Unit - 5

Overloading and Templates

Q1) Explain function overloading with examples

A1)

Function Overloading

Function Overloading is defined as the process of having two or more function with the same name, but different in parameters is known as function overloading in C++. In function overloading, the function is redefined by using either different types of arguments or a different number of arguments. It is only through these differences compiler can differentiate between the functions.

The advantage of Function overloading is that it increases the readability of the program because you don't need to use different names for the same action.

Function Overloading Example

Let's see the simple example of function overloading where we are changing number of arguments of add() method.

// program of function overloading when number of arguments vary.

1. #include <iostream>    
2. Using namespace std;    
3. Class Cal {    
4.    Public:    
5. Static int add(int a,int b){      
6.        Return a + b;      
7.    }      
8. Static int add(int a, int b, int c)      
9.    {      
10.        Return a + b + c;      
11.    }      
12. };     
13. Int main(void) {    
14.    Cal C;                                                    //     class object declaration.   
15.    Cout<<C.add(10, 20)<<endl;      
16.    Cout<<C.add(12, 20, 23);     
17.   Return 0;    
18. }    

Output:

30

55

Let's see the simple example when the type of the arguments vary.

// Program of function overloading with different types of arguments.

1. #include<iostream>  
2. Using namespace std;  
3. Int mul(int,int);  
4. Float mul(float,int);  
5.  
6.  
7. Int mul(int a,int b)  
8. {  
9.    Return a*b;  
10. }  
11. Float mul(double x, int y)  
12. {  
13.    Return x*y;  
14. }  
15. Int main()  
16. {  
17.    Int r1 = mul(6,7);  
18.    Float r2 = mul(0.2,3);   
19.    Std::cout << "r1 is : " <<r1<< std::endl;  
20.    Std::cout <<"r2 is : "  <<r2<< std::endl;  
21.    Return 0;  
22. }  

Output:

r1 is : 42

r2 is : 0.6 

Q2) Explain Function Overloading and Ambiguity in detail with examples

A2)

When the compiler is unable to decide which function is to be invoked among the overloaded function, this situation is known as function overloading.

When the compiler shows the ambiguity error, the compiler does not run the program.

Causes of Function Overloading:

• Type Conversion.
• Function with default arguments.
• Function with pass by reference.

• Type Conversion:

Let's see a simple example.

1. #include<iostream>  
2. Using namespace std;  
3. Void fun(int);  
4. Void fun(float);  
5. Void fun(int i)  
6. {  
7.    Std::cout << "Value of i is : " <<i<< std::endl;  
8. }  
9. Void fun(float j)  
10. {  
11.    Std::cout << "Value of j is : " <<j<< std::endl;  
12. }  
13. Int main()  
14. {  
15.    Fun(12);  
16.    Fun(1.2);  
17.    Return 0;  
18. }  

The above example shows an error "call of overloaded 'fun(double)' is ambiguous". The fun(10) will call the first function. The fun(1.2) calls the second function according to our prediction. But, this does not refer to any function as in C++, all the floating point constants are treated as double not as a float. If we replace float to double, the program works. Therefore, this is a type conversion from float to double.

• Function with Default Arguments

Let's see a simple example.

1. #include<iostream>  
2. Using namespace std;  
3. Void fun(int);  
4. Void fun(int,int);  
5. Void fun(int i)  
6. {  
7.    Std::cout << "Value of i is : " <<i<< std::endl;  
8. }  
9. Void fun(int a,int b=9)  
10. {  
11.    Std::cout << "Value of a is : " <<a<< std::endl;  
12.    Std::cout << "Value of b is : " <<b<< std::endl;  
13. }  
14. Int main()  
15. {  
16.    Fun(12);  
17.   
18.    Return 0;  
19. }  

The above example shows an error "call of overloaded 'fun(int)' is ambiguous". The fun(int a, int b=9) can be called in two ways: first is by calling the function with one argument, i.e., fun(12) and another way is calling the function with two arguments, i.e., fun(4,5). The fun(int i) function is invoked with one argument. Therefore, the compiler could not be able to select among fun(int i) and fun(int a,int b=9).

• Function with pass by reference

Let's see a simple example.

1. #include <iostream>  
2. Using namespace std;  
3. Void fun(int);  
4. Void fun(int &);   
5. Int main()  
6. {  
7. Int a=10;  
8. Fun(a); // error, which f()?  
9. Return 0;  
10. }  
11. Void fun(int x)  
12. {  
13. Std::cout << "Value of x is : " <<x<< std::endl;  
14. }  
15. Void fun(int &b)  
16. {  
17. Std::cout << "Value of b is : " <<b<< std::endl;  
18. }  

The above example shows an error "call of overloaded 'fun(int&)' is ambiguous". The first function takes one integer argument and the second function takes a reference parameter as an argument. In this case, the compiler does not know which function is needed by the user as there is no syntactical difference between the fun(int) and fun(int &).

