zakSyed zakSyed - 12 days ago 9
Java Question

How is an overloaded method chosen when a parameter is the literal null value?

I came across this question in a quiz,

public class MoneyCalc {

public void method(Object o) {
System.out.println("Object Verion");
}

public void method(String s) {
System.out.println("String Version");
}

public static void main(String args[]) {
MoneyCalc question = new MoneyCalc();
question.method(null);
}
}


The output of this program is "String Version". But I was not able to understand why passing a null to an overloaded method chose the string version. Is null a String variable pointing to nothing ?

However when the code is changed to,

public class MoneyCalc {

public void method(StringBuffer sb) {
System.out.println("StringBuffer Verion");
}

public void method(String s) {
System.out.println("String Version");
}

public static void main(String args[]) {
MoneyCalc question = new MoneyCalc();
question.method(null);
}
}


it gives a compile error saying "The method method(StringBuffer) is ambiguous for the type MoneyCalc"

Answer

Is null a String variable pointing to nothing ?

A null reference can be converted to an expression of any class type. So in the case of String, this is fine:

String x = null;

The String overload here is chosen because the Java compiler picks the most specific overload, as per section 15.12.2.5 of the JLS. In particular:

The informal intuition is that one method is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time type error.

In your second case, both methods are still applicable, but neither String nor StringBuffer is more specific than the other, therefore neither method is more specific than the other, hence the compiler error.