Develop Reference

Java: try/catch with InputMismatchException creates infinite loop while reading of a file

I have a task to make a program that will add up all the valid integers in a file and to ignore anything that isn’t a valid int. I have to use Try and Catch.
File Numbers = new File("Numbers.txt");
Scanner readFile = null;
int i = 0;
int total= 0;
boolean success = false;
while(!success){
try {
readFile = new Scanner(Numbers);
while(readFile.hasNext()){
i = readFile.nextInt();
System.out.println(i);
total = i + total;
};
success = true;// Ends The loop
} catch (FileNotFoundException e1) {
System.err.println(Numbers.getName()+" does not exist");
}
catch(InputMismatchException e2){
System.err.println("Data incorrect type expecting an int found: " + readFile.nextLine());
readFile.next();
}
System.out.println("total is: " + total);
};
The problem is that the program gets caught in an infinite loop, where instead of going past the exception it just starts again.The task seems pretty straight forward, yet i don't know why it wont work?

You fall into infinite loop because when exception happens, the success variable didn't change its value to true. In order to do some action even when exception happens you should add the finnaly block. It could look like this:
try {
// do some stuff
} catch (Exception e) {
// catch the exception
} finally {
if (!readFile.hasNext()) success = true;
}
And by the way, never do this: catch (Exception e), I did it just for example sake. Instead always catch the specific exception. Because Exception is the most basic class in the exception hierarchy, so it will catch up all the exceptions, and unless you re-throw it you could have false feeling of "safiness". When you want to catch all the exceptions, you should do this:
try {
// do stuff
} catch (RuntimeException e) {
throw e;
} catch (Exception e) {
e.printStackTrace(); // or other approptiate action, i.e. log it.
}

Assume any of the following FileNotFound or InputMismatchException exceptions will raise, then your program wont change success to true. Thus it returns to the outer while loop and read the same file. Because nothing has changed the same Exception will be thrown again.
==> Endless loop.
To fix that I suggest to move the try/catch block to the inner while.

How to throw an exception if parsing with Long.parseLong() fails?

I am trying to write some kind of a stack calculator.
Here is a part of my code where I am handling a push command. I want to push integers only, so I have to get rid of any invalid strings like foobar (which cannot be parsed into integer) or 999999999999 (which exceeds the integer range).
strings in my code is a table of strings containing commands like POP or PUSH, numbers, and random clutter already split by white characters.
Main problem:
I've got difficulties with throwing an exception for long parseNumber = Long.parseLong(strings[i]); - I don't know how to handle the case, when strings[i] cannot be parsed into a long and subsequently into an integer.
while (i < strings.length) {
try {
if (strings[i].equals("PUSH")) {
// PUSH
i++;
if (strings[i].length() > 10)
throw new OverflowException(strings[i]);
// How to throw an exception when it is not possible to parse
// the string?
long parseNumber = Long.parseLong(strings[i]);
if (parseNumber > Integer.MAX_VALUE)
throw new OverflowException(strings[i]);
if (parseNumber < Integer.MIN_VALUE)
throw new UnderflowException(strings[i]);
number = (int)parseNumber;
stack.push(number);
}
// Some options like POP, ADD, etc. are omitted here
// because they are of little importance.
}
catch (InvalidInputException e)
System.out.println(e.getMessage());
catch (OverflowException e)
System.out.println(e.getMessage());
catch (UnderflowException e)
System.out.println(e.getMessage());
finally {
i++;
continue;
}
}

Long.parseLong(String str) throws a NumberFormatException if the string cannot be parsed by any reason. You can catch the same by adding a catch block for your try, as below:
catch ( NumberFormatException e) {
System.out.println(e.getMessage());
}

No need to worry. Long.parseLong() throws a NumberFormatException if it got other than a Number.

After reading your comments and answers I was able to come up with such a solution (this code is embedded inside the outside try-catch.)
if (strings[i].equals("PUSH")) {
// PUSH
i++;
if (strings[i].length() > 10) {
throw new OverflowException(strings[i]);
}
try{
parseNumber = Long.parseLong(strings[i]);
if (parseNumber > Integer.MAX_VALUE) {
throw new OverflowException(strings[i]);
}
if (parseNumber < Integer.MIN_VALUE) {
throw new UnderflowException(strings[i]);
}
number = (int)parseNumber;
stack.push(number);
}
catch (NumberFormatException n){
throw new InvalidInputException(strings[i]);
}
}

Throwing exception error

My program is not compiling and showing me this error:
System.out.println("Error: "+ e.getMessage());
Scanner scanner = new Scanner(System.in);
try {
int num = scanner.nextInt();
if (num != 0) {
throw new Exception("Not zero");
}
System.out.println("I'm happy with the input.");
} catch (InputMismatchException e) //InputMismatchException is never thrown in body of corresponding try statement
{
System.out.println("Invalid Entry");
} catch (Exception e) {
System.out.println("Error: "+ e.getMessage());
}

The error message is quite clear:
InputMismatchException is never thrown in body of corresponding try statement
You're trying to catch an exception which is guaranteed not to be thrown from the try block. This is useless, and even invalid. Remove the catch (InputMismatchException e) block.
Actually, the try block can throw a java.util.InputMismatchException. So I guess that you're in fact catching an InputMismatchException of another package. Check your imports, and make sure you import java.util.InputMismatchException and not some other com.foo.bar.InputMismatchException.
EDIT:
the error message confirms what I thought. You're catching javaapplication9.InputMismatchException, instead of java.util.InputMismatchException. I'm not sure why you defined your own InputMismatchException.