Q3) Explain Operators Overloading

A3)

Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type. For example, C++ provides the ability to add the variables of the user-defined data type that is applied to the built-in data types.

The advantage of Operators overloading is to perform different operations on the same operand.

Operator that cannot be overloaded are as follows:

• Scope operator (::)
• Sizeof
• Member selector(.)
• Member pointer selector(*)
• Ternary operator(?:)

Syntax of Operator Overloading

1. Return_type class_name  : : operator op(argument_list)  
2. {  
3.     // body of the function.  
4. }  

Where the return type is the type of value returned by the function.

Class_name is the name of the class.

Operator op is an operator function where op is the operator being overloaded, and the operator is the keyword.

Rules for Operator Overloading

• Existing operators can only be overloaded, but the new operators cannot be overloaded.
• The overloaded operator contains atleast one operand of the user-defined data type.
• We cannot use friend function to overload certain operators. However, the member function can be used to overload those operators.
• When unary operators are overloaded through a member function take no explicit arguments, but, if they are overloaded by a friend function, takes one argument.
• When binary operators are overloaded through a member function takes one explicit argument, and if they are overloaded through a friend function takes two explicit arguments.

Q4) Give Operators Overloading Example

A4)

Let's see the simple example of operator overloading in C++. In this example, void operator ++ () operator function is defined (inside Test class).

// program to overload the unary operator ++.

1. #include <iostream>    
2. Using namespace std;    
3. Class Test    
4. {    
5.   Private:    
6.      Int num;    
7.   Public:    
8.       Test(): num(8){}    
9.       Void operator ++()         {     
10.          Num = num+2;     
11.       }    
12.       Void Print() {     
13.           Cout<<"The Count is: "<<num;     
14.       }    
15. };    
16. Int main()    
17. {    
18.    Test tt;    
19.    ++tt;  // calling of a function "void operator ++()"    
20.    Tt.Print();    
21.    Return 0;    
22. }    

Output:

The Count is: 10

Let's see a simple example of overloading the binary operators.

// program to overload the binary operators.

1. #include <iostream>  
2. Using namespace std;  
3. Class A  
4. {  
5.    
6.    Int x;  
7.      Public:  
8.      A(){}  
9.    A(int i)  
10.    {  
11.       x=i;  
12.    }  
13.    Void operator+(A);  
14.    Void display();  
15. };  
16.  
17. Void A :: operator+(A a)  
18. {  
19.     
20.    Int m = x+a.x;  
21.    Cout<<"The result of the addition of two objects is : "<<m;  
22.  
23. }  
24. Int main()  
25. {  
26.    A a1(5);  
27.    A a2(4);  
28.    a1+a2;  
29.    Return 0;  
30. }  

Output:

The result of the addition of two objects is : 9

Q5) What is Function Template?

A5)

The general form of a template function definition is shown here −

Template <class type> ret-type func-name(parameter list) {

   // body of function

}

Here, type is a placeholder name for a data type used by the function. This name can be used within the function definition.

The following is the example of a function template that returns the maximum of two values

#include <iostream>

#include <string>

Using namespace std;

Template <typename T>

Inline T const& Max (T const& a, T const& b) {

   return a < b ? b:a;

}

Int main () {

   int i = 39;

   int j = 20;

   cout << "Max(i, j): " << Max(i, j) << endl;

   double f1 = 13.5;

   double f2 = 20.7;

   cout << "Max(f1, f2): " << Max(f1, f2) << endl;

   string s1 = "Hello";

   string s2 = "World";

   cout << "Max(s1, s2): " << Max(s1, s2) << endl;

   return 0;

}

If we compile and run above code, this would produce the following result −

Max(i, j): 39

Max(f1, f2): 20.7

Max(s1, s2): World

Q6) What is Class Template?

A6)

Just as we can define function templates, we can also define class templates. The general form of a generic class declaration is shown here −

Template <class type> class class-name {

   .

   .

   .

}

Here, type is the placeholder type name, which will be specified when a class is instantiated. You can define more than one generic data type by using a comma-separated list.

