index.

• Answer:

  values[0] = 10;
  values[9] = 10;
       or better:
  values[values.length - 1] = 10;

Self Check 7.5

Declare an array called

that can hold ten elements of type

.

• Answer: String[] words = new String[10];

Self Check 7.6

Declare an array containing two strings,

, and

.

• Answer: String[] words = { "Yes", "No" };

Self Check 7.7

Can you produce the output on page 312 without storing the inputs in an array, by using an algorithm similar to the algorithm for finding the maximum in Section 6.7.5?

• Answer: No. Because you don’t store the values, you need to print them when you read them. But you don’t know where to add the <= until you have seen all values.

Self Check 7.8

Declare a method of a class

that returns a combination of

numbers. You don’t need to implement the method.

• Answer:

  public class Lottery
  {
     public int[] getCombination(int n) { . . . }
     . . .
  }

Make Parallel Arrays into Arrays of Objects

• Don't do this

  int[] accountNumbers;
  double[] balances;

• Don't use parallel arrays
  
  

Figure 4

Make Parallel Arrays into Arrays of Objects

Avoid parallel arrays by changing them into arrays of objects:

BankAccount[] accounts;

Figure 5

The Enhanced for Loop

• You can use the enhanced for loop to visit all elements of an array.
• Totaling the elements in an array with the enhanced for loop

  double[] values = . . .;
  double total = 0;
  for (double element : values)
  {
     total = total + element;
  }

• The loop body is executed for each element in the array values.
• Read the loop as for each element in values.
• Traditional alternative:

  for (int i = 0; i < values.length; i++)
  {
     double element = values[i];
     total = total + element;
  }

The Enhanced for Loop

• Not suitable for all array algorithms.
• Does not allow you to modify the contents of an array.
• The following loop does not fill an array with zeros:

  for (double element : values)
  {
     element = 0; // ERROR: this assignment does not modify array elements
  }

• Use a basic for loop instead:

  for (int i = 0; i < values.length; i++)
  {
     values[i] = 0; // OK
  }

• Use the enhanced for loop if you do not need the index values in the loop body.
• The enhanced for loop is a convenient mechanism for traversing all elements in a collection.
  
  

Syntax 7.2 The Enhanced for Loop

Self Check 7.9

What does this enhanced

loop do?

int counter = 0;
for (double element : values)
{
   if (element == 0) { counter++; }
}

• Answer: It counts how many elements of values are zero.

Self Check 7.10

Write an enhanced

loop that prints all elements in the array

.

• Answer:

  for (double x : values)
  {
     System.out.println(x);
  }

Self Check 7.11

Write an enhanced

loop that multiplies all elements in a

array named

, accumulating the result in a variable named

.

• Answer:

  double product = 1;
  for (double f : factors)
  {
     product = product * f;
  }

Self Check 7.12

Why is the enhanced

loop not an appropriate shortcut for the following basic

loop?

for (int i = 0; i < values.length; i++) { values[i] = i * i; }

• Answer: The loop writes a value into values[i]. The enhanced for loop does not have the index variable i.

Common Array Algorithm: Filling

• Fill an array with squares (0, 1, 4, 9, 16, ...):

  for (int i = 0; i < values.length; i++) 
  { 
     values[i] = i * i; 
  }

Common Array Algorithm: Sum and Average

• To compute the sum of all elements in an array:

  double total = 0; 
  for (double element : values) 
  { 
     total = total + element; 
  }

• To obtain the average:

  double average = 0;
  if (values.length > 0) { average = total / values.length; }

Common Array Algorithm: Maximum or Minimum

• Finding the maximum in an array

  double largest = values[0];
  for (int i = 1; i < values.length; i++)
  {
     if (values[i] > largest)
     {
        largest = values[i];
     }
  } 
          

  • The loop starts at 1 because we initialize largest with values[0].
    
• Finding the minimum: reverse the comparison.
• These algorithms require that the array contain at least one element.

Common Array Algorithm: Element Separators

• When you display the elements of an array, you usually want to separate them:

  Ann | Bob | Cindy

• Note that there is one fewer separator than there are elements
• Print the separator before each element except the initial one (with index 0):

  for (int i = 0; i < names.size(); i++) 
  { 
     if (i > 0) 
     { 
        System.out.print(" | "); 
     } 
     System.out.print(names.value[i]); 
  }

• To print five elements, you need four separators.
  
