Java异常处理不返回任何值

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.

Answer 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.

Answer 2

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();
}

Answer 3

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];
}