Welcome to Lesson 15!

Learning Objectives
By the end of today's class, you should know...
  • How do you use loops to prevent type mismatch exception?
  • How do you use a for loop, the length method for a String and the charAt method to iterate through a String, character-by-character?
  • How do you use the substring() method for a String?
    • What does it return to you?
  • What is the problem caused by newline characters when mixing next(), nextInt() and nextDouble() with nextLine()?


  • Last day to drop with a "W" is this Friday.
  • Quiz 6 on Monday
    • Return Quiz 5
  • Midterm 2 one week from today - lesson 9 - lesson 16
    • Same format as Midterm 1: ~15 multiple choice/short answer/true-false questions and 1 full program
    • Study old quizzes and review activities
    • Review your homework assignments and get help on any you struggled with
  • Don't forget to complete Lab 7 on Friday

Review Activity:

With a partner, answer the following questions:

  • What is the advantage of using a do-while loop instead of a while loop?
  • Change the following while loop into a do-while loop:

System.out.print("Please enter your 9 digit password: ");

String password = input.next();

while (password.length() != 9) {

    System.out.print("Please enter your 9 digit password: ");

    password = input.next();


Loops and Error Checking

Using Loops to Check for input mismatch

  • A problem arises if the user of a our programs enters a String when we expect a number
  • For instance, in the following, we get incorrect results if the user enters "seven"
    double value = 0.0;
    value = input.nextDouble(); //user enters "seven" not 7
  • The problem is that Scanner cannot convert the word "seven" into the number 7
  • When this happens the code crashes with an error message about an input mismatch exception:
  • When the program suddenly crashes, the user will be confused.
  • Also the rest of the program cannot proceed.
  • A better approach would be to anticipate the possibility of the above user error and handle it in our code.
  • We can detect the failure condition using code like the following:

            double value = 0.0;
       Scanner input = new Scanner(System.in);
       System.out.print("Enter a positive number: ");
       if(input.hasNextDouble()) { //Check to see if the user entered a double
           value = input.nextDouble(); //read in the double
       else {
           System.out.println("Error! Please enter a number not text!");
           input.next(); //read in the String value from the user to clear it from buffer

  • Depending on the expected input, we can use one of the following functions to test for input mismatch:
hasNextInt() //returns true if the user entered an int
hasNextDouble() //returns true if the user entered a double
  • Both functions will return true if the user entered the correct type and false otherwise.
  • However, there is a problem with the above approach?
    • What happens if the user enters the incorrect type more than once?
  • Let's use a loop instead!

Example Program with Input Validation


package loopy;

* @author Jennifer Parrish
import java.util.Scanner;

public class Loopy {

* @param args the command line arguments
public static void main(String[] args) {
double value = 0.0;
Scanner input = new Scanner(System.in);
System.out.print("Enter a number: ");

while(!input.hasNextDouble()) { //Check to see if the user entered a double
System.out.println("Error! Please enter a number not text!");
input.nextLine(); //read in the String value from the user to clear it from buffer

System.out.print("Enter a number: "); //prompt again
value = input.nextDouble(); //read in the double
System.out.println("You entered " + value);
More information on hasNextDouble() and hasNextInt() methods

Activity 15.1: Scores (10 pts)

  • In this exercise we use indefinite loops to process user input and to ensure correct user input.
  • Find a partner for pair programming.
  • Open a new Eclipse project named Scores.
  • Declare two variables of type double named sumScores and nextScore and initialize the variables to 0. In addition, declare an integer variable named count and initialize it to 0. The following is the psuedocode for these steps:
    set sumScores to 0
    set nextScore to 0

    set count to 0
  • Compile your code to make sure you declared the variables correctly.

  • Now we want to use a loop to enter a series of scores. Since we do not know how many scores to enter, we use an indefinite loop like the following:

  • In addition, add a statement to display sumScores after the loop.

    Compile your code to make sure you added the loop correctly. To exit the loop you will need to enter a negative number.