  

Common Array Algorithm: Linear Search

• To find the position of an element:
  • Visit all elements until you have found a match or you have come to the end of the array
• Example: Find the first element that is equal to 100

  int searchedValue = 100;
  int pos = 0;
  boolean found = false;
  while (pos < values.length && !found)
  {
     if (values[pos] == searchedValue)
     {
        found = true;
     }
     else
     {
        pos++;
     }
  }
  if (found) { System.out.println("Found at position: " + pos); }
  else { System.out.println("Not found"); }

Common Array Algorithm: Linear Search

• This algorithm is called a linear search.
• A linear search inspects elements in sequence until a match is found.
• To search for a specific element, visit the elements and stop when you encounter the match.
  
  

Common Array Algorithm: Removing an Element

Problem: To remove the element with index

from the array

with number of elements

.

• Unordered
  1. Overwrite the element to be removed with the last element of the array.
  2. Decrement the currentSize variable.

  values[pos] = values[currentSize - 1];
  currentSize--;



Figure 6 Removing an Element in an Unordered Array

Common Array Algorithm: Removing an Element

• Ordered array
  1. Move all elements following the element to be removed to a lower index.
  2. Decrement the variable holding the size of the array .

  for (int i = pos + 1; i < currentSize; i++)
  {
     values[i - 1] = values[i];
  }
  currentSize--;



Figure 7 Removing an Element in an Ordered Array

Common Array Algorithm: Inserting an Element

• If order does not matter
  1. Insert the new element at the end of the array.
  2. Increment the variable tracking the size of the array.

  if (currentSize < values.length) 
  {
     currentSize++ 
     values[currentSize -1 ] = newElement; ; 
  }



Figure 8 Inserting an Element in an Unordered Array

Common Array Algorithm: Inserting an Element

• If order matters
  1. Increment the variable tracking the size of the array.
  2. Move all elements after the insertion location to a higher index.
  3. Insert the element .

  if (currentSize < values.length)
  {
     currentSize++;
     for (int i = currentSize - 1; i > pos; i--)
     {
        values[i] = values[i - 1];
     }
     values[pos] = newElement;
  }



Figure 9 Inserting an Element in an Ordered Array

Common Array Algorithm: Swapping Elements

• To swap two elements, you need a temporary variable.
  
  
• We need to save the first value in the temporary variable before replacing it.
  

  double temp = values[i];
  values[i] = values[j];

• Now we can set values[j] to the saved value.

  values[j] = temp;

Common Array Algorithm: Swapping Elements

Figure 10 Swapping Array Elements

Common Array Algorithm: Copying an Array

Common Array Algorithm: Growing an Array

Reading Input

• To read a sequence of arbitrary length:
  • Add the inputs to an array until the end of the input has been reached.
  • Grow when needed.

  double[] inputs = new double[INITIAL_SIZE];
  int currentSize = 0;
  while (in.hasNextDouble())
  {
     // Grow the array if it has been completely filled
     if (currentSize >= inputs.length)
     {
        inputs = Arrays.copyOf(inputs, 2 * inputs.length); // Grow the inputs array
     }
     inputs[currentSize] = in.nextDouble();
     currentSize++;
  }

  • Discard unfilled elements.

  inputs = Arrays.copyOf(inputs, currentSize);

section_3/LargestInArray.java

Program Run:

• Please enter values, Q to quit:
  34.5 80 115 44.5 Q
  34.5
  80
  115 <== largest value
  44.5
  
  

Self Check 7.13

Given these inputs, what is the output of the

program?

• Answer:

  20 <== largest value
  10
  20 <== largest value

Self Check 7.14

Write a loop that counts how many elements in an array are equal to zero.

• Answer:

  int count = 0;
  for (double x : values)
  {
     if (x == 0) { count++; }
  }

Self Check 7.15

Consider the algorithm to find the

double largest = 0;
for (int i = 0; i < values.length; i++)
{
   if (values[i] > largest)
   {
      largest = values[i];
   }
}

• Answer: If all elements of values are negative, then the result is incorrectly computed as 0.

