Review (background knowledge):   how Java determine which method to use

  • Java determines which method to call/use by match the method signature with the the actual parameters

    Example: 

       public class Overload
   {
      public static void main(String[] args)
      {
         meth(1);          // Calls meth( int x )
	 meth(1.0);        // Calls meth( double x )
	 meth(1, 1);       // Calls meth( int x, int y )
      }

      public static void meth( int x )
      {
   	 System.out.println("Running: meth( int x ) ");
      }
    
      public static void meth( double x )
      {
   	 System.out.println("Running: meth( double x ) ");
      }
    
      public static void meth( int x, int y )
      {
   	 System.out.println("Running: meth( int x, int y ) ");
      }
   }

DEMO: 04-inheritance/11-overload/Demo2.java

Overloading vs. overriding

  • Overloading = defining different methods with the same (method) name but with different (method) signatures

    Example: 

    public class NewClass extends SomeClass
{
    NewClass(int a)
    {
        super(a);
    }

    // Inherits: method1()  and method2()

    public void method1(int a) // Overloads method1()
    {
        System.out.println("NewClass.m1(int)");
    }

    public void method3()
    {
        method1();   // Invokes SomeClass method1
	method1(22); // Invoked NewClass method1(int)  
    }
} 
    public class SomeClass
{
    public int x;

    public SomeClass(int a)
    {
        x = a;
    }

    public void method1( )
    {
        System.out.println("SomeClass.method1()");
    }
    
    public void method2( )
    {
        System.out.println("SomeClass.method2()");
    }
}

DEMO: demo/04-inheritance/11-overload    (Trace with BlueJ)

Overloading vs. overriding

  • Overriding = replacing an inherited method by defining a method with the same (method) signature

    Example: 

    public class NewClass extends SomeClass
{
    NewClass(int a)
    {
        super(a);
    }

    // Inherits: method1()  and method2()

    public void method1()      // Overrides method1()
    {
        System.out.println("NewClass.m1()");
    }

    public void method3()
    {
        method1();       // Invokes NewClass method1   
	super.method1(); // Invokes SomeClass method1
    }
} 
    public class SomeClass
{
    public int x;

    public SomeClass(int a)
    {
        x = a;
    }

    public void method1( )
    {
        System.out.println("SomeClass.method1()");
    }
    
    public void method2( )
    {
        System.out.println("SomeClass.method2()");
    }
}

DEMO: demo/04-inheritance/12-override    (Trace with BlueJ)

Quiz 1: on overloading vs. overriding

  • What is printed by the following program that invokes a.method3(): 

    public class NewClass extends SomeClass
{
    // Inherits: method1(double) 

    public void method1(int x)    
    {
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0);
	method1(4);
    }
} 

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: ???
    }
}

  • Note:   be careful... did we overload or did we override method method1(double) ?

 

Quiz 1: on overloading vs. overriding

  • Answer: 

    public class NewClass extends SomeClass
{
    // Inherits: method1(double) 

    public void method1(int x)    
    {   // Overloading
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0); // method1(double)
	method1(4);   // method1(int)
    }
} 
    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // SomeClass.m1()
    }                   **********
}

  • Overloaded method:   method1(int) is a different method than method1(double)

DEMO: demo/04-inheritance/13-quiz1/

Quiz 2: on overloading vs. overriding

  • What is printed by the following program that invokes a.method3(): 

    public class NewClass extends SomeClass
{
    // Inherits: method1(double) 

    public void method1(double x)    
    {
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0);
	method1(4);
    }
} 

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: ???
    }
}

  • Note:   be careful... did we overload or did we override method method1(double) ?

 

Quiz 2: on overloading vs. overriding

  • Answer: 

    public class NewClass extends SomeClass
{
    // Inherits: method1(double) 

    public void method1(double x)    
    {   // Override
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0); // method1(double)
	method1(4);   // method1(double)
    }                 //         ^^^^^ 
}                     // int --> double

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: *********
    }                           *********
}

  • Note:   since there is no method with signature method1(int), Java will cast int --> double to match

DEMO: demo/04-inheritance/14-quiz2/

Java's indication for overriding

  • You can add the special override annotation @Override before an overriding method for clarity: 

    public class NewClass extends SomeClass
{
    // Inherits: method1(double) 
    @Override
    public void method1(double x)    
    {
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0);
	method1(4);
    }
} 

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: ???
    }
}

  • The Java compiler will report an error if the defined method does not override any inherited methods !

DEMO: demo/04-inheritance/14-quiz2/ --- edit NewClass and re-compile

Additional conditions on overriding methods

  • The overriding method must have the same return type as the overridden method 

    public class NewClass extends SomeClass
{
    // Overrides method1(double)  
    // must have same return type 
    public void method1(double x)    
    {
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0);
	method1(4);
    }
} 

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: ???
    }
}

  • You will get a type incompatible error when you use different return types

DEMO: demo/04-inheritance/14-quiz2/ --- change type void to int in NewClass method1 and re-compile

Additional conditions on overriding methods

  • The overriding method should have the same accessibility modifier as the overridden method 

    public class NewClass extends SomeClass
{
    // Overrides method1(double)   
    // should use same access spec 
    public void method1(double x)    
    {
        System.out.print("***********"); 
        System.out.println("x = " + x );

    }

    public void method3()
    {
        method1(4.0);
	method1(4);
    }
} 

    public class SomeClass
{
    public void method1(double x)
    {
        System.out.print("SomeClass.m1(). ");
        System.out.println("x = " + x);
    }
} 

public class myProg
{
    public static void main(String[] args)
    {
        NewClass a = new NewClass();

        a.method3(); // Output: ???
    }
}

  • Complicated errors can result when you do not use the same accessibility,

DEMO: demo/04-inheritance/14-quiz2/ --- change public to private in NewClass method1 --> compile error