  • The loop includes statements to collect the sum of the scores in the variable named sumScores. Add a statement after the loop to print sumScores to the console. When you run the program after adding this code, the output should look like:
    Enter score #1: 60
    Enter score #2: 50
    Enter score #3: 20
    Enter score #4: 99
    Enter score #5: 100
    Enter score #6: 0
    Enter score #7: 45
    Enter score #8: 86
    Enter score #9: -1

    Sum of scores: 460.0

    The loop uses the sumScores variable to accumulate scores during each repetition of the loop. 

  • We could write our indefinite loop using a do-while loop instead. Replace your current loop with the following:

  • Note that the statements inside the loop did not change, only the loop statement itself. To make sure you made the changes correctly, compile and run your code and check to see if it works the same. The difference between a while and do-while loop is that a do-while ensures the body of the loop is executed at least once. 

  • One problem with our program is the user can still enter letters instead of digits. We can prevent this error by checking input.hasNextDouble() and looping until the user enters a correct value. Replace the current if statement with the following and move the nextScore = input.nextDouble()inside of an else statement as shown below:
The completed program

  • Compile and run your program to make sure you added the changes correctly. Try entering letters instead of digits and verify you see an error message.

Enter score #1: seventy
Error! Please enter a number, not text.
Enter score #2: 79
Enter score #3: 80
Enter score #4: twenty
Error! Please enter a number, not text.
Enter score #5: Error! Please enter a number, not text.
Enter score #6: 20
Enter score #7: 0
Enter score #8: -1

Sum of scores: 179.0
  • Note that when you report the error message and ask the user to try again, the count variable gets off track. There are two approaches to fixing this problem.
    • With your partner, try to find at least one solution to this problem.
  • When finished, upload your Scores.java file to Canvas.

Strings Continued

Review: Strings Versus Characters

  • Remember that a string is a series of characters enclosed in double quotes such as:
    "Hello"  "b"  "3.14159"  "$3.95"  "My address is 378 Eastbrook Dr"
  • We can store text in a variable of type string, like:
    String firstName;             // declaration
    firstName = "Jennifer";       // assignment
    String lastName = "Parrish";  // declaration + assignment
    String fullName = firstName + " " + lastName; // concatenation (+) of 2 stings
  • On the other hand, a character is a single letter, number or special symbol
  • We enclose characters in a single quote, rather than a double quote, like:
    'a'   'b'   'Z'   '3'   'q'   '$'   '*'
  • Also, we can store a a single character using a variable of type char, such as:
    char letterA = 'A';
    char letterB = 'B';
  • By declaring a char variable or using single quotes, Java knows to treat the number as a character
  • Thus, when we print a character, we see a letter rather than a number:
    char letter = 'A';
    System.out.println(letter + 'B');
  • As we can see, a string is made up of characters and characters are numerical codes
  • We can use this information to work with characters and strings

Indexing a String

  • Strings have a built-in indexing system with each stored in a character sequence starting at 0 (zero)

String sport = "Basketball";

  • We can access any individual character of a String variable using the String method charAt(i)
  • The general syntax is:
  • Where:
    • stringVariable: the name of your string variable
    • index: the number of the character position
  • For example:
    String str = "abcdef";
    char firstLetter = str.charAt(0);
    System.out.println(""+firstLetter + str.charAt(1));
  • The above code displays:
  • Notice that the charAt method returns a char data type
  • What will the following print?
public static void main(String[] args) {
      String sport = "Basketball";
      System.out.println("The league: " + sport.charAt(1) + sport.charAt(7));

      System.out.println("Abbreviation for this sport: " + sport.charAt(0) + sport.charAt(6);


Using a For Loop to Iterate Strings
  • Recall that member method length() returns the number of characters in a string variable:
    String s = "abcdef";
    int n = s.length();
  • After we know the length, it is easy to iterate through the individual characters of a string using a counting loop:
    System.out.println("Enter a message: ");
    String msg;
    msg = input.nextLine();
    for (int i = 0; i < msg.length(); i++) {
        System.out.println("Index " + i + ": " + msg.charAt(i));

Group Activity: Iterating Strings

  • Open a new Java project called Loopy or alter an existing one.
  • Add the following for-loop code to the main method.
    String msg = "Hello, world!";

    for (int i = 0; i < msg.length(); i++) {
        System.out.println("Index " + i + ": " + msg.charAt(i));

  • Compile and run your code. What do you see when you compile?

