How do overloaded methods work?

Question

public class Test1  {

    public static void main(String[] args)   {
        Test1 test1 = new Test1();
        test1.testMethod(null);
    }

    public void testM         


        

Answer 1

Adding on to an existing reply, I'm not sure if this is because of a newer Java version since the question, but when I tried to compile the code with a method taking an Integer as a parameter instead of an Object, the code still did compile. However, the call with null as the parameter still invoked the String parameter method at run-time.

For example,

public void testMethod(int i){
    System.out.println("Inside int Method");     
}

public void testMethod(String s){
    System.out.println("Inside String Method");     
}

will still give the output:

Inside String Method

when called as:

test1.testMethod(null);

The main reason for this is because String does accept null as a value and int doesn't. So null is classified as a String Object.

Coming back to the question asked, The type Object is encounter only when a new object is created. This is done by either type casting null as an Object by

test1.testMethod((Object) null);

or using any type of object for a primitive data type such as

test1.testMethod((Integer) null);
    or
test1.testMethod((Boolean) null);

or by simply creating a new object by

test1.testMethod(new  Test1());

It should be noted that

test1.testMethod((String) null);

will again invoke the String method as this would create an object of type String.

Also,

test1.testMethod((int) null);
    and
test1.testMethod((boolean) null);

will give a compile time error since boolean and int do not accept null as a valid value and as int!=Integer and boolean!=Boolean. Integer and Boolean type cast to Objects of type int and boolean.