Method ambiguity

[pramod talekar]

Hello,

Can anyone tell me why 'ss' and not 'oo' is printed after below code's execution

public class ambg {
	public void aa(Object o) {
		System.out.println("oo");
	}
	
	public void aa(String s) {
		System.out.println("ss");
	}
	
	public static void main(String[] args) {
		ambg a = new ambg();
		String test="gg";
		a.aa(test);
	}
}

[Joseph Mokenela]

It is because you are sending a String as an argument (test), so the method that takes the string parameter will be the one that gets called, hence will print oo.

[Campbell Ritchie]

The rules about overloading are not easy to understand; look at the §15.12 link in that page I quoted. As an over‑simplification, if there are two alternatives, the compiler chooses the “more specific” parameter type. So String takes precedence over Object because it as a sub-class and therefore more specific.

[Nitish Bijalwan]

Campbell Ritchie wrote:The rules about overloading are not easy to understand; look at the §15.12 link in that page I quoted. As an over‑simplification, if there are two alternatives, the compiler chooses the “more specific” parameter type. So String takes precedence over Object because it as a sub-class and therefore more specific.

That is correct , first complier or jvm (not sure) chooses method that has more speific parameter as argument if it doesn't matches than it chooses next alternativei.e next paramenter in which argumanet can be fit

[Campbell Ritchie]

Overloading: version chosen by compiler at compile time.
Overriding: version chosen by JVM at run-time.

[pramod talekar]

Thanks Campbell.

This question was asked in an interview and the interviewer said that the compiler will throw an error due to ambiguity.

[Campbell Ritchie]

pramod talekar wrote:Thanks Campbell.

This question was asked in an interview . . .

You’re welcome

Why didn’t you say it was an interview question? I thought it was a real-life question

[pramod talekar]

I replicated it in real time and found no ambiguity [:-p]

[Aditya Jha]

pramod talekar wrote:This question was asked in an interview and the interviewer said that the compiler will throw an error due to ambiguity.

The ambiguity will come only when you call the method with a null literal.

// Compiler is not happy
a.aa(null);

Please note that it has to be the literal. Even passing a null String reference will enable compiler to select the right method (and hence it will not report any problem).

// No problems here
String test = null;
a.aa(test);

Lastly, sorry for being too finicky (programmer's brain's OCD), but "compiler will throw an error" does not make sense at all. Compiler can only report compilation problems. It can not throw any Error or Exception.

[pramod talekar]

Thanks Aditya.

Even if I pass in 'null' and run the program in Eclipse, it prints ss and not oo.
If I put in a String literal with null value, still it prints ss.

Only when I comment out the String argument method, it prints oo.
Does running the program through command line print something different ?

[Panagiotis Kalogeropoulos]

When you invoke an overloaded method, the compiler tries to find which of these methods is "closer" (or more specific, as Campbell already mentioned) to the arguments that you are passing. So when you are passing null as argument, the compiler is thinking: "Hmmmm, he gave me a null argument and I have a method that accepts a String and an Object. String is more specific that Object, so I will invoke the method that accepts String as an argument."

For instance, try to execute the following program:

public class Test {

    public static void main(String[] args)  {
       String s = "test";
        Integer i = new Integer(10);//not a String, but an Object
        Object o = new Object();
        Test t = new Test();
        System.out.print("1)");
        t.aa(s);
        System.out.print("2)");
        t.aa(i);
        System.out.print("3)");
        t.aa(o);
        

    }    
    
    public void aa(Object o) {  
        System.out.println("oo");  
    }  
      
    public void aa(String s) {  
        System.out.println("ss");  
    }
}

The output will be:

1)ss
2)oo
3)oo

Result 1) comes from the invocation of the aa(String s) and 3) from the aa(Object o). The compiler tried to find the most specific method related to the arguments. The same applies to result 2) where we have an Intreger as an argument. The most specific method that can be found is aa(Object o) -- remember that an Integer is not a String but it is an Object --. As an exercise, try to experiment with different methods and arguments. For example, can you find what will be printed when we execute the following program? (try to guess before you run it and see the actual results)

public class Test {

    public static void main(String[] args)  {
        String s = "test";
        Integer i = new Integer(10);
        Double d = new Double(10d);
        char c = 's';
        Object o = new Object();
        Test t = new Test();
        t.aa(s);
        t.aa(i);
        t.aa(d);
        t.aa(c);
        t.aa(o);


    }

    public void aa(Object o) {
        System.out.println("oo");
    }

    public void aa(String s) {
        System.out.println("ss");
    }
    
    public void aa(Number n) {
        System.out.println("nn");
    }
    
    public void aa(Integer i) {
        System.out.println("ii");
    }
}