Why is Java picking this overload over the other

Question

Could anyone explain to me why Java is picking the second overload instead of the first?

public static void foo (int a, double b, double...c) {}
public static void foo (double...a) {}
public static void bar ()
{
    // this is the second
    foo(1);
}

I thought when I pass 1 as the argument, Java would have picked the first argument because int is more specific than double?

Thanks

Ok if `1` was passed to `int`, what would go into the `double` type which is 2nd argument? — Rohit Jain, Oct 17 '13 at 20:06

score 5 · Accepted Answer · answered Oct 17 '13 at 20:06

The second method is the only one that can match. You have only 1 argument. The first foo has at least two required: an int and a double, so that can't match.

The second foo matches because any number of numbers can match. Java will implicitly promote your int 1 to a double so it can match, with method invocation conversion.

Why is Java picking this overload over the other

1 Answers1