How the Compiler Chooses the Most Specific Method from Several Versions of an Overloaded Method

Description

Demo Code

public class Main {
  public double add(int a, int b) {
    System.out.println("Inside add(int a, int b)");
    double s = a + b;    
    return s;/*from   w ww . j a  va2  s  .co  m*/
  }
  
  public double add(double a, double b) {
    System.out.println("Inside add(double a, double b)");
    double s = a + b;    
    return s;
  }

  public void test(Employee e) {
    System.out.println("Inside test(Employee e)");
  }

  public void test(Manager e) {
    System.out.println("Inside test(Manager m)");
  }
  
  public static void main(String[] args) {
    Main ot = new Main();

    int i = 10;
    int j = 15;
    double d1 = 1.4;
    double d2 = 2.5;
    float f1 = 1.3F;
    float f2 = 1.5F;
    short s1 = 2;
    short s2 = 6;
    
    ot.add(i, j);
    ot.add(d1, j);
    ot.add(i, s1);
    ot.add(s1, s2);
    ot.add(f1, f2);
    ot.add(f1, s2);
    
    Employee emp = new Employee();
    Manager mgr = new Manager();
    ot.test(emp);
    ot.test(mgr);
    
    emp = mgr;
    ot.test(emp);
  }
}

class Employee {
  private String name = "Unknown";

  public void setName(String name) {
    this.name = name;
  }

  public String getName() {
    return name;
  }

  public boolean equals(Object obj) {
    boolean isEqual = false;

    // We compare objects of the Employee class with the objects of
    // Employee class or its descendants
    if (obj instanceof Employee) {
      // If two have the same name, consider them equal.
      Employee e = (Employee) obj;
      String n = e.getName();
      isEqual = n.equals(this.name);
    }

    return isEqual;
  }
}

class Manager extends Employee {
  // No code is needed for now
}

How the Compiler Chooses the Most Specific Method from Several Versions of an Overloaded Method - Java Object Oriented Design

Description

Demo Code

Result

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