W6: Exceptions and I/O

Please submit your answers to the questions as comments in a W6.md file you’ll be writing in this lecture.

Grading rubric and submission

When you are done, submit your W6.md file to BB.

You will be graded on the following:

Item	Points
Answers are correct	100

Questions

1

What does the following code print to the terminal?

int[] x = {};
int y = 0;
System.out.println(x[y]);

Answer

It will crash and print a stack trace to the terminal because an ArrayIndexOutOfBoundsException was raised when doing x[y]

2

What does the following code print to the terminal?

int[] x = {};
int y = 0;

try {
    System.out.println("before");
    System.out.println(x[y]);
    System.out.println("after");
} catch (Exception e){
    System.out.println("there was something wrong");
}

Answer

before
there was something wrong

3

Does the following program crash? What gets printed to the terminal?

int[] x = {};
int y = 0;

try {
    System.out.println("before");
    System.out.println(x[y]);
    System.out.println("after");
} catch (Exception e){
    System.out.println("there was something wrong");
    e.printStackTrace();
}

Answer

before
there was something wrong
java.lang.ArrayIndexOutOfBoundsException: 0
    at [filename:line number]

4

What does the following code print to the terminal?

int[] x = {};
int y = 0;

try {
    System.out.println("before1");
    System.out.println(x[y]);
    System.out.println("after1");
} catch (ArrayIndexOutOfBoundsException e){
    System.out.println("there was something wrong1");
}

try {
    System.out.println("before2");
    System.out.println(x[y]);
    System.out.println("after2");
} catch (Exception e){
    System.out.println("there was something wrong2");
} catch (ArrayIndexOutOfBoundsException e){
    System.out.println("there was something wrong3");
}

Answer

before1
there was something wrong1
before2
there was something wrong2

5

How do you manually raise an exception in Java? Why are you allowed to do this?

Answer

You can call the exception’s constructor to create a new exception object, and then raise that exception with the throw keyword inside a method. This has the same behavior as a “naturally occuring” exception (such as a division by zero or a null pointer exception).

When a method explicitly raises an exception like this (and it’s not caught elsewhere in the method), the method signature should be annotated with a throws clause.

You’re allowed to do this in Java because sometimes you’ll want your own custom exception types (i.e. not “naturally occurring”) to be raised and caught elsewhere in your code.

6

What does the following code print to the terminal for the cases below? Assume the code compiles.

public static void main(String[] args){

    int x = 3;
    String y = mystery();

    try {
        System.out.println("here 1");
        foo(x, y);
        System.out.println("here 2");
        int z = 3 / 0;
    } catch (Exception e){
        System.out.println("error 1");
    } catch (ArithmeticException e){
        System.out.println("error 2");
    }

    System.out.println("here 3");
}

public static void foo(int x, String y) throws ThingException{
    
   int y_int = -4;
   try{
         y_int = Integer.parseInt(y);
        check(y_int);
    }
    catch (Exception e){
        System.out.println("bad y");
        throw new ThingException();
    }

    System.out.println(y_int);
    int z = 4 / y_int;
    System.out.println(y_int);
}

public static void check(int y){
    if (y < 0)
        throw new ArithmeticException();
    System.out.println("in check");
    y = y / 0;
    System.out.println("end check");
}

Case 1: mystery() returns "1"

Case 2: mystery() returns "0"

Case 3: mystery() returns "oops"

Answer

Case 1:

here 1
in check
bad y
error 1
here 3

Case 2:

here 1
in check
bad y
error 1
here 3

Case 3:

here 1
bad y
error 1
here 3

7

Give an example of when you would use exception handling over an if statement. #### Answer An if statement only allows an error state to be handled locally. Exceptions allow error states to “bubble up” the call stack to be handled elsewhere.

For example, if you are trying to open a file that doesn’t exist, the actual code that interfaces with the OS and file system to open the file is very low-level, deep in some library. If the filename is incorrect, it’s not the place to ask the user to re-enter it; we don’t know if the filename was typed in, came from a GUI, or was read from some other file – we don’t know where this filename came from. By raising an exception that can bubble up the call stack until it is caught, we have a chance to catch it in the location where the user actually specifies a file (whether by keyboard, command line, GUI window, or reading a file).

8

Give a reason why it is not a good idea to only wrap your entire main method inside one try-catch block.

Answer

If you only have one try-catch block in main your code may never crash, but you also lose the ability to appropriately handle exceptions where they happen.

9

What is the difference between a checked versus an unchecked exception?

Answer

Unchecked exceptions are not required to be caught in a try-catch block (eventually). Checked exceptions must eventually be caught; this can be done locally, or the method that raises a checked exception must declare this exception type can be raised with a throws clause.

10

Write code that opens a file called data.txt and prints out its contents, line by line.

Answer

try {
    LineNumberReader reader = new LineNumberReader(new InputStreamReader(new FileInputStream("data.txt")));
    String line = reader.readLine();
    while( line != null ) {
      System.out.println(line);
      line = reader.readLine();  
    }
    reader.close();
} catch (IOException e){
    e.printStackTrace();
}

11

Write code that prints Hello World! and How are you? on two separate lines of a file called data.txt

Answer

try {
        PrintWriter pw = new PrintWriter(new File("output.txt")); 
        pw.println("Hello World!");
        pw.println("How are you?");
        pw.close(); 

} catch (IOException e){
    e.printStackTrace();
}

12

What happens if you don’t close an input/outputstream buffer in your code?

Answer

Usually, nothing, if your program is short and runs quickly; once the main method exits, all these resources are returned anyway…

However, if you are opening multiple buffers (and not closing them) in a loop that runs for a long time (this opening too many of these buffers) you can run out of memory and have other bad things happen (ask me how I know). It’s good practice to always close the buffers you open. Newer versions of Java even do this automatically.

Old Java that uses a finally clause to wrap everything up at the end regardless if an exception was raised or not:

static String readFirstLineFromFileWithFinallyBlock(String path) throws IOException {
   
    FileReader fr = new FileReader(path);
    BufferedReader br = new BufferedReader(fr);
    try {
        return br.readLine();
    } finally {
        br.close();
        fr.close();
    }
}

versus the newer version of Java that closes these resources automatically using a try-with-resources statement:

static String readFirstLineFromFile(String path) throws IOException {
    try (FileReader fr = new FileReader(path);
        BufferedReader br = new BufferedReader(fr)) {
        return br.readLine();
    }
}