Input and Output

• Many useful programs save and retrieve data between executions of the program
  • Microsoft Word, Bluej, many games
• Need to be able to read and write data to files
  • called input/output or I/O
• We will learn about reading and writing text data
• Use PrintWriter for output, Scanner for input

File object

• Represents the path to a file
  • file may or may not exist
• To create a file

  File file = new File ("object.dat");

• File has many useful methods

Text Output

Use PrintWriter to write text files

• Construct a PrintWriter from a File object

  new PrintWriter(new File("output.txt"))

  • uses an anonymous object
• Convienence constructor

  PrintWriter out = new PrintWriter("output.txt"));

• If the file does not exist, an empty file is created
• If the file does exist, it is emptied

Writing text files

• Use the familiar print() and println() methods

  out.println(45.71);
  out.println(new Rectangle(10, 20, 100, 50));
  out.print("Hello, SJSU");

• You must close the file when you are done
  • Otherwise, not all the output may be written to the file

  out.close()

Text Input

Use Scanner to read text files

• Construct a Scanner from a File object

  Scanner in = new Scanner( new File("input.txt"));

Reading text files

Use the same methods you used for keyborad input

• in.next()
  • skips whitespaces, gets the text up to the next whitespace
  • leaves the trailing whitespace in the imput stream
• in.nextInt() or in.nextDouble()
  • calls next() and converts to int or double
  • leaves the trailing whitespace in the input stream
• in.nextLine()
  • reads the next line up to and including the new line characte

You should also close the Scanner when done

in.close();

Mixing next and nextLine

• Input file: contains IDs and names

  1729
  Harry
  1730
  Wilma

• Initially, the input contains
  
• Call

  int studentID = in.nextInt(); 

• Now the input contains
  
• If you call nextLine, you don't read Harry. Reads an empty string!
• Remedy:  call nextLine after reading the ID:

  int studentID = in.nextInt(); 
  in.nextLine(); // Consume the \n 
  String name = in.nextLine();

Class Question #1

Suppose the input contains the characters 2nd place. What is the value of number and word after these statements?

int number = in.nextInt(); 
String word = in.next();

1. number is 2 and word is "nd"
2. number is 2 and word is "nd place"
3. number is 2, then an exception is thrown
4. an exception is thrown because a "n" is not a valid part of an integer

FileNotFoundException

• When the file does not exist, a FileNotFoundException can be thrown
• To handle the exception, add the throws clause to the method header

  public static void main(String[] args) 
                       throws FileNotFoundException

• More about exceptions later

Tricks and techniques for text input:
    hasNext...

• To read all the lines in a file

  while (in.hasNextLine())
  {
     String line = in.nextLine();
     
     //processing
  }

• If the next word is not always the same type
  • use hasNext(), hasNextInt(), etc. to avoid InputMismatchException

    if (in.hasNextInt())
    {
        int number = in.nextInt();
        
        //processing
     }

• . . .

Tricks and techniques for text input:
     words

• a word is sequence of characters that does not contain a delimiter
  • default delimiter is white space
• To use a different delimiter to determine words
  • use Scanner's useDelimiter() method
• To have a word boundary be anything that is not a letter

  Scanner in = new Scanner(. . .); 
  in.useDelimiter("[^A-Za-z]+");
  ...

• The stuff in quotes is called a regular expression

Tricks and techniques for text input:
     processing lines

When the file is line oriented but contains mixed data:

China 1330044605
United States 303824646
...

• Read a line at a time, then make another Scanner to process each line

  String line = in.nextLine();
  Scanner lineIn = new Scanner(line);
  String country = lineIn.next();
  while (!in.hasNextDouble()) 
     country = country + " " + lineIn.next(); 
  double area = lineIn.nextDouble(); 

• The while keeps concatenating words until a double is found

Tricks and techniques for text input:
     processing lines

When the numeric data contain non-numeric characters like $

China $13344605 N/A
United States $303824646 $46,000
...

• Use indexOf or lastIndexOf to find the character
• Then use substring to extract the country name
• This is the kind of thing you will need to do in your homework

Class question #2

An input file contains lines such as

Fred 40
Wilma 28
Mary Ann 30

Consider this code  to process a line:

String line = in.nextLine();
int n = line.lastIndexOf(' ');
String name = line.substring(0, n);
int age = Integer.parseInt(line.substring(n));

What is the problem with this code?

1. The name includes the trailing space
2. The call to Integer.parseInt throws an exception
3. There is no probelm. The code works correctly
4. The code can not handle a two name with two words

What is the solution?

• The problem is the space
  • n is 4 and the character at 4 is a space
  • line.substring(n) contains the leading pace at index 4

    F	r	e	d		4	0
    0	1	2	3	4	5	6

• Solution 1

  int age = Integer.parseInt(line.substring(n).trim());

• Solution 2

  int age = Integer.parseInt(line.substring(n + 1));

Throwing an Exception

• Exception handling: transfer control from error location to competent handler
• When you detect an error, just throw an exception

  public class BankAccount
  {
     public void withdraw(double amount)
     {
        if (amount > balance)
        {
           throw new IllegalArgumentException("Amount exceeds balance");;
        }
        balance = balance - amount;
     }
     . . .
  }

Command line arguments

Special Topic 11.3 on the web

• You can run Java from the command line (cygwin on the PC, Terminal on the Mac ...)
• To execute a class entered in NetBeans
  • You need to change to the project directory and then give the class path to the .class file

     java -classpath build/classes programName 

• Any data after the programName is bundled up and put into an array and passed as the parameter to the main method
• This is useful for automating processing

Arguments array

• To specify arguments on the command line:

   java -classpath build/classes programName string1 string2 ...

• If the parameter to the main method is called args, then
  • args[0] is string1
  • args[1] is string2
• Your can process the parameters like this

   if (args.length > 0)
     {
        fileName = args[0]
        ...
     }
     else
        System.out.println("Usage java programName string1");

• You will use this in your homework to specify the file name