Dr. P.S. Ahluwalia (Course Director) In Service Course 2009 For PGT (Comp. Sci) K.V.No.2, Jaipur

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1 Study Material Class XII (Comp.Sc.) :: ::

2 Rapid and major metamorphoses in the realm of the Computer world affected every dimension of our life. Today we are living into an Information age and modern IT practices is shaping our children. Children today need free access to technology and guidance in the responsible and effective use of the technology. The influence of technology forced CBSE to introduce Computer Science, Informatics Practices and Web Technologies at Senior Secondary Level. This booklet is specially designed tool for slow learners to cope with the complexity of the subject. It covers all the most relevant topics and questionnaires in the light of CBSE Examination. This book is outcome of In-service Course-2009 for PGT (Computer Science) conducted at K.V.No.2, Jaipur. I am very much thankful to Dr. K.P. Chamola, Assistant Commissioner, KVS, and Sri H.C. Chawala, Education Officer, Kendriya Vidyalaya of Jaipur Region for providing the opportunity to conduct In-service Course for PGT (Computer Science) in this Vidyalaya. I am also thankful to Mr. Aslam Parvez (Associate Director), Mr. Sanjay Gupta, Mr. Y. Rohilla, being the Resource Persons and all the PGT Comp.Sci who have attended the In-Service Course-2009, without whose contribution and assistance this material would not have seen the light of the day. Dr. P.S. Ahluwalia (Course Director) In Service Course 2009 For PGT (Comp. Sci) K.V.No.2, Jaipur Study Material Class XII (Comp.Sc.) :: 2::

3 Table of Contents. C++ Revision Tour 3 2. Object Oriented Programming 0 3. Function Overloading 3 4. Classes and Objects 5 5. Constructors and Destructors 2 6. Inheritance Arrays 3 8. Database Concepts and SQL Boolean Algebra Data Communication & Network Concepts 55 Study Material Class XII (Comp.Sc.) :: 3::

4 Chapter : C++ Revision Tour Key Points Variables:-Variables represent named storage location whose values can be manipulated during program run. For eg. int A; A is a variable name that will be use to store an integer value. Keywords: The keywords are also identifiers but can not be user defined since they are reserved words. For eg: int,float, void etc are keywords that have special meaning. Data types:- Data types are means to identify the type of data and associates operation for handling it. There are two types of data types: Fundamental data types, Derived data types Fundamental data types are int, float, char, double Derived Data types:- From the fundamental types other can be derived by using the declaration operators. Ex: Array, structure and Pointer. Operators in c++ The symbols that perform some predefined operation on operands. There are five types of operators: Arithmetic Operators: +, -, *, /, % Relational Operators: >, <, >=, <=,!=, == Assignment Operators: = Increment/Decrement Operators: ++, -- Logical operator : &&,,! Flow of Control To control the flow of the program there are three types of statements are used these are Selection Statements, Iteration Statements, Jumping Statements Selection Statements: - These types of statements are only used for decision making. There are two types of selection statements named if(two way branching), switch(multiway branching. Iteration Statements: - The iteration statement allows a set of instructions to be performed until a certain condition is fulfilled. It is also called loop. There are three types of loops as follows:- for, while, do For loop:-for loop is used to repeat a set of statements for specified no. of times. All the three parts i.e, initialization, condition checking, increment and decrement operations of the loop are listed in one line. While loop: while loop is used when we want to repeat a set of statements till a particular condition is true. In case of while we have to put all the three parts in different lines. Do..while loop: In case of Do while loop first the body of the loop will be executed and then the expression will be evaluated. Now if the expression is evaluate to true then the body of the loop will be executed again otherwise the control will move out of the loop body. Jumping: The jump statements unconditionally transfer control anywhere in the program. There are three types of jumping statements as follows: goto, break, continue goto:-the goto statement is used to alter the program sequence by transferring the control to some other part of the program. break: The break statement is used to break the loop statements or switch-case statement and the control comes out of the loop. Study Material Class XII (Comp.Sc.) :: 4::

