Correct version of a method.

[Tejaswini Shirkhedkar]

public class AQuestion
{
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[])
{
AQuestion question = new AQuestion();
question.method(null);
}
}
Answers
1 The code does not compile.
2 The code compiles cleanly and shows "Object Version".
3 The code compiles cleanly and shows "String Version"
4 The code throws an Exception at Runtime.

The answer given is 3.
The default value for object reference is also null as it is for an uninitialized string reference.Why does, then, the compiler does not complain about ambiguous method call?
Thanks!

Teju

[arnel nicolas]

From my understanding, both methods are applicable not actually ambigous as you have mentioned. The compiler will look for the most specific method that can accomodate the value passed in the parameter method.

Example on an ambigous call:


public class Ambigous{


void setNumber(long n) {
System.out.println(n);
}

void setNumber(int n) {
System.out.println(n);
}

public static void main(String args[]) {
new Ambigous().setNumber(2);
}
}

--Hope this help

[arnel nicolas]

Sorry , i need to make a few correction with my example for ambiguous class above. Those codes will work fine. See my example below:

public class Ambiguous{

class AnotherObject{}

void aMethod(Object s) {
System.out.println("Object");
}

void aMethod(String s) {
System.out.println("String");
}


void aMethod(AnotherObject s) {
System.out.println("AnotherObject");
}

public static void main(String args[]) {
new Ambiguous.aMethod(null);
}
}

Hope this will help you.

[Jong Limb]

String is derived from Object and is therefore, more specific.