overloading method output need explanation

[Athira Vineeth]

Hi

in the below example

public class OverLoadTest {
public static void main(String[] args) {
		OverLoadTest overLoadTest=new OverLoadTest();
		overLoadTest.printValue(null);
	}
public void printValue(Object ob){
	System.out.println("Object");
}
public void printValue(String ob){
	System.out.println("String");
}
}

The output of this program is "String".
Why so?Why it is not executing the printValue(Object ob) method?

[Panagiotis Kalogeropoulos]

Although both methods are applicable to your situation (both can accept null as a parameter), the most specific method will be called, or in other words, the method that is "closer" to the argument that you specify. Since a parameter in printValue(String ob) can be passed to printValue(Object ob) also, but not vice versa, then the printValue(String ob) is more specific and that is why it is called. The same rule applies to primitives too. For instance, check the following code:

public static void main(String args[]) {
        int num1 = 1;
        long num2 = 1l;
        method(num1);
        method(num2);
    }
    
    static void method(int num) {
        System.out.println("doIt int");
    }
    
    static void method(long num) {
        System.out.println("doIt long");
    }

The most specific method is called depending on the arguments.

Hope this helps!

[Campbell Ritchie]

It’s in the Java Language Specification (JLS); look for the links to §15.12, too. Note this is a complicated subject, even by the standards of the JLS.

[Athira Vineeth]

yes its help me a lot.

thank you so much for your time.