5 Arrays continue: continue statement is used to skip the current iteration and continue with next iteration Array is a collection of homogenous data type under the same name and stored in the continue manner. There are two types of array as follows Structure Single Dimensional Array: The simplest form of array is single dimensional arrays. An array definition specifies a variable name along with size to specify how many data items the array will contains. Syntax: datatype array-name[size]; Mutiple Dimensional Array: Where we specify two or more dimension. Ex: A 2D array is used to store matrix. Structure is a collection of variables under a single name. Variables can be of any data type: int, float, char etc. The main difference between structure and array is that arrays are collections of the same data type and structure is a collection of variables may be of different types under a single name. Example: Functions A function is a named unit of a group of programs statements. This unit can be invoked from other parts of the program as and when required. The general form of a fiunction definition is as given below: type function-name (parameter list) body of the function Parameters can be passed to function in two manners: call by value and call by reference. The call by value method copies the values of actual parameters into the formal parameters, that is, the function creates its own copy of argument values and then uses them. When a function is called by reference, then, the formal parameters become references to the actual parameters in the calling function. Study Material Class XII (Comp.Sc.) :: 5::

6 Questions and Answers 2 Mark Questions Q. Difference between Global and Local variable. Ans Local variable:- The local variable are the variable that are only accessible by the block of code in which it declared. These variables are also called private variables. Global Variable:-Global variables are the variables that are accessible in entire program. Global variable have global scope. Q.2 Difference between entry control and exit control loop. Ans Entry control Loop: In the case of entry control Loop first the loop expression will be checked and then the body of the loop will be executed. Example: for Loop, while Loop Exit control Loop: in the case of exit control loop first the body of the loop will be executed then the given expression will be checked. It means that the body of the loop will execute atleast once. Example: do..while loop. Q.3 what is the difference between # define and Const? Ans Both #define and const define constants, however #define can define simple constants without considering datatype, whereas const can define almost any type of constant, including constant objects of structure and classes. Q.4 Define the typedef Command with example. Ans typedef command defines a new name or an alias name for an existing type. For example, all transactions generally involve amounts which are of double type, so for a bank application, we can safely provide an alias name as amount to predefine double type, we will write typedef double amount. Now we can define any amount using the data type amount as amount loan,balance; Q.5 Differentiate between Logical error and Syntax error, with suitable example. Ans Logical Error: Logical error is an error which occurs because of wrong interpretation of logic. For Example: if by mistake we divide a number with zero. Syntax Error: A syntax error is the error that occurs when statements are wrongly written violating rules of the programming language. For example: Missing Semicolon. wrong use of Keywords. Q.6 What is ternary operator? Is there any ternary operator available in C++? Which? Ans A ternary operator requires three operands. C++ has only one ternary operator or conditional operator i.e?: The syntax of conditional operator is : (Conditional expression)? expression : expression 2; it replaces simple if else statement. Ex: large=(a>b)?a:b; Q.7 Differentiate between break and continue statements. Ans break: The break statement terminates the loop and come out of the loop. For this purpose break keyword is used. continue: continue statement forces the next iteration of the loop to take place, skipping the current iteration. For this purpose continue keyword is used. Study Material Class XII (Comp.Sc.) :: 6::

7 Q. 8(a) Name the Header file(s) that shall be needed for successful compilation of the following C++ code void main() int a[0]; for (int i=0;i<0;i++) cin>>a[i]; if(a[i]%2==0) a[i]=pow(a[i],3); else a[i]=sqrt(a[i]); if(a[i]>32767) exit(0); getch(); Ans iostream.h, math.h, process.h, conio.h Q. 9. Name the header files to which the following belong : (i) puts ( ) (ii) isalnum ( ) (iii) abs( ) (iv) strcmp( ) Ans (i) stdio.h (ii) ctype.h (iii) math.h (iv) string.h Q.0. Name the header file of C++ to which following functions belong: (i) strcat() (ii) scanf() (iii) getchar() (iv) clrscr()ans (i) string.h (ii) stdio.h (iii) stdio.h (iv) conio.h Q.. Name the header files that shall be needed for the following void main( ) char String[ ] = Peace ; cout << setw(2)<<string; Ans (i) iomanip.h (ii) iostream.h Q2. Write a structure specification that includes two structure variables-distance and time. The distance includes two variables both of type float called feet and inches. The time includes three variables all of type int called hrs, mins and secs. Intialise such a structure with values feet, 8.5 inches, 0 hrs, 5 mins, 7 secs. Ans struct distance float feet; float inches; struct time ; int hrs; int mins; int secs; ; struct tour distance d; time t; ; tour t=345.00,8.5,0,5,7; Q3. Write structure definition for structure containing the following: (i) rollno, name, grade. (ii) bookno, bookname, auther, price. Ans struct stud Study Material Class XII (Comp.Sc.) :: 7::

