Some problems with passing arguments to overloaded functions

Question

package java_coding;

public class String_Null {
public static void main(String[] args) {
    // TODO Auto-generated method stub
    String_Null string_Null = new String_Null();
    string_Null.method(null);
}

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

public void method( Object object) {

    System.out.println("Object");
}


}

The result is String. Explain why?

If you wanted "Object", you'd have to cast the `null` to something incompatible with `String`. — Mad Physicist, Sep 07 '18 at 13:32

score 2 · Answer 1 · edited Sep 07 '18 at 13:32

2

Because the most specific method is chosen, the most specific is String (compared to Object). This is in the JLS btw... I'll dig this up shortly.

According to the JLS

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run- time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

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 error.

edited Sep 07 '18 at 13:32

Andy Turner

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answered Sep 07 '18 at 13:27

Eugene

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Could this be a comment?? – miiiii Sep 07 '18 at 13:27
It could; but it is the correct answer. – Andy Turner Sep 07 '18 at 13:30
2

@MadMan. It actually is the answer. There's really no other reason – Mad Physicist Sep 07 '18 at 13:31
@Eugene added the next sentence too, because I think it is useful to understand what "most specific" actually means. – Andy Turner Sep 07 '18 at 13:33
@AndyTurner I wanted to say thank you, but you you *were not present* here to tag. thank you – Eugene Sep 07 '18 at 13:34
got it...!! the quotes says it all..!! – miiiii Sep 07 '18 at 13:34

Some problems with passing arguments to overloaded functions

1 Answers1