Varargs in method overloading in Java

Question

The following code doesn't compile.

package varargspkg;  public class Main {      public static void test(int... i) {         for (int t = 0; t < i.length; t++) {             System.out.println(i[t]);         }          System.out.println("int");     }      public static void test(float... f) {         for (int t = 0; t < f.length; t++) {             System.out.println(f[t]);         }          System.out.println("float");     }      public static void main(String[] args) {         test(1, 2);  //Compilation error here quoted as follows.     } } 

A compile-time error is issued.

reference to test is ambiguous, both method test(int...) in varargspkg.Main and method test(float...) in varargspkg.Main match

It seems to be obvious because the parameter values in the method call test(1, 2); can be promoted to int as well as float

If anyone or both of the parameters are suffixed by F or f, it compiles.

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If we however, represent the receiving parameters in the method signature with respective wrapper types as follows

public static void test(Integer... i) {     System.out.println("Integer" + Arrays.asList(i)); }  public static void test(Float... f) {     System.out.println("Float" + Arrays.asList(f)); } 

then the call to the method test(1, 2); doesn't issue any compilation error. The method to be invoked in this case is the one that accepts one Integer varargs parameter (the first one in the preceding snippet).

Why is in this case the error as in the first case not reported? It appears that auto-boxing and automatic type promotion are both applied here. Is auto-boxing applied first so that the error is resolved?

The Oracle docs says,

Generally speaking, you should not overload a varargs method, or it will be difficult for programmers to figure out which overloading gets called.

The last sentence in this link. It's however for the sake of better understanding varargs.

Also to add below code compiles just fine.

public class OverLoading {      public static void main(String[] args) {         load(1);     }      public static void load(int i) {         System.out.println("int");     }      public static void load(float i) {         System.out.println("float");     } } 

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EDIT:

The following is the snap shot that indicates the compilation error. I have created a new application therefore the package name is different.

I'm using JDK 6.

Answer 1

You can either Widen or Box but you cannot do both, unless you are boxing and widening to Object (An int to Integer(Boxing) and then Integer to Object(Widening) is legal, since every class is a subclass of Object, so it is possible for Integer to be passed to Object parameter)

Similarly an int to Number is also legal (int -> Integer -> Number) Since Number is the super class of Integer it is possible.

Let's see this in your example: -

public static void test(Integer...i)  public static void test(Float...f) 

There are some rules that are followed when selecting which overloaded method to select, when Boxing, Widening, and Var-args are combined: -

1. Primitive widening uses the smallest method argument possible
2. Wrapper type cannot be widened to another Wrapper type
3. You can Box from int to Integer and widen to Object but no to Long
4. Widening beats Boxing, Boxing beats Var-args.
5. You can Box and then Widen (An int can become Object via Integer)
6. You cannot Widen and then Box (An int cannot become Long)
7. You cannot combine var-args, with either widening or boxing

So, based on the above given rules: -

When you pass two integers to above functions,

• according to rule 3, it will have to be first Widened and then Boxed to fit into a Long, which is illegal according to rule 5 (You cannot Widen and then Box).
• So, it is Boxed to store in Integer var-args.

But in first case, where you have methods with var-args of primitive types: -

public static void test(int...i) public static void test(float...f) 

Then test(1, 2) can invoke both the methods (Since neither of them is more suitable for rule 1 to apply) : -

• In first case it will be var-args
• In second case, it will be Widening and then Var-args (which is allowed)

Now, when you have methods with exactly one int and one flost: -

public static void test(int i) public static void test(float f) 

Then on invoking using test(1), rule 1 is followed, and smallest possible widening (i.e. the int where no widening is needed at all) is chosen. So 1st method will be invoked.

For more information, you can refer to JLS - Method Invocation Conversion

Answer 2

In Java, 1 is how you represent an int. It can be either auto-boxed to an instance of Integer or promoted to float, and this explains why the compiler can't decide on the method it should call. But it will never be auto-boxed to Long or Float (or any other type).

On the other hand, if you write 1F, it is the representation of a float, that can be auto-boxed to a Float (and, in the same spirit, will never be auto-boxed to an Integer or anything else).