8 int rollno; char name[20]; char grade; ; struct book int bookno; char bookname[30]; char author[20]; int price; ; Q4 Ans What is the difference between call by value and call by reference. call by value method copies the values of actual parameters into the formal parameters, that is, the function creates its own copy of argument values and then uses them. So any change in formal parameters won t be visible in calling function. Whereas When a function is called by reference, then the formal parameters become references to the actual parameters in the calling function. Q5 Find the output of the following program : #include<iostream.h> struct MyBox int length, breadth, height; ; void Dimension(MyBox M) cout<<m.length<< x <<M.breadth<< x <<M.height; void main() MyBox B=0,5,5,B2,B3; ++B.height; Dimension(B); B3=B; ++B3.lenght; B3.breadth++; Dimension(B3); B2=B3; B2.height+=5; B2.lenght--; Dimension(B2); Ans 0 x 5 x 6 x 6 x 6 0 x 6 x Q6 find the output of the following program: #include<iostream.h> void main() int U=0,V=20 for(int i=;i<=2;i++) cout<< [] <<U++<< & <<V-5<<endl; cout<< [2] <<++V<< & <<U+2<<endl; Study Material Class XII (Comp.Sc.) :: 8::

9 Ans. [] 0 & 5 [2] 2 & 3 [3] & 6 [4] 22 & 4 Q6 Find the correct possible output from the option: #include<stdlib.h> #include<iostream.h> void main() Randomize(); Char Area[][0]= North, South, East, West ; int ToGo; for(int i=0;i<3;;i++) ToGo=random(2)+; Cout<<Area[ToGo]<< : ; Outputs: i. South : East : South : ii. North : South : East : iii. South : East : West: iv. South : East : East : Ans. (i),(iv). Q7 Rewrite the following program after removing the syntactical error(s) if any. Underline each correction. #include<iostream.h> Void main() First=0, Second=20; Jumpto(First, Second); Jumpto(second); Void Jumpto(int N, int N2=20) N=N+N2; Cout<<N>>N2; Ans. #include<iostream.h> Void main() int First=0, Second=20; Jumpto(First, Second); Jumpto(Second); Void Jumpto(int N, int N2=20) N=N+N2; Cout<<N<<N2; Q8. Find the output of the following program : #include<iostream.h> void exec(int a, int &b); void main() int a=0; Study Material Class XII (Comp.Sc.) :: 9::

10 Ans: int b=20; cout<< a<< \t <<b; exec(a,b) cout<< a<< \t <<b; void exec(int a, int &b) a=a+00; b=b+50; cout<< a<< \t <<b; Study Material Class XII (Comp.Sc.) :: 0::

