How JVM finds method (parameter with the closest matching) to call in case of function overloading

Question

The JVM decides which overloaded method to call at compile time. I have one example:

public class MainClass{

  public static void go(Long n) {System.out.pr         


        

Answer 1

Java looks for the closest match first. It tries to find the following:

1. Exact match by type
2. Matching a super class type
3. Converting to a larger primitive type
4. Converting to an auto boxed type
5. Varargs