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

See JLS Section 15.12.2, for rules compiler follows to determine which method to invoke. Compiler always chooses the most specific method in case your methods are overloaded:

There may be more than one such method, in which case the most specific one is chosen. The descriptor (signature plus return type) of the most specific method is one used at run time to perform the method dispatch.

Compiler first tries to resolve the method without boxing or unboxing as quoted there:

The first phase (§15.12.2.2) performs overload resolution without permitting boxing or unboxing conversion, or the use of variable arity method invocation. If no applicable method is found during this phase then processing continues to the second phase.

Emphasis mine.

So, in your 1^st code, since short can be used as argument for parameter of type int. Compiler will not use the method with parameter Short, as that requires boxing. While in case of long type, since it cannot be used as argument for type int, it goes for boxing it to Long. Remember Widening is preferred over Boxing.

In your 2^nd, there is no other way than boxing int to Integer. So, it calls method with Integer parameter.