CODEWITHSUNDEEP
javaintbytevariadic-functionsshort
Frosted Cupcake
Frosted Cupcake

Reputation: 1970

Method Overloading with varargs

I am a bit confused about this topic, reason being in this code:

public class File
{
    public static void main(String[] args)
    {
        numbers();
    }

    static void numbers(int...x)
    {
        System.out.println("Integers");
    }
    static void numbers(byte...x)
    {
        System.out.println("Byte");
    }
    static void numbers(short...x)
    {
        System.out.println("Short");
    }
}

The output of this code is "Byte", and the reason is the most specific type is chosen, since among byte, short and int the most specific type is byte, that's why it is chosen.

But if I modify my code to-

public class File
{
    public static void main(String[] args)
    {
        numbers(1, 4);
    }

    static void numbers(int...x)
    {
        System.out.println("Integers");
    }
    static void numbers(byte...x)
    {
        System.out.println("Byte");
    }
    static void numbers(short...x)
    {
        System.out.println("Short");
    }
}

The output is "Integers", and I'm unable to figure out why? I know in case of arithmetic instructions byte and short are implicitly promoted to int, but here we are calling a method with values which is within the range of byte and short, then how the method with int arguments is invoked?

And also, if I comment out the method with int arguments, then the code shows an error of no suitable method found. Why???

Upvotes: 2

Views: 55

Answers (1)

Kayaman
Kayaman

Reputation: 73558

1 and 4 are integer literals. So the int... version is called.

There are no byte or short literals, but if you were to call Numbers((byte)2, (byte)3); the byte... version would be called.

Upvotes: 4

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