Self Check 7.16

When printing separators, we skipped the separator before the initial element. Rewrite the loop so that the separator is printed after each element, except for the last element.

• Answer:

  for (int i = 0; i < values.length; i++)
  {
     System.out.print(values[i]);
     if (i < values.length - 1)
     {
        System.out.print(" | ");
     }
  }

  Now you know why we set up the loop the other way.

Self Check 7.17

What is wrong with these statements for printing an array with separators?

System.out.print(values[0]);
for (int i = 1; i < values.length; i++)
{
   System.out.print(", " + values[i]);
}

• Answer: If the array has no elements, then the program terminates with an exception.

Self Check 7.18

When finding the position of a match, we used a

loop, not a

loop. What is wrong with using this loop instead?

for (pos = 0; pos < values.length && !found; pos++)
{
   if (values[pos] > 100)
   {
      found = true;
   }
}

• Answer: If there is a match, then pos is incremented before the loop exits.

Self Check 7.19

When inserting an element into an array, we moved the elements with larger index values, starting at the end of the array. Why is it wrong to start at the insertion location, like this?

for (int i = pos; i < currentSize - 1; i++)
{
   values[i + 1] = values[i];
}

• Answer: This loop sets all elements to values[pos].

Problem Solving: Adapting Algorithms

• By combining fundamental algorithms, you can solve complex programming tasks.
• Problem: Compute the final quiz score by dropping the lowest and finding the sum of all the remaining scores.
• Related algorithms:
  • Calculating the sum
  • Finding the minimum value
  • Removing an element
• A plan of attack

  Find the minimum.
  Remove it from the array.
  Calculate the sum.

• Try it out. The minimum is 4
  
  
• We need to use a linear search to find the position of the minimum.

Problem Solving: Adapting Algorithms - continued

• Revise the plan of attack

  Find the minimum value.
  Find its position.
  Remove that position from the array.
  Calculate the sum.

• Try it out
• Compute the sum: 8 + 7 + 8.5 + 9.5 + 7 + 10 = 50.
• This walk-through demonstrates that our strategy works.

Problem Solving: Adapting Algorithms - continued

• Inefficient to find the minimum and then make another pass through the array to obtain its position.
• Modify minimum algorithm to remember the position of the smallest element

  int smallestPosition = 0;
  for (int i = 1; i < values.length; i++)
  {
     if (values[i] < values[smallestPosition])
     {
        smallestPosition = i;
     }
  }

• Final Strategy

  Find the position of the minimum.
  Remove it from the array.
  Calculate the sum.

Self Check 7.20

Section 7.3.6 has two algorithms for removing an element. Which of the two should be used to solve the task described in this section?

• Answer: Use the first algorithm. The order of elements does not matter when computing the sum.

Self Check 7.21

It isn’t actually necessary to remove the minimum in order to compute the total score. Describe an alternative.

• Answer:

  Find the minimum value.
  Calculate the sum.
  Subtract the minimum value.

Self Check 7.22

How can you print the number of positive and negative values in a given array, using one or more of the algorithms in Section 6.7?

• Answer: Use the algorithm for counting matches (Section 6.7.2) twice, once for counting the positive values and once for counting the negative values.

Self Check 7.23

How can you print all positive values in an array, separated by commas?

• Answer: You need to modify the algorithm in Section 7.3.4.

  boolean first = true;
  for (int i = 0; i < values.length; i++)
  {
     if (values[i] > 0))
     {
        if (first) { first = false; }
        else { System.out.print(", "); }
     }
     System.out.print(values[i]);
  }

  Note that you can no longer use i > 0 as the criterion for printing a separator.

Self Check 7.24

Consider the following algorithm for collecting all matches in an array: int matchesSize = 0;

for (int i = 0; i < values.length; i++)
{
   if (values[i] fulfills the condition)
   {
      matches[matchesSize] = values[i];
      matchesSize++;
   }
}

How can this algorithm help you with Self Check 23?

• Answer: Use the algorithm to collect all positive elements in an array, then use the algorithm in Section 7.3.4 to print the array of matches.

Problem Solving: Discovering Algorithms by Manipulating Physical Objects

• Manipulating physical objects can give you ideas for discovering algorithms.
  