How do I display an exception?

I am checking if number the user entered is Zeckendorf and I want to display an exception if it is not, how do i do that? Also how do I convert the Zeckondorf to its decimal equivalent?
import java.util.Scanner;
public class IZeckendorfNumberConvertor {
static String number;
int zeckonderEquivalent;
static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
convertToZeckonderNumber();
isTrueZeckonderNumber();
}
private static boolean isTrueZeckonderNumber() {
System.out.println("Enter a Zeckonder Number:");
number = scanner.nextLine();
if (number.equals("10100"))
{
return true; }
else if (number.equals("010011") || number.equals("010100"))
{
return false; }
return false;
}
private static void convertToZeckonderNumber() {
}}

I advise you not to display an exception (i.e. trace and such) as it is very user Unfriendly.
You can use the throw syntax to throw a proper exception :
throw new Exception("Given number is not a Zeckendorf number");
but be sure to catch it and display a nice and clean message :
try {
// Your input code goes here
} catch (Exception e) {
System.out.println(e.getMessage());
}
Another easier option will be to just check the return value of the method and print the results accordingly.
I will recommend to use the latter solution as exceptions are used when something bad and unexpected happens in your program and you want to handle it gracefully. In your case the behavior is expected (i.e. user giving a wrong number) so checking the return value will be much clean and easier.

Use try catch block for catch an exception
try {
} catch (Exception e) {
e.printStackTrace();
}
Also use throw for throw a new exception

Assuming to really do want to display the exception, and not a more user friendly message, the first step is probably to get the exception as a string. Then you can do what you like with that string (echo to console, place in a javax.swing.JTextArea, email it, etc).
If you just want the message from the exception, then getMessage() will do:
try { ... }
catch(FooException e) {
String msg = e.getMessage();
}
However, if you want the whole exception, stack trace and all, you'll want something like this:
public static String stackTrace(Exception e) {
StringWriter w = new StringWriter();
e.printStackTrace(new PrintWriter(w));
return w.toString();
}
// ...
try { ... }
catch(FooException e) {
String st = stackTrace(e);
}
If you just want to echo the full stack trace to the console, there is the printStackTrace(), no-arg method:
try { ... }
catch(FooException e) {
e.printStackTrace();
}
If you want to take more control of the presentation of the stack trace you can get the details with:
try { ... }
catch(FooException e) {
StackTraceElement[] stes = e.getStackTrace();
// Do something pretty with 'stes' here...
}

You can just print a error message to the user saying that the input is wrong using a simple if.
if(yourCondition){
// Happy scenario
// Go shead
}else{
// Error Scenario
System.out.println("Error. Invalid Input.");
// If you persist to throw an exception, then you can do something like this
// throw new Exception("Exception Explanation"); // I've commented this, but you can uncomment it if needed
// But the advice is to display an error message, than throw a exception.
}
And regarding the conversion, you can convert binary to decimal like this
int i = Integer.parseInt(binaryString, 2); // 2 indicates the radix here, since you want to convert from binary.

With this code snippet you can convert the String into an integer :
int numberAsInt;
try {
numberAsInt = Integer.parseInt(number);
} catch (NumberFormatException e) {
//Will throw an Exception
}
If you want to create your own Exception class, you can do it like shown here or just throw a RuntimeException with
throw new RuntimeException("Your Message");

My opinion, you can try some thing like following
public static void main(String[] args) {
if(!isTrueZeckonderNumber()){
// your message should goes here
System.out.println("Your message");
}
}
If you really want to throws an exception do following
private static boolean isTrueZeckonderNumber() throws Exception{
System.out.println("Enter a Zeckonder Number:");
number = scanner.nextLine();
if (number.equals("10100")) {
return true;
} else{
throw new Exception("your message");
}
}

What do you mean you want to display an exception?
I would suggest just giving the user feedback instead, as exceptions are used more commonly for EXCEPTIONAL actions that are not supposed to happen.
However if you do want to, you can print a message explaining what happened.
try {
} catch (Exception e) {
System.out.println(e.getMessage());
}

try, catch and if error in a combo box with java and netbeans

I assure you all that this is not homework.
I have a combo box:
try {
double input = Double.valueOf(input1.getText());
double output = 0;
My combo box code here:
if (output < 2.50){
answerField.setText("£2.50");
}
else{
answerField.setText("£" + String.valueOf (fmt.format(output)));
catch (Exception e) { JOptionPane.showMessageDialog(this,
"Please enter numbers in the fields.
Numbers are 1 2 3 4 5 etc.",
"Oooops!!!",
JOptionPane.ERROR_MESSAGE);
}
The problem I have is that when I run it like this it won't work.
if I take the if statement out and leave try catch in, it works
and also if I take the try and catch out it and leave only the if, it also works
When I run it using the above, I get this error:
catch without try
; expected
Can anyone help?

you must close your else bracket, and then your try bracket:
try {
if (..) {
...
} else {
...
}
} catch (Exception e) {..}