Reason that Java casts to long instead of using the varargs paramter with the correct type?

Asked Oct 24 '19 at 17:15

Active Oct 24 '19 at 17:15

Viewed 26 times

I have a method with parameters (byte... a) and have it overloaded with (long a, long b). Calling the method with two bytes casts it them to long rather than using the provided (byte... a) method with the correct type. I would expect it to use the method with the correct type rather than casting it. My question is what make Java choose long over byte for this ambiguous case?

public static void  main(String[] args) {

        byte b = 3;
        foo(b);
        foo(b,b);
        foo(b,b,b);     
    }
    static void foo(byte... a) {

        for(byte elem:a) {
            System.out.print(elem);
        }
        System.out.println();
    }
    static void foo(long a, long b) {
        System.out.println("Java casts these as longs");
    }

The output of this is

3
Java casts these as longs
333

asked Oct 24 '19 at 17:15

danwardvs

1

yes because you don't have `foo` method that accepts `1` and `3` arguments – Ryuzaki L Oct 24 '19 at 17:17
varargs have the lowest execution priority. It should also be understood that unboxing is less priority than expanding to a wider primitive type. – Arthur Kupriyanov Oct 24 '19 at 17:24

Reason that Java casts to long instead of using the varargs paramter with the correct type?

0 Answers0