The substring() Method

  • The substring(begin, end) method also uses index numbers to select part of a String:
    String greeting = "Hello, World!";
    String sub = greeting.substring(0, 4);
    System.out.println(sub); // does this print "Hell" or "Hello"?
  • The begin argument starts with the index number of the first character
  • The end argument is always one past the last character to extract
    H e l l o ,
    W o r l d !
    0 1 2 3 4 5 6 7 8 9 10 11 12
  • As another example, we can extract "World" using:
    String w = greeting.substring(7, 12);
  • Note from the example above that the substring() method returns a String not a char.
  • Which extracts the characters highlighted in the following diagram:
    H e l l o ,
    W o r l d !
    0 1 2 3 4 5 6 7 8 9 10 11 12
  • Note that you get a Runtime Exception (error) if you try to select an index lower than 0 or beyond the end of a string
  • For example, the code:
    char bogus = greeting.charAt(15);
    produces an error like:

    Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 15
            at java.lang.String.charAt(String.java:687)
            at StringMethods.main(StringMethods.java:13)

Activity 15.2: How many words in your sentence? (10 pts)

  • Let's write a program that counts the number of words in a sentence input by the user.
  • Find a partner and open up a new Java project called Words
  • Alter the block comments to contain your names and section information.
  • Next declare a String variable at the top of your program named sentence:
String sentence;
  • Welcome the user to the program with the following message:
Give me a sentence, and I will count the number of words.
  • Prompt the user to input a sentence and store the user input as the sentence variable.
    Please enter your sentence.
  • Don't forget to use input.nextLine() here to read in the entire sentence.
  • How can we determine how many words are in a sentence?
  • We need to look at the whitespace.
  • Next, we will use a for loop to scroll through the sentence looking for blank spaces. 
  • Each time we encounter a new blank space, we will add one to our total for the number of words in the sentence.
  • Create a new variable at the top of main to store our counter for the number of words in the sentence. Assign it a value of 1. Why do we want to give it an initial value of 1 not 0?
int numWords = 1;
  • Now, create a for loop to iterate through the sentence. Don't forget to use the length() method here.
for (int i = 0; i < sentence.length(); i++) {
  • Now, run your program and verify that you get the following output.

  • However, this is not the purpose of our program.
  • We want to count the number of words.
  • Remove the System.out.println() statement from your for loop.
  • Replace the System.out.print statement with an if statement to check the value of the the sentence at index i to determine if it is a blank space.

if (sentence.charAt(i) == ' ') //Why am I using == here and not equals()?

  • When we encounter a blank space, we need to add one to the numWords variable.
  • Your for loop should now look like this:

for (int i = 0; i < sentence.length(); i++) {
    if (sentence.charAt(i) == ' ') {
  • Finally, outside of your for loop, add a statement to print out the number of words.
System.out.println("There are " + numWords + " words in \"" + sentence + "\"");

  • Your program should now look like this:
  • Let's see if we can also use the substring() method to print the individual words in the sentence.
  • Add a print statement above the for loop to print the following message:

The words in your sentence are:

  • Notice that the starting index of each word occurs at the index following the blank spaces in the sentence:




  • In the above example, the word "love" begins at index 2, the word "my" begins at index 7 and the word "pet" begins at index 10, one index after each blank space.
  • Also the previous word ends at the index right before each blank space.
  • We already know the location of the blank spaces due our if statement.

    if (sentence.charAt(i) == ' ')

    //i will be the index of a blank space if the condition is true

  • Now, let's put it all together to print out each word using the substring() method.
  • First, add a new variable declaration at the top of your program:

int start_index = 0;

  • Next, inside the if statement of the for loop, add a print statement like the following to print out each word:

System.out.println(sentence.substring(start_index, i));

  • Finally, beneath that print statement, let's reset the starting index for the next word in the sentence:

        start_index = i+1; //Why do we use i+1 here?

  • Try running your code. You should notice a problem here. Why does the last word not print out?
    • Hint: The last character of the sentence is a \n not a ' '.
  • Let's add one last print statement, outside the loop to print the last word:

System.out.println(sentence.substring(start_index, sentence.length()));

//why do I use sentence.length()?

