Method overloading Problem

[Mohit G Gupta]

TestCafe 3.0 by sahir shah

  class Rectangle{
           public int area(int length , int width) {
             return  length * width; 
           }
          }

         class Square extends Rectangle{
            public int area(long length , long width) {
            return  (int) Math.pow(length ,2);
           }
         }
        class Test{
          public static void main(String args[]) {
           Square r = new Square();
           System.out.println(r.area(5 , 4)); 
          }   
        }  

ANswer:

This code will fail to compile because the
compiler cannot resolve the method call here.

why cannot resolve method call ??
can't it call the method with int as arguments as they much narrower er

[Ram Narayan.M]

Mohit... Your answer is right... The answer given above is wrong... Better you execute the program and come to conclusion... Believing the answer theortically does not lead to exploration of concepts...

[Abimaran Kugathasan]

This will compile fine, but, the answer is some what different. The concept behind this is called as choose the most specific method invocation.

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

The informal intuition is that one method is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time type error.

Have a look on this JLS

[Trivikram Kamat]

Hey, the code does compile and displays 20 in the output which is returned by method public int area(int length , int width);
Java version 1.6.0_21