Java 7 overloading with varargs

Question

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bug with varargs and overloading?

could anyone explain me how this one works:

class Vararg {
    static void vararg(int... x) {
        System.out.println("Integer..."); 
    }

    static void vararg(long... x) { 
        System.out.println("long..."); 
    }

    public static void main(String [] args) {
        int s = 0;
        vararg(s,s);
    }
}

Get compile time error

class Vararg {
    static void vararg(Integer... x) {
        System.out.println("Integer..."); 
    }

    static void vararg(long... x) {
        System.out.println("long..."); 
    }

    public static void main(String [] args) {
        int s = 0;
        vararg(s,s);
    }
}

Also get compile time error. What is the mechanism when we use overloading with varargs? Is it a bug with overloading varargs methods?

your runtime parameters to the method are ambigous – TheWhiteRabbit Feb 05 '13 at 13:31 — TheWhiteRabbit, Feb 05 '13 at 13:31

assylias · Answer 1 · 2013-02-05T13:59:10.677

In substance, to determine which method is applicable, the compiler runs a few steps:

first trying to find a method without using boxing/unboxing or varargs
second trying to find a method, allowing boxing / unboxing, but without varargs
third allowing boxing, unboxing and varargs

In your case, the third step is applicable to all methods.

The compiler then determines which methods are applicable and if one is more specific than another. The detailed rules are in the JLS 15.12.2.4. In short, it looks at the types and checks if one is more specific than another (i.e. one type is a subclass of the other for references or one type is narrower than another for primitives).

In your case:

in example 1, both methods are applicable but int is more specific than long so the vararg(int...) is chosen
in example 2, Integer has no specificity relationship with long (one is a reference the other is a primitve), so both are maximally specific and there is an ambiguity which leads to a compile error.

EDIT

I thought you were saying that your first example compiles but not the second (which is the expected behaviour). You seem to be saying that neither compiles, in which case it is probably due to a bug in you version of javac, which has been fixed with Java 7. See details in the release notes, section called "Changes in Most Specific Varargs Method Selection".

The specific RFE is http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6199075 — tucuxi, Feb 05 '13 at 14:11

score 0 · Answer 2 · answered Feb 05 '13 at 13:40

As you have sent int value in vararg(type... var) method and the overloaded methods contain one Integer and one long so, int value gets automatically casted to long and hence create ambiguity. use different parameters type in overloaded methods.

class Vararg {
static void vararg(Integer... x) {
    System.out.println("Integer..."); 
}

static void vararg(String... x) {
    System.out.println("String..."); 
}

public static void main(String [] args) {
    int s = 0;
    vararg(s,s);
}

}

Java 7 overloading with varargs

2 Answers2

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