i am confusing about Overloading

[YUE ZHU]

K&B p240:compiler choose old style before it choose new one,
but this get me confusing,how about this output?

public class Confusing {
     private Confusing(Object o) {
         System.out.println("Object");
     }

// private Confusing(double[] dArray) {
// System.out.println("double array");
// }
     private Confusing(double...ds ){
         System.out.println("double ...");
     }

     public static void main(String[] args) {
         new Confusing(null);
     }
}

old beat vararg? No.
//output
double ...

Somebody help!

Best,
Yue

[ December 28, 2006: Message edited by: YUE ZHU ]

[ December 28, 2006: Message edited by: YUE ZHU ]
[ January 10, 2007: Message edited by: YUE ZHU ]

[Aniket Patil]

A method is said to be maximally specific for a method invocation if it is applicable and accessible and there is no other applicable and accessible method that is more specific.

If there is exactly one maximally specific method, then it is in fact the most specific method; it is necessarily more specific than any other method that is applicable and accessible.

It is possible that no method is the most specific, because there are two or more maximally specific method declarations. It's not possible to identify a most-specific method from among the applicable methods. In this case, we say that the method invocation is ambiguous, and a compile-time error occurs.

The precise definition of the most specific property is as follows. Let m be a name and suppose that there are two declarations of methods named m, each having n parameters. Suppose that one declaration appears within a class or interface T and that the types of the parameters are T1, . . . , Tn; suppose moreover that the other declaration appears within a class or interface U and that the types of the parameters are U1, . . . , Un . Then the method m declared in T is more specific than the method m declared in U if and only if both of the following are true:

T can be converted to U by method invocation conversion.
Tj can be converted to Uj by method invocation conversion, for all j from 1 to n.

A slight change to your program,

  
public class Confusing {
private Confusing(Object o) {
System.out.println("Object");
}
 
private Confusing(Double dArray) {
System.out.println("double array");
}
private Confusing(double...ds ){
System.out.println("double ...");
}
 
public static void main(String[] args) {
new Confusing(null);
}
}  

And an ambiguity error results:
javac Confusing.java
Confusing.java:14: reference to Confusing is ambiguous, both method Confusing(java.lang.Double) in Confusing and method Confusing(double...) in Confusing match new Confusing(null);
^
1 error

[YUE ZHU]

Thank you Aniket. Your explanation is clear.

Best,
Yue

[YUE ZHU]

I run this fine,and outputs:
long

class ConfusingAgain {
static void f(int i,Integer integer,long l){
System.out.println("Integer");
}

static void f(int i,long ll,        long l){
System.out.println("long");
}
 
/**
 * @param args
 */
public static void main(String[] args) {
f(5,5,5);
 
}
 
}    

but,when I add a parameter more, it's go wrong:

class ConfusingAgain {
static void f(int i,Integer integer,long l,Integer ii){
System.out.println("Integer");
}

static void f(int i,long ll,        long l,Integer ii){
System.out.println("long");
}
 
/**
 * @param args
 */
public static void main(String[] args) {
f(5,5,5,5);
 
}
 
}    

It shows ambiguous, I don't know why? Anyone help?

Best,
Yue

[Anton Uwe]

mhhh... if you change the calling code to f(5,5,5,new Integer(5)) it works... Could someone explain this behavior please? I'm confused, too.

[YUE ZHU]

Bug of Java Compiler? :-)

Anyone help?

Best,
Yue