Most specific Method-Overloading??

[Joe San]

Guys,

I tried the code below,

public class Main {
  static void m1(boolean b1) {System.out.print("boolean ");}
  static void m1(byte b1) {System.out.print("byte ");}
  static void m1(short s1) {System.out.print("short ");}
  static void m1(char c1) {System.out.print("char ");}
  static void m1(int i1) {System.out.print("int ");}
    public static void main (String[] args)  {
     short b1; m1(b1 = 1); m1(b1); m1(b1 == 1);
  }
}

And it gave me short short boolean as the o/p. From the definition of the most specific method, it is one in which the arguments that we pass can be passed to the other methods without a compile time error. Now in the above case, anything that can be passed to a short cannot be passed to a byte, so shouldn't it give a compile time error??

[Tommaso Nuccio]

Hi,

1. short b1;
2. m1(b1 = 1);
3. m1(b1);
4. m1(b1 == 1);

in line 2. b1 is assigned 1, so b1 is a short with a value of 1.
in line 3. you call m1() with a short again, so it calls m1(short s1) and prints "short".
The point is, b1 doesn't become a byte in that code.

Did I get your problem right?

Ciao,
Tommaso

[Arpit Khandelwal]

i guess this is the most basic case of methode overloading , if you are passing specificaly a "short" type of variable you would expect
"static void m1(short s1)" function to be called and not a function that takes a byte or a integer parameter , of course would there have been no any m1(short s1) kind of function then static void m1(int i1) might be called.

I hope i have understood the question correctly.