  
• The Problem: You are given an array whose size is an even number, and you are to switch the first and the second half.
• Example

Problem Solving: Discovering Algorithms by Manipulating Physical Objects

Problem Solving: Discovering Algorithms by Manipulating Physical Objects

Problem Solving: Discovering Algorithms by Manipulating Physical Objects

• Two more swaps
  
  
• The pseudocode

  i = 0
  j = size / 2
  While (i < size / 2)
     Swap elements at positions i and j
     i++
     j++

Self Check 7.25

Walk through the algorithm that we developed in this section, using two paper clips to indicate the positions for i and j. Explain why there are no bounds errors in the pseudocode.

• Answer: The paperclip for i assumes positions 0, 1, 2, 3. When i is incremented to 4, the condition i < size / 2 becomes false, and the loop ends. Similarly, the paperclip for j assumes positions 4, 5, 6, 7, which are the valid positions for the second half of the array.
  

Self Check 7.26

Take out some coins and simulate the following pseudocode, using two paper clips to indicate the positions for i and j.

i = 0
j = size - 1
While (i < j)
   Swap elements at positions i and j
   i++
   j--

What does the algorithm do?

• Answer: It reverses the elements in the array.

Self Check 7.27

Consider the task of rearranging all elements in an array so that the even numbers come first. Otherwise, the order doesn’t matter. For example, the array

1 4 14 2 1 3 5 6 23

could be rearranged to

4 2 14 6 1 5 3 23 1

Using coins and paperclips, discover an algorithm that solves this task by swapping elements, then describe it in pseudocode.

• Answer: Here is one solution. The basic idea is to move all odd elements to the end. Put one paper clip at the beginning of the array and one at the end. If the element at the first paper clip is odd, swap it with the one at the other paper clip and move that paper clip to the left. Otherwise, move the first paper clip to the right. Stop when the two paper clips meet. Here is the pseudocode:

  i = 0
  j = size - 1
  While (i < j)
     If (a[i] is odd)
        Swap elements at positions i and j.
        j--
     Else
        i++

Self Check 7.28

Discover an algorithm for the task of Self Check 27 that uses removal and insertion of elements instead of swapping.

• Answer: Here is one solution. The idea is to remove all odd elements and move them to the end. The trick is to know when to stop. Nothing is gained by moving odd elements into the area that already contains moved elements, so we want to mark that area with another paper clip.

  i = 0
  moved = size
  While (i < moved)
     If (a[i] is odd)
        Remove the element at position i and add it at the end.
     moved--

Self Check 7.29

Consider the algorithm in Section 6.7.5 that finds the largest element in a sequence of inputs—not the largest element in an array. Why is this algorithm better visualized by picking playing cards from a deck rather than arranging toy soldiers in a sequence?

• Answer: When you read inputs, you get to see values one at a time, and you can’t peek ahead. Picking cards one at a time from a deck of cards simulates this process better than looking at a sequence of items, all of which are revealed.

Two-Dimensional Arrays

• An arrangement consisting of rows and columns of values
  • Also called a matrix.
• Example: medal counts of the figure skating competitions at the 2010 Winter Olympics.


Figure 13 Figure Skating Medal counts

• Use a two-dimensional array to store tabular data.
• When constructing a two-dimensional array, specify how many rows and columns are needed:

  final int COUNTRIES = 7;
  final int MEDALS = 3;
  int[][] counts = new int[COUNTRIES][MEDALS];

Two-Dimensional Arrays

• You can declare and initialize the array by grouping each row:

  int[][] counts =
  {
     { 1, 0, 1 },
     { 1, 1, 0 },
     { 0, 0, 1 },
     { 1, 0, 0 },
     { 0, 1, 1 },
     { 0, 1, 1 },
     { 1, 1, 0 }
  };

• You cannot change the size of a two-dimensional array once it has been declared.

