Why does ambiguity not raised for ArithmeticException parameter method

Question

Below code executed with out any compilation error of ambiguity, and output is ArithmeticException. Can you help me to know the reason?

class Test {

    public static void main(String[] args) throws UnknownHostException, IOException {
        testMetod(null);
    }

    // Overloaded method of parameter type Object
    private static void testMetod(Object object) {
        System.out.println("Object");
    }

    // Overloaded method of parameter type Exception
    private static void testMetod(Exception e) {
        System.out.println("Exception");
    }

    // Overloaded method of parameter type ArithmeticException
    private static void testMetod(ArithmeticException ae) {
        System.out.println("ArithmeticException");
    } 
}

Answer 1

In cases like this the rule is to match the most specific method. Since ArithmeticException extends Exception and ArithmeticException extends Object, there is no ambiguity: ArithmeticException is more specific than any of the others.

If you add this though:

private static void testMetod(String string) {
    System.out.println("String");
}

You will get a compilation error because neither ArithmeticException extends String is true, nor the other way around: there is no single most specific parameter class.

It's probably important to say at this point that all this is happening at compile time. Once the target method is resolved and the code is compiled, a call to an overloaded method is just like a call to any other method. This is in contrast with method overriding.