About Question enthuware.ocajp.i.v7.2.1100 :

[zel_bl]

This text helped
http://javabypatel.blogspot.com/2016/05 ... ading.html

[asi-aal]

void probe(long x) { System.out.println("In long"); } //3

void probe(Long x) { System.out.println("In LONG"); } //4

public static void main(String[] args){
Integer a = 4; new TestClass().probe(a); //5
int b = 4; new TestClass().probe(b); //6
}

Both will result "in long" can you explain me the idea behind the first one "Integer a = 4; new TestClass().probe(a)" ?

[admin]

Integer is not a subclass of Long, so probe(a) cannot be bound to probe(Long x).

But an Integer can be unboxed to an int and int can be promoted to long (not Long). Therefore, probe(long x) will be invoked.

You might want to go through a book to understand how method selection works in case of overloading.

Section 10.2.3 of this book explains it very clearly: https://amzn.to/2PucBeT