Java exception handling returning no value

Question

I'm trying to understand how I can use java exception handling to return just the exception message when the input is invalid. As far as I understand I have to use return out of try catch, to get this compiled (or in both). But actually I don't want to return anything if the input parameter is invalid.

If I was dealing with strings, there would be null. But that doesn't seem to work with int.

Is there no way doing this?

public class Arrays {

public static int[] intArray = {1,2,3,4,5,60,7,8,9,10};


public static int arrayGet(int[] array, int i){
    try{
        return intArray[i];

    }
    catch (ArrayIndexOutOfBoundsException e){
        System.out.println("Please enter number between 0 and "+i);
    }
}


public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println(arrayGet(intArray,11));
}
}

The code my not make a lot of sense, but I'd like to understand how to deal with the general situation.

you want to return a number if successful, otherwise string. Correct me if that is not the case. — Mritunjay
– Mritunjay, Commented Dec 11, 2016 at 14:42
I think, you should think of escalating the exception to main and catch in main and print. — Mritunjay
– Mritunjay, Commented Dec 11, 2016 at 14:48
If you want to return the exception message, why catch the exception at all? Just let the caller deal with it. You could wrap it in a checked exception and re-throw too, so you force the caller to deal with it. — Jorn Vernee
– Jorn Vernee, Commented Dec 11, 2016 at 14:51
that makes sense. But, this is a student assignment and we're asked to provide a code for a method, where there's a handling of boundary. Am I right that I can't do this within a method except printing out some message and let the method run into an exception. — Zin Yosrim
– Zin Yosrim, Commented Dec 11, 2016 at 14:55

Gurwinder Singh · Accepted Answer · 2016-12-11 15:01:24Z

1

At present your arrayGet is not compilable. If you want to return a String in case of error you can do this in catch block of arrayGet like

} catch (ArrayIndexOutOfBounds e) {
    throw new Exception("my message");
}

And in the main method

try {
    int i = arrayGet(11);
} catch (Exception e) {
    String msg = e.getMessage();
}

answered Dec 11, 2016 at 15:01

Gurwinder Singh

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1 Comment

Zin Yosrim Over a year ago

I had to change the method declaration a bit public static int arrayGet(int[] array, int i) throws Exception and your code worked fine for me. thx

itssme · Accepted Answer · 2016-12-11 14:47:49Z

1

You will have to return some value (which must be an int). With Strings you don't have such a problem because String is not a Standard data type and therefore can be Null.But here you will have to return some value in the catch statement.

answered Dec 11, 2016 at 14:47

itssme

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Jorn Vernee · Accepted Answer · 2016-12-11 15:08:08Z

The common super type of int and String is Object (after boxing). That means you can return a String or int if you declare the return type as Object:

public static Object arrayGet(int[] array, int i){
    try{
        return intArray[i];    
    }
    catch (ArrayIndexOutOfBoundsException e){
        System.out.println("Please enter number between 0 and "+i);
        return e.getMessage();
    }
}

But the caller will not know which type was returned, so they can only really use it as an Object.

Object o = arrayGet(array, 11);
// o is maybe an int, or maybe a String. But it's definitely an Object.

In this case the arguments of the method are the culprit. A way to let the caller know is by throwing an IllegalArgumentException:

public static int arrayGet(int[] array, int i){
    if(i < 0 || i >= array.length)
        throw new IllegalArgumentException("Please enter number between 0 and " + i);

    return intArray[i];
}

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