Syntax 7.3 Two-Dimensional Array Declaration

Accessing Elements

• Access by using two index values, array[i][j]

  int medalCount = counts[3][1];

• Use nested loops to access all elements in a two-dimensional array.
• Example: print all the elements of the counts array
  

  for (int i = 0; i < COUNTRIES; i++)
  {
     // Process the ith row
     for (int j = 0; j < MEDALS; j++)
     {
        // Process the jth column in the ith row
        System.out.printf("%8d", counts[i][j]);
     }
     System.out.println(); // Start a new line at the end of the row
  }



Figure 14 Accessing an Element in a Two-Dimensional Array

Accessing Elements

• Number of rows: counts.length
• Number of columns: counts[0].length
• Example: print all the elements of the counts array

  for (int i = 0; i < counts.length; i++)
  {
     for (int j = 0; j < counts[0].length; j++)
     {
        System.out.printf("%8d", counts[i][j]);
     }
     System.out.println();
  }

Locating Neighboring Elements

Figure 15 Neighboring Locations in a Two-Dimensional Array

• Watch out for elements at the boundary array
  • counts[0][1] does not have a neighbor to the top

Accessing Rows and Columns

• Problem: To find the number of medals won by a country
  • Find the sum of the elements in a row
• To find the sum of the ith row
  • compute the sum of counts[i][j], where j ranges from 0 to MEDALS - 1.
    

• Loop to compute the sum of the ith row

  int total = 0;
  for (int j = 0; j < MEDALS; j++)
  {
     total = total + counts[i][j];
  }

Accessing Rows and Columns

• To find the sum of the jth column
  • Form the sum of counts[i][j], where i ranges from 0 to COUNTRIES - 1
  
  

  int total = 0;
  for (int i = 0; i < COUNTRIES; i++)
  {
     total = total + counts[i][j];
  }

• 

section_6/Medals.java

Program Run:

•         Country    Gold  Silver  Bronze   Total
           Canada       1       0       1       2
            China       1       1       0       2
          Germany       0       0       1       1
            Korea       1       0       0       1
            Japan       0       1       1       2
           Russia       0       1       1       2
    United States       1       1       0       2

Self Check 7.30

What results do you get if you total the columns in our sample data?

• Answer: You get the total number of gold, silver, and bronze medals in the competition. In our example, there are four of each.

Self Check 7.31

Consider an 8 × 8 array for a board game:

int[][] board = new int[8][8];

Using two nested loops, initialize the board so that zeros and ones alternate, as
on a checkerboard:

0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
. . .
1 0 1 0 1 0 1 0

Hint: Check whether i + j is even.

• Answer:

  for (int i = 0; i < 8; i++)
  {
     for (int j = 0; j < 8; j++)
     {
        board[i][j] = (i + j) % 2;
     }
  }

Self Check 7.32

Declare a two-dimensional array for representing a tic-tac-toe board. The board has three rows and columns and contains strings

,

, and

.

• Answer: String[][] board = new String[3][3];

Self Check 7.33

Write an assignment statement to place an

in the upper-right corner of the tic-tac-toe board in Self Check 32.

• Answer: board[0][2] = "x";

Self Check 7.34

Which elements are on the diagonal joining the upper-left and the lower-right corners of the tic-tac-toe board in Self Check 32?

• Answer: board[0][0], board[1][1], board[2][2]

Array Lists

• An array list stores a sequence of values whose size can change.
• An array list can grow and shrink as needed.
• ArrayList class supplies methods for many common tasks, such as inserting and removing elements.
• An array list expands to hold as many elements as needed.
  
  

Syntax 7.4 Array Lists

Declaring and Using Array Lists

• To declare an array list of strings

  ArrayList<String> names = new ArrayList<String>();
      

• To use an array list

  import java.util.ArrayList;

• ArrayList is a generic class
• Angle brackets denote a type parameter
  • Replace String with any other class to get a different array list type

Declaring and Using Array Lists

• ArrayList<String> is first constructed, it has size 0
• Use the add method to add an object to the end of the array list:

  names.add("Emily"); // Now names has size 1 and element "Emily"
  names.add("Bob"); // Now names has size 2 and elements "Emily", "Bob"
  names.add("Cindy"); // names has size 3 and elements "Emily", "Bob", and "Cindy"

• The size method gives the current size of the array list.
  • Size is now 3


Figure 17 Adding an Array List Element with add

Declaring and Using Array Lists

• To obtain an array list element, use the get method
  • Index starts at 0
• To retrieve the name with index 2:

  String name = names.get(2); // Gets the third element of the array list

• The last valid index is names.size() - 1
  • A common bounds error:

  int i = names.size();
  name = names.get(i); // Error

• To set an array list element to a new value, use the set method:

  names.set(2, "Carolyn");

Declaring and Using Array Lists

• An array list has methods for adding and removing elements in the middle.
  