11 CHAPTER-2 : Object Oriented Programming Key Points: OOP: It means Object oriented programming which views a problem in terms of object rather than procedure. Object: An object is an identifiable entity with some characteristics and behavior. Class: A class is group of similar objects that share common properties and relationship. Data Abstraction: Abstraction refers to the act of representing essential features without including the background details or explanations. Encapsulation: The wrapping up of data members and member functions (that operate on the data) into a single unit (called class) in known as Encapsulation. Modularity: The act of dividing a complete program into different individual components (functions or modules) is called modularity. Inheritance: Inheritance is the capability of one class (sub class) of things to inherit characteristics or properties from another class (base class). Polymorphism: Polymorphism is the ability for a message or data to be processed in more than one form. 2 Mark Questions Questions and Answers Q What is programming paradigms? Give names of some popular programming paradigms? Ans:- It means an approach to programming or it defines the methodology of designing and implementing programs using building blocks of a programming language. Ex: Object Oriented Programming, Procedural Programming Q 2 What is the difference between Object Oriented Programming and Procedural Programming? Ans:- Object Oriented Programming Procedural Programming. It views a problem in term of objects.. It views a problem in terms of procedure 2. Features like data encapsulation, rather than objects. polymorphism, inheritance etc. are 2. Features like data encapsulation, present. polymorphism, inheritance etc. are not 3. It follows Bottom- Up approach in present. program design. 3. It follows Top- Down approach in program design. Q3 Define an object and a class with one example of each? Ans:- An object is an identifiable entity with some characteristics and behaviour. Eg. Orange is an object. Its characteristics are: it is spherical shaped and its colour is orange & its behaviour is: it is juicy and sweet- sour taste. A class is group of similar objects that share common properties and relationship. Eg. Parrot, sparrow and pigeon are objects that belongs to the class Bird. Q4 Define Data Abstraction Concept of OOP with one example? Ans:- Data Abstraction: Abstraction refers to the act of representing essential features without including the background details or explanations (i.e. Hiding of Data). Eg. While driving a car, you have to only know the essential features to drive a car such as gear & steering handling, use of clutch, accelerator and brakes etc. and not to know about how the engine working and internal wiring etc. Q5 Define Encapsulation Concept of OOP with one example? Study Material Class XII (Comp.Sc.) :: ::

12 Ans:- Encapsulation: The wrapping up of data members and member functions (that operate on the data) into a single unit (called class) in known as Encapsulation. Eg.The data members like rollno, name, marks & members functions like getdata ( ) and putdata ( ) are enclosed in a single unit called student class. Q6 Define Inheritance Concept of OOP with one example? Ans:- Inheritance: Inheritance is the capability of one class of things to inherit characteristics or properties from another class. The Class whose properties of data members are inherited, is called Base Class or Super Class and the class that inherits these properties, is called Derived Class or Sub Class. Eg. The class Human inherits certain properties such as ability to speak, breathe, eat etc from the class Mammal as these properties are not unique to humans & class Mammal again inherits some of its properties from another class Animal. Animal Mammal Human Q7 Define Polymorphism Concept of OOP with one example? Ans:- Polymorphism (poly +morph) means many forms i.e. ability for a message or data to be processed in more than one form. It is implemented as Overloading and Virtual functions in C++. Eg. Human is a subclass of Mammal. Similarly Dog, Cat are also subclasses of Mammal. If a message see through daylight is passed to all mammals, they all will behave alike. But if a message see through darkness is passed to all mammals, then humans and dogs will not be able to see where as cats will be able to see during night also. Q8 How Data Abstraction and Encapsulation are interrelated? Ans:- Data Abstraction focuses upon the observable behaviour of an object where as Encapsulation focuses upon the implementation that gives rise to this behaviour. So, we can say that Encapsulation is a way to implement Data Abstraction. Q9 What are the advantages of inheritance? Ans:- i) It ensures the closeness with the real- world models. ii) Reusability: One can derive a sub class from an existing base class and add new features in the sub class without modifying the inherited features. iii) Transitive nature: If a class C has been declared as a sub class of B which itself is a subclass of A then C must also inherits the properties of A. i.e. A B, B C Then A C. Q0 Give two advantages and disadvantages of OOP? Ans:- Advantages: Study Material Class XII (Comp.Sc.) :: 2::

