Couldn’t understand the output of this code?

Question

     public class Test {

   public void method(String param)
   {
       System.out.println("String version");
   }

   public void method(StringBuffer param)
   {
       System.out.println("String Buffer");
   }

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

This code result is compilation error says “reference to method is ambiguous”

    public class Test
{
    public void method1(Object param)
    {
        System.out.println("Object Version ");
    }

    public void method1(String param)
    {
        System.out.println("String Version ");
    }

    public static void main(String[] args)
    {
        Test test=new Test();
        test.method1(null);
    }
}

This code result is “String Version”

Actually I can’t understand the result of second piece of code. Why don’t both piece of code has the same result

score 2 · Accepted Answer · answered Aug 10 '13 at 13:24

2

In first case,

null is a subtype of every other reference type. So, compiler finds ambiguity in deciding which method to call..

In second case, it finds more specific object for null which happens to be String. Hence it calls method1 and prints String Version

answered Aug 10 '13 at 13:24

Vikas V

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But also Object accept null value. Why does compiler choose String?? – Yasmin Aug 10 '13 at 13:30
1

String extends object and thus is more specific – Andrew Martin Aug 10 '13 at 13:31
@Yasmin [Java Docs](http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.12.2) tells that compiler always chooses most specifi method. String is more specific that Object. – Vikas V Aug 10 '13 at 13:32
and as String and StringBuffer both of them extend from Object class the compiler counters ambiguous Right?? – Yasmin Aug 10 '13 at 13:37
@Yasmin: Exactly. The compiler wouldn't know which was more specific so would throw up an error. – Andrew Martin Aug 10 '13 at 13:54

Couldn’t understand the output of this code?

1 Answers1