  
• This statement adds a new element at position 1 and moves all elements with index 1 or larger by one position.

  names.add(1, "Ann")

• The remove method,
  • removes the element at a given position
  • moves all elements after the removed element down by one position
  • and reduces the size of the array list by 1.

  names.remove(1);

• To print an array list:

  System.out.println(names); // Prints [Emily, Bob, Carolyn]

Declaring and Using Array Lists

Figure 18 Adding and Removing Elements in the Middle of an Array List

Using the Enhanced for Loop with Array Lists

• You can use the enhanced for loop to visit all the elements of an array list

  ArrayList<String> names = . . . ;
  for (String name : names)
  {
     System.out.println(name);
  }

• This is equivalent to:

  for (int i = 0; i < names.size(); i++)
  {
     String name = names.get(i);
     System.out.println(name);
  }

Copying Array Lists

• Copying an array list reference yields two references to the same array list.
• After the code below is executed
  • Both names and friends reference the same array list to which the string "Harry" was added.

  ArrayList<String> friends = names;
  friends.add("Harry");



Figure 19 Copying an Array List Reference



• To make a copy of an array list, construct the copy and pass the original list into the constructor:

  ArrayList<String> newNames = new ArrayList<String>(names);

Working with Array Lists

ArrayList<String> names =
   new ArrayList<String>();

names.add("Ann");
names.add("Cindy");

System.out.println(names);

names.add(1, "Bob");

names.remove(0);

names.set(0, "Bill");

String name = names.get(i);

String last =
   names.get(names.size() - 1);

ArrayList<Integer> squares =
   new ArrayList<Integer>();
for (int i = 0; i < 10; i++)
{
   squares.add(i * i);
}

Wrapper Classes

Wrapper Classes

• To collect double values in an array list, you use an ArrayList<Double>.
• if you assign a double value to a Double variable, the number is automatically “put into a box”
• Called auto-boxing:
  • Automatic conversion between primitive types and the corresponding wrapper classes:

  Double wrapper = 29.95; 

  • Wrapper values are automatically “unboxed” to primitive types

  double x = wrapper;
     



Figure 20 A Wrapper Class Variable

Using Array Algorithms with Array Lists

• The array algorithms can be converted to array lists simply by using the array list methods instead of the array syntax.
• Code to find the largest element in an array:

  double largest = values[0];
  for (int i = 1; i < values.length; i++)
  {
     if (values[i] > largest)
     {
        largest = values[i];
     }
  }

• Code to find the largest element in an array list

  double largest = values.get(0);
  for (int i = 1; i < values.size(); i++)
  {
     if (values.get(i) > largest)
     {
         largest = values.get(i);
     }
  }

Storing Input Values in an Array List

• To collect an unknown number of inputs, array lists are much easier to use than arrays.
• Simply read the inputs and add them to an array list:

  ArrayList<Double> inputs = new ArrayList<Double>();
  while (in.hasNextDouble())
  {
     inputs.add(in.nextDouble());
  }

Removing Matches

• To remove elements from an array list, call the remove method.
• Error: skips the element after the moved element

  ArrayList<String> words = ...;
  for (int i = 0; i < words.size(); i++)
  {
     String word = words.get(i);
     if (word.length() < 4)
     {
        Remove the element at index i.
     }
  }

• Concrete example
  
  
• Should not increment i when an element is removed

Removing Matches

• Pseudocode
  

  If the element at index i matches the condition
  Remove the element.
  Else
  Increment i.

• Use a while loop, not a for loop

  int i = 0;
  while (i < words.size())
  {
     String word = words.get(i);
     if (word.length() < 4)
     {
        words.remove(i);
     }
     else
     {
        i++;
     }
  }

Choosing Between Array Lists and Arrays

• For most programming tasks, array lists are easier to use than arrays
  • Array lists can grow and shrink.
  • Arrays have a nicer syntax.
• Recommendations
  • If the size of a collection never changes, use an array.
  • If you collect a long sequence of primitive type values and you are concerned about efficiency, use an array.
  • Otherwise, use an array list.

