null as a parameter to method

[Sandip Shinde]

I have following code

public class Tester
{

//main method
public static void main(String[] args)
{
painter(null);///look here
}

public static void painter(Object obj)
{
System.out.println("Object Method");
}

public static void painter(String str)
{
System.out.println("String method");
}

}


now I see "String method" as output for this program , on which basis String parameter method is called while I am passing null as parameter?

[Abdulla Mamuwala]

Here you have method painter() which you have overloaded in the same class Tester. Generally the term overloaded is used in case of inheritance, that is when we have a base class and a sub class, and than in the sub class we overlaod the method from the base class. Anyways,

Generally in the case of an overloaded method, the more specific of the two methods is selected. In your case the parameter which you are passing to your methods is "null".

Now, String is a subclass of Object, therefore "null" is more applicable to String rather than Object class.

Lets take another example to clarify it, lets replace "null" by a literal "8". Refer to the code below:

public class Tester
{
 
//main method
public static void main(String[] args)
{
painter(8);///look here
}
 
public static void painter(long l)
{
System.out.println("Long selected");
}
 
public static void painter(int i)
{
System.out.println("Integer selected");
}
 
} 

The output of this code will be Integer selected, although both the painter methods are applicable, the integer method is selected becuase it is more specific of the two.

[Joel McNary]

Note that the "most specific" method is chosen. That (essentially) means that the class that is most removed from the Object class. What happens if you have two classes that are equally specific?

public class Tester {
    //main method
    public static void main(String[] args) {
        painter(null); ///look here
    }
 
    public static void painter(Object obj) {
        System.out.println("Object Method");
    }
 
    public static void painter(String str) {
        System.out.println("String method");
    }
 
    //New Method; StringBuffer is just as specific as String
    public static void painter(StringBuffer str) {
        System.out.println("StringBuffer method");
    }
}

You get this error:

Tester.java:4: reference to painter is ambiguous, both method
painter(java.lang.String) in Tester and method 
painter(java.lang.StringBuffer) in Tester match
        painter(null); ///look here
        ^
1 error

[Roger Chung-Wee]

Generally the term overloaded is used in case of inheritance, that is when we have a base class and a sub class, and than in the sub class we overlaod the method from the base class. Anyways,

Replace the words "overloaded" with "overridden" and tell us if what you originally said is correct.

[Joel McNary]

The output of this code will be Integer selected, although both the painter methods are applicable, the integer method is selected becuase it is more specific of the two

Actually, it is because you are passing an integer. Specificality doesn't apply to primitives, as they do not extend one another. There's no confusion to the compiler because a literal 8 is always and forever an integer. To get a literal 8 as a long, you owuld have to use 8L.

[Abdulla Mamuwala]

Catch your point guys, thanks, sorry for using the wrong terminology Sandeep.

[Sandip Shinde]

Thanks for great reply from u all,
Got the exact basis for overloded methods invocation(more specific one will be called .. clear enough)

Thanks again