Var arg methods causing error in client code

Question

I was trying to figure out the ins-and-outs of varargs and wrote the following code

public class VarArgTest {

/**
 * @param args
 */
public static void main(String[] args) {
    VarArgTest v = new VarArgTest();
    //Code 1
    System.out.println("haha");
}

public void m1(Integer a, Integer... ints) {
    System.out.println("1");
}

public void m1(Integer... ints) {
    System.out.println("2");
}
}

Now if I replace Code 1 by

v.m1(new Integer(1), new Integer(2));

I get the following message

The method m1(Integer, Integer[]) is ambiguous for the type VarArgTest

My question is, why throw the error at client code? Should it not be caught by compiler while defining the two m1 methods?

this could be a help: http://stackoverflow.com/questions/12879910/varargs-in-method-overloading-in-java — Habib, Apr 12 '13 at 05:17
@Habib Thanks for the link. But my question still stands, can the compiler not issue a warning at least? — Chetter Hummin, Apr 12 '13 at 05:22

gk5885 · Answer 1 · 2013-04-12T05:21:45.577

4

It's not an error in the overloads because there are still unambiguous invocations of those two methods.

v.m1(new Integer[] {1, 2})

and

v.m1(1, new Integer[] {2})

would both work just fine. It only becomes ambiguous when the compiler has to figure out how to turn the arguments in the invocation into arrays.

edited Apr 12 '13 at 05:21

answered Apr 12 '13 at 05:16

gk5885

3,742
21
16

Fair enough, but surely the compiler can give a warning? – Chetter Hummin Apr 12 '13 at 05:23
Take that part up with javac. :-) – gk5885 Apr 12 '13 at 05:26
Heh. May just need to do that. – Chetter Hummin Apr 12 '13 at 05:46

score 0 · Answer 2 · answered Apr 12 '13 at 05:15

0

That means that those two methods cannot be distinguished when you would call

 v.m1(1,2,3,4)

compilator does not know which method should be used.

You need to have different parameters (method signatures) when you overload.

e.g. this would be ok for compilator :

   public void m1(String a, Integer... ints) {
    System.out.println("1");
   }

   public void m1(Integer... ints) {
    System.out.println("2");
   }

answered Apr 12 '13 at 05:15

Martin V.

3,560
6
31
47

Agreed. I would not recommend my code above. The question is why Java compiler allows that without even a warning. – Chetter Hummin Apr 12 '13 at 05:24

Var arg methods causing error in client code

2 Answers2