Following is the example to define class Stack<> and implement generic methods to push and pop the elements from the stack −

#include <iostream>

#include <vector>

#include <cstdlib>

#include <string>

#include <stdexcept>

Using namespace std;

Template <class T>

Class Stack {

   private:

      vector<T> elems;    // elements

   public:

      void push(T const&);  // push element

      void pop();               // pop element

      T top() const;            // return top element

      bool empty() const {      // return true if empty.

         return elems.empty();

      }

};

Template <class T>

Void Stack<T>::push (T const& elem) {

   // append copy of passed element

   elems.push_back(elem);   

}

Template <class T>

Void Stack<T>::pop () {

   if (elems.empty()) {

      throw out_of_range("Stack<>::pop(): empty stack");

   }

   // remove last element

   elems.pop_back();        

}

Template <class T>

T Stack<T>::top () const {

   if (elems.empty()) {

      throw out_of_range("Stack<>::top(): empty stack");

   }

   // return copy of last element

   return elems.back();     

}

Int main() {

   try {

      Stack<int>         intStack;  // stack of ints

      Stack<string> stringStack;    // stack of strings

      // manipulate int stack

      intStack.push(7);

      cout << intStack.top() <<endl;

      // manipulate string stack

      stringStack.push("hello");

      cout << stringStack.top() << std::endl;

      stringStack.pop();

      stringStack.pop();

   } catch (exception const& ex) {

      cerr << "Exception: " << ex.what() <<endl;

      return -1;

   }

}

If we compile and run above code, this would produce the following result −

7

Hello

Exception: Stack<>::pop(): empty stac

Q7) What is polymorphism explain with examples?

A7)

The term "Polymorphism" is the combination of "poly" + "morphs" which means many forms. It is a greek word. In object-oriented programming, we use 3 main concepts: inheritance, encapsulation, and polymorphism.

Real Life Example Of Polymorphism

Let's consider a real-life example of polymorphism. A lady behaves like a teacher in a classroom, mother or daughter in a home and customer in a market. Here, a single person is behaving differently according to the situations.

There are two types of polymorphism in C++:

• Compile time polymorphism: The overloaded functions are invoked by matching the type and number of arguments. This information is available at the compile time and, therefore, compiler selects the appropriate function at the compile time. It is achieved by function overloading and operator overloading which is also known as static binding or early binding. Now, let's consider the case where function name and prototype is same.

1.   Class A                                  //  base class declaration.  
2.  {  
3.       Int a;  
4.       Public:  
5.       Void display()  
6.       {   
7.             Cout<< "Class A ";  
8.        }  
9.  };  
10. Class B : public A                       //  derived class declaration.  
11. {  
12.    Int b;  
13.    Public:  
14.   Void display()  
15.  {  
16.        Cout<<"Class B";  
17.  }  
18. };  

In the above case, the prototype of display() function is the same in both the base and derived class. Therefore, the static binding cannot be applied. It would be great if the appropriate function is selected at the run time. This is known as run time polymorphism.

• Run time polymorphism: Run time polymorphism is achieved when the object's method is invoked at the run time instead of compile time. It is achieved by method overriding which is also known as dynamic binding or late binding.

Q8) Differences b/w compiles time and run time polymorphism.

A8)

Q9) Give example of run time polymorphism

A9)

Let's see a simple example of run time polymorphism in C++.

// an example without the virtual keyword.

1. #include <iostream>    
2. Using namespace std;    
3. Class Animal {    
4.    Public:    
5. Void eat(){      
6. Cout<<"Eating...";      
7.    }        
8. };     
9. Class Dog: public Animal      
10. {      
11. Public:    
12. Void eat()      
13.    {           cout<<"Eating bread...";      
14.    }      
15. };    
16. Int main(void) {    
17.   Dog d = Dog();      
18.   d.eat();    
19.   Return 0;    
20. }    

Output:

Eating bread...

Q10) Give example of run time polymorphism in C++ where we are having two derived classes.

A10)

Let's see another example of run time polymorphism in C++ where we are having two derived classes.

// an example with virtual keyword.

1. #include <iostream>    
2. Using namespace std;    
3. Class Shape {                                        //  base class  
4.    Public:    
5. Virtual void draw(){                             // virtual function  
6. Cout<<"drawing..."<<endl;      
7.    }        
8. };     
9. Class Rectangle: public Shape                  //  inheriting Shape class.  
10. {      
11. Public:    
12. Void draw()      
13.   {      
14.       Cout<<"drawing rectangle..."<<endl;      
15.    }      
16. };    
17. Class Circle: public Shape                        //  inheriting Shape class.  
18.  
19. {      
20. Public:    
21. Void draw()      
22.   {      
23.      Cout<<"drawing circle..."<<endl;      
24.   }      
25. };    
26. Int main(void) {    
27.    Shape *s;                               //  base class pointer.  
28.    Shape sh;                               // base class object.  
29.       Rectangle rec;    
30.        Circle cir;    
31.      s=&sh;    
32.     s->draw();     
33.        s=&rec;    
34.     s->draw();      
35.    s=?    
36.    s->draw();     
37. }    

Output:

Drawing...

Drawing rectangle...

Drawing circle...