Choosing Between Array Lists and Arrays

section_7/LargestInArrayList.java

Program Run:

• Please enter values, Q to quit:
  35 80 115 44.5 Q
  35
  80
  115 <== largest value
  44.5
  

Self Check 7.35

Declare an array list

of integers that contains the first five prime numbers (2, 3, 5, 7, and 11).

• Answer:

  ArrayList<Integer> primes = new ArrayList<Integer>();
  primes.add(2);
  primes.add(3);
  primes.add(5);
  primes.add(7);
  primes.add(11);

Self Check 7.36

Given the array list primes declared in Self Check 35, write a loop to print its elements in reverse order, starting with the last element.

• Answer:

  for (int i = primes.size() - 1; i >= 0; i--)
  {
     System.out.println(primes.get(i));
  }

Self Check 7.37

What does the array list names contain after the following statements?

ArrayList<String> names = new ArrayList<String>;
names.add("Bob");
names.add(0, "Ann");
names.remove(1);
names.add("Cal");

• Answer: "Ann", "Cal"

Self Check 7.38

What is wrong with this code snippet?

ArrayList<String> names;
names.add(Bob);

• Answer: The names variable has not been initialized.

Self Check 7.39

Consider this method that appends the elements of one array list to another:

public void append(ArrayList<String> target, ArrayList<String> source)
{
   for (int i = 0; i < source.size(); i++)
   {
      target.add(source.get(i));
   }
}

What are the contents of

and

after these statements?

ArrayList<String> names1 = new ArrayList<String>();
names1.add("Emily");
names1.add("Bob");
names1.add("Cindy");
ArrayList<String> names2 = new ArrayList<String>();
names2.add("Dave");
append(names1, names2);

• Answer: names1 contains "Emily", "Bob", "Cindy", "Dave"; names2 contains "Dave"

Self Check 7.40

Suppose you want to store the names of the weekdays. Should you use an array list or an array of seven strings?

• Answer: Because the number of weekdays doesn’t change, there is no disadvantage to using an array, and it is easier to initialize:

  String[] weekdayNames = { "Monday", "Tuesday", "Wednesday", "Thursday", “Friday”,
      "Saturday", "Sunday" };

Self Check 7.41

The

directory of your source code contains an alternate implementation of the problem solution in How To 7.1 on page 334. Compare the array and array list implementations. What is the primary advantage of the latter?

• Answer: Reading inputs into an array list is much easier.

Regression Testing

• Test suite: a set of tests for repeated testing
• Cycling: bug that is fixed but reappears in later versions
• Regression testing: involves repeating previously run tests to ensure that known failures of prior versions do not appear in new versions

Regression Testing - Two Approaches

• Organize a suite of test with multiple tester classes: ScoreTester1, ScoreTester2, ...

  public class ScoreTester1
  {
     public static void main(String[] args)
     {
        Student fred = new Student(100);
        fred.addScore(10);
        fred.addScore(20);
        fred.addScore(5);
        System.out.println("Final score: " + fred.finalScore());
        System.out.println("Expected: 30");
     }
  }
      

• Provide a generic tester, and feed it inputs from multiple files.

Input and Output Redirection

Section_8/input1.txt

30
10
20
5

• Type the following command into a shell window
  • Input redirection

  java ScoreTester < input1.txt

• Program Run:

  Final score: 30
  Expected: 30

• Output redirection:

  java ScoreTester < input1.txt > output1.txt

Self Check 7.42

Suppose you modified the code for a method. Why do you want to repeat tests that already passed with the previous version of the code?

• Answer: It is possible to introduce errors when modifying code.

Self Check 7.43

Suppose a customer of your program finds an error. What action should you take beyond fixing the error?

• Answer: Add a test case to the test suite that verifies that the error is fixed.

Self Check 7.44

Why doesn't the

program contain prompts for the inputs?

• Answer: There is no human user who would see the prompts because input is provided from a file.

Which code snippet prints out the elements in a partially filled array of integers assuming currentSize?

Which code will define the size of an array?

When the order of elements is unimportant what is the most efficient way to remove an element from an array?

What is the valid range of index values for an array of size 10?