Try-Catch-Throw in the same Java class

Question

Is it possible to catch a method in the current class the try-catch block is running on? for example:

  public static void arrayOutOfBoundsException(){
      System.out.println("Array out of bounds");
  }

    .....

  public static void doingSomething(){
    try
    {
       if(something[i] >= something_else);
    }
    catch (arrayOutOfBoundsException e)
    {
       System.out.println("Method Halted!, continuing doing the next thing");
    }
  }

If this is possible how will it be the correct way to call the catch method?

If this is not possible, could anyone point me in the right direction, of how to stop an exception from halting my program execution in Java without having to create any new classes in the package, or fixing the code that produces ArrayOutOfBoundsException error.

Thanks in Advance,

A Java Rookie

Answer 1

I think you mean to have the arrayOutOfBoundsException method throw the exception caused... If i understand correctly:

public static void arrayOutOfBoundsExceptionMethod() throws ArrayIndexOutOfBoundsException  {
      \\...
      \\ code that throws the exception
  }

    .....

  public static void doingSomething(){
    try
    {
       if(something[i] >= something_else);
       arrayOutOfBoundsExceptionMethod();
    }
    catch (ArrayIndexOutOfBoundsException e)
    {
       System.out.println("Method Halted!, continuing doing the next thing");
    }
  }

Is that what you are asking?

Answer 2

What you are wanting to do is handle an Exception.

public static void doingSomething(){
    try {
        if (something[i] >= something_else) { ... }
    } catch (ArrayIndexOutOfBoundsException e) {
        System.out.println("Method Halted!, continuing doing the next thing");
    }
}

That's all you need. No extra classes, no extra methods.

An Exception is a special type of class that can be "thrown" (you can throw it yourself by using the throw keyword, or Java may throw one for you if, for example, you try to access an array index that does not exist or try to perform some operation on a null). A thrown exception will "unwrap" your call stack ("escaping" from each function call) until your program finally terminates. Unless you catch it, which is exactly what the syntax above does.

So if you were writing a function a() that called a function b() that called a function c() and c() threw an exception, but the exception was not caught in b() or c(), you could still catch it in a():

void a() {
    try {
        b();
    catch (SomeExceptionClass e) {
        // Handle
    }
}

That said, if it is possible to prevent an exception from being thrown in the first place, that is often a better idea. In your particular case, this would be possible since all arrays in Java know their own length:

public static void doingSomething(){
    if (i >= something.length) {
        System.out.println("Method Halted!, continuing doing the next thing");
    } else {
        if (something[i] >= something_else) { ... }
    }
}

Answer 3

Hmm.. I think you are a little bit confused. You don't catch a method with try catch blocks. You catch Exceptions. Exceptions are more like things.

This alone (note the capital "A" of ArrayOutOfBoundsException) will prevent your program to terminate even if that Exception is thrown. You don't need to declare a "arrayOutOfBoundsException()" method.

public static void doingSomething(){
    try
    {
       if(something[i] >= something_else);
    }
    catch (ArrayOutOfBoundsException e)
    {
       System.out.println("Method Halted!, continuing doing the next thing");
    }
  }

Note the little "e"? That's the reference you can use to refer to this Exception. So, you can ask things to this Exceptions using this little local variable. For example, if you want to print where the exception happened, you can do e.printStackTrace(); .

Answer 4

I'm trying to understand what you're asking, your terminology is wrong in places :) If you try and set an index out of bounds, you'll get a built-in exception anyway, ArrayIndexOutOfBoundsException. You can catch this and use this inside the method, or throw it using a class XX throws ArrayIndexOutOfBoundsException declaration in the header of the method, so that the invoking method can catch it (its not required to catch an ArrayIndexOutOfBoundsException as it's a RuntimeException -- i.e., nothing happens if it goes uncaught, it's unchecked.

See this wiki article for more details: http://en.wikibooks.org/wiki/Java_Programming/Throwing_and_Catching_Exceptions.

Answer 5

You catch Exceptions, with a try-catch block. Not sure if you can catch methods with this.

public static void myMethod(){
    try {
        // check for some given condition
    }
    catch (conditionNotSatisfied e) {
        System.out.println("Condition was not satisfied, you tried to do something which is not allowed");
    }
}

The 'e' in the catch() is the reference used for referring to an Exception. Exception refers to this when it wishes to check your given condition and return you a warning or error about the condition not getting satisfied.

Answer 6

It is not clear what you mean by "Is it possible to catch a method". In your case, you should write your code as :

   if(i < something.length && something[i] >= something_else);

No ArrayOutOfBoundsException is needed.