Why Do I Get the "Unhandled Exception Type Ioexception"

Why do I get the Unhandled exception type IOException?

You should add "throws IOException" to your main method:

public static void main(String[] args) throws IOException {

You can read a bit more about checked exceptions (which are specific to Java) in JLS.

Why I got Unhandled exception type IOException

Because Files.createDirectory() may throw java.io.IOException and you did neither catch it nor declare to throw it.

Catch the exception to handle errors

import java.nio.file.*;

public class JavaIO {
    public static void main(String[] args) {

        String dirString = "C:/Users/USER/Desktop/Test/Files";
        Path dirPath = Paths.get(dirString);
        if(Files.notExists(dirPath)){
            try{
                Files.createDirectory(dirPath);
            } catch(java.io.IOException e){
                System.out.println("createDirectory failed:" + e);
            }
        }
        System.out.println("Err");
        System.exit(1);
    }
}

or add declaration to throw it to ignore its possibility to be thrown.

import java.nio.file.*;

public class JavaIO {
    public static void main(String[] args) throws java.io.IOException {

        String dirString = "C:/Users/USER/Desktop/Test/Files";
        Path dirPath = Paths.get(dirString);
        if(Files.notExists(dirPath)){
            Files.createDirectory(dirPath);
        }
        System.out.println("Err");
        System.exit(1);
    }
}

Unhandled exception type IOException in Java

Alessio, In Java, you have 2 types of Exceptions :

• Exception: they are checked
• Runtime exceptions: not checked

When a method can trigger a "checked" exception, you have to handle it.

To do so you have 2 ways:

• add a "throws IOException" to the end of your method's signature (which is what the link you posted is suggesting)
• handle your exception with a try catch block.

In your case, IOException is a "checked" exception and thus you should add a "throws IOException" to your getR method or handle it with a try catch.

PS: If you are learning java, it would be better to use an IDE such as eclipse, you will understand the mistakes easier because it will point you to it with some text higlighting and it will provide you with solutions.

Unhandled Exception java.io.iOException

try {
      startRecording();
    }catch(IOException ex) {
    //Do something with the exception
}

Unhandled Exception Type IOException

Yes, IOException is a checked exception, which means you either need to catch it, or declare that your method will throw it too. What do you want to happen if the exception is thrown?

Note that you should generally be closing the reader in a finally block anyway, so that it gets closed even in the face of another exception.

See the Java Tutorial lesson on exceptions for more details about checked and unchecked exceptions.

Unhandled exception type Exception in Eclipse

You call teste.parse(someString), where teste is an expression which has type Parser. That means this is a call to the method parse(String) in your Parser type....

and that is declared with throws Exception.

Exceptions are a mechanism to convey alternate return options. The parse method can run its course in one of two ways: It can 'return', in which case it returns nothing (void), or, it can 'throw'. What it can throw is limited by its throws line - in this case, it can throw just about anything (Exception is the supertype of almost all things you can throw).

The way java handles this is that your code needs to handle every possible way a method can conclude.

So, you need a 'path' for your code when the parser() method returns (this is trivial; it's a void method, you get that 'for free', you don't need to write anything special for this), but you also need a path for that other exit scenario: When it throws something. You get handling of RuntimeException for free, but for others, you have two options:

catch it:

try {
    teste.parse(someString);
    // this code runs in the 'return' case.
} catch (Exception e) {
   // this code runs in the 'throws' case.
}

this would imply you know what to do when your parse method decided to exit via the throws path.

Alternatively, you fix this by having your main method also 'fork', and decree that it has two ways to finish: Either via the return route or the throw route:

public static void main(String[] args) throws Exception {
    teste.parse(someString);
}
// this main method has declared that it has two separate
// exit routes. 'return', and 'throws something'.

java will start an application by running its main method, and java can deal with a main that has two alternate exit routes (return, or throw something). It handles the 'return' route by doing nothing. It handles the 'throw something' route by printing the type of the exception, the message, the stack trace, and the entire causal chain. That is an excellent default, and you should not attempt to come up with a different one by e.g. catching that exception and attempting to 'log it'.

This: Just add throws Exception to your main method declaration. Put the throws Exception back on your parse method, ignore @Eritrean's advice.

NB: All methods are inherently declared as if they said throws RuntimeException, Error (as in, any error and any runtimeexception can be thrown without writing a throws clause for it, as all methods implicitly have that clause baked in already), this is why I said earlier that RuntimeExceptions are 'handled for free'. The idea is that all exceptions that subclass RuntimeException are things that are so universal or so unlikely, it would be unwieldy to force management of this onto the programmer. That's why you never need to write throws NullPointerException or throws InternalError.

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