  • Run it again and make sure you get the correct output:

  • Submit your program to Canvas when you are finished.
  • Your code should now look like the following:

The Problem with Newlines

  • When you press the Enter key, a newline character ('\n') is inserted as part of the input
  • The newline character can cause problems when you mix input.next(), input.nextInt() or input.nextDouble() with input.nexLine()
  • Recall that input.next(), as well as nextInt() and nextDouble():
    1. Skips whitespace
    2. Reads non-whitespace characters into the variable
    3. Stops reading when whitespace is found
  • Since whitespace includes newline characters, using input.next() will leave a newline character in the input stream
  • However, input.nextLine() just stops reading when it first finds a newline character
  • This can lead to mysterious results in code like the following:
    System.out.print("Enter your age: ");
    int age = input.nextInt();
    System.out.print("Enter your full name: ");
    String name = input.nextLine();
    System.out.println("Your age: " + age + "\n"
         + "Your full name: " + name);
  • To correct this problem we add an additional input.nextLine() just before input.nextLine()

System.out.print("Enter your full name: "); 
input.nextLine(); //clear out the \n
String name = input.nextLine();

  • We can see how to use this fix in the following example

Example Using an additional input.nextLine()

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    System.out.print("Enter your age: ");
    int age = input.nextInt();

    System.out.print("Enter your full name: ");
    input.nextLine(); //clear out the \n
    String name = input.nextLine();     System.out.println("Your age: " + age + "\n"     + "Your full name: " + name); }

Activity 15.3: Sentence Continued (10 pts)
  • Let's add to our program involving sentences from the last activity. We will also calculate the number of letters in the sentence and take in a user guess for a number of letters.
  • At the end of the program, we will confirm whether their guess was correct or not.
  • Declare a new variable beneath the other two at the top of the program. This variable will be used to store our calculation for the number of letters. 
  • Also, add a variable to store the user input for their guess. 
  • The variable declaration section of your program should now look like this:
String sentence;
int numWords = 1;
int numLetters = 0;
int start_index = 0;
int guess;
Scanner input = new Scanner(System.in);
  • Now alter your first System.out.println statement to reflect the additional uses of this program. Remove the original message in your System.out.println statement and replace it with the one below:
Think of a sentence in your mind.
Later I will tell you how many words and letters are in your sentence.

  • Now, ask the user to enter a guess for how many letters are in the sentence. We want the user to guess without counting the number of letters.
Enter a guess for the number of letters in your sentence (don't count!): _
  • Store the user guess as the guess variable using input.nextInt().
  • Next, prompt the user to enter the sentence. Your prompt should remain the same from the last exercise and, as before, you should use input.nextLine() to store the user input as the sentence variable.
Please enter your sentence: _
  • Verify that your code inside main looks like identical to the code below:

  • Now, run your program. You should notice a problem.
  • How can we fix this problem?
  • Add an additional input.nextLine() above your sentence = input.nextLine();
sentence = input.nextLine();
  • Compile and run your code again and verify that it is now working properly.
  • Now let's alter the code inside the for loop to calculate how many letters are in the sentence.
  • Since we don't want to count any blank spaces, we only want to increment the numLetters variable when we are NOT incrementing the numWords variable. 
  • Therefore, we need to add an else statement to our for loop. Make sure your if-else in the for loop looks like this:
if (sentence.charAt(i) == ' ') {
    System.out.println(sentence.substring(start_index, i));
    start_index = i+1; //new starting index
} else {
  • Now, let's add another print statement below the for loop to print out the number of letters in the sentence,
System.out.println("And, " + numLetters + " letters.");
  • Did our user guess the number of letters correctly? Now is the time to let him or her know. Add the following if-else block at the bottom of main.
if (guess == numLetters) {
    System.out.println("You guessed right!");
} else {
    System.out.println("You guessed wrong!");
  • Run your program again and you should get the following output. Note: if you don't get the output below, compare your program to the final version at the end of this exercise. When you are finished, upload to Canvas.

  • The revised portion of your code should look like the following:

Wrap Up

  • With your partner, answer the questions from today's learning objectives

Upcoming Assignments
  • Assignment 15 due Wednesday at 9:20am on Canvas
  • Lab 7 due Friday at midnight