Function overloading concepts

Question

I am a little doubtful about the output of the following code:
public class Test {
     void c(Object a){}
     void c(Float a){}
     void c(Number a){}
     void m(){
          c(null);
     public static void main(String[] args) {
          Test t = new Test();
          t.m();
I an not able to find out the reason why the Float version of the c() will be called?

Answer 1

DeltaGeek wrote:
Hendawy wrote:
The reason behind this is that a null value is a void reference to, i.e. a pointer to nothing. Since it is a pointer (address), it is actually a number. That's why the float version is best matching in your code.Wow...just....not even close.
When determining what method to call, the compiler will find the most specific method it can that matches the arguments you pass in. Since you're passing in an uncast null, it can match any reference type.
In this case, Float happens to be the most specific reference type it could use. This is also why adding an overload for Integer will cause a compilation error, as you now have multiple methods with the maximum level of specificity.
Also note that casting the null to one of Float or Integer would resolve the compile error but hide the design problem.But i have a doubt; if i follow your approach then this code must work:
public class Test {
     void c(Object a){}
     void c(Float a){}
     void c(Number a){}
     void c(DerivedInteger a){}
     void m(){
          c(null);
     public static void main(String[] args) {
          Test t = new Test();
          t.m();
class MyInteger extends Number{
     public double doubleValue() {
          return 0;
     public float floatValue() {
          return 0;
     public int intValue() {
          return 0;
     public long longValue() {
          return 0;
class DerivedInteger extends MyInteger{
because U is the most specific one,