13 i) Re-use of code. ii) Its code is near to real world models. Disadvantages: i) One needs to do proper planning, skills & design for OOP programming. ii) With OOP, classes tend to overly generalized. Q How does a class enforce data-hiding and abstraction? Ans:- A class binds data members and its associated functions together into a single unit thereby enforcing encapsulation. A class groups its members into three sections : private, protected and public. The private and protected members remain hidden from outside world. Thus through private and protected members, a class enforces data-hiding. The outside world is given only the essential and necessary information through public members, Rest of the things remain hidden, which is nothing but abstraction. Q2 Give an example in C++ to show polymorphism implementation in C++. Ans:- In C++ polymorphism is implemented through function overloading. A function name having several definitions that are differentiable by the number of types of their arguments is known function overloading. Eg. float area(float a) return a*a; float area (float a, float b) return a*b; Study Material Class XII (Comp.Sc.) :: 3::

14 CHAPTER-3 : Function Overloading Key Points: Function overloading is one of the most powerful features of C++ programming language. It forms the basis of polymorphism (compile-time polymorphism).function overloading is the practice of declaring the same function with different signatures. The same function name will be used with different number of parameters and parameters of different type. But overloading of functions with different return types are not allowed. For example let us assume an Add and Display function with different types of parameters. // Sample code for function overloading void AddAndDisplay(int x, int y) cout<<"integer result: "<<(x+y); void AddAndDisplay(double x, double y) cout<< " Double result: "<<(x+y); void AddAndDisplay(float x, float y) cout<< " float result: "<<(x+y); 2 Mark Questions Questions and Answers Q: What is function overloading? Give an example illustrating is use in C++ program? Ans: A function having several definitions that are differentiable by the number or types of their arguments is known as function overloading. Example Float area (float a) Return a * a; Float area (float a, float b) Return a*b; Q2 Ans: Q3 Ans: Q4 How would you compare default arguments and function overloading? In case of default arguments, default values can be provided in the function prototype itself and the function may be called even if an argument is missing(provided the default value for the missing argument is present in the prototype).but there is one limitation if you want to default a middle argument, then all the arguments on its right must also be defaulted. We can overload a function for all possible argument combination. It not only overcome the limitation of default arguments but also the compiler is saved from the trouble of testing the default value. How does function overloading implement polymorphism? Polymorphism is the ability for a message or data to be processed in more than one form. Function overloading is implement to function having (one name and)more than one distinct meanings. Answer the questions (i) and (ii) after going through the following class: Study Material Class XII (Comp.Sc.) :: 4::

15 class player int health; int age; public: player() health=6; age=8 //Constructor player(int s, int a) health =s; age = a ; //Constructor2 player( player &p) //Constructor3 ~player() cout<< Memory Deallocate ; //Destructor ; void main() player p(7,24);//statement player p3 = p; //Statement3 (i) When p3 object created specify which constructor invoked and why? (ii) Write complete definition for Constructor3? (iii) Which feature of object oriented programming is demonstrated using costructor,constructor2, constructor3 in the above class player? Ans: (i) Copy constructor is invoked when p3 object is created, because statement3 is assignment statement in which object p3 is initialize with object p. (ii) player( player &p) health = p.health age= p.age (iii) Function overloading (constructor overloading) Study Material Class XII (Comp.Sc.) :: 5::

16 CHAPTER- 4 : Classes and Objects Key Points: Class: A class is a way to bind the data describing an entity and its associated functions together. A class makes a data type that is used to create objects of its type. Objects: An instance of a class type is known as object. Method: It is a function that is part of a class. It is used to do things. Inline function: All functions defined inside a class and the functions whose definition starts with keyword inline are inline functions. Inline functions expand its code at the function call instead of jumping to the function code. Friend function: A friend function is non member function that is granted access to class private and protected members whereas, a friend class is a class whose member functions can access another class private and protected members. Access Specifiers: A class in C++ represents a group of data and associated functions divided into one or more of these parts: Public, Private and Protected. Nesting of member functions: When a member function is called by another member function, it is called nesting of member function. 2 Mark Questions Questions and Answers Q. What is class? What is its significance? Ans. A class is a way to bind the data describing an entity and its associated functions together. Class facilitates user defined data type or a type representing similar objects. Significance: The significance of classes lies in the fact that they can model real model entities effectively in the programming form. Real world entities not only posses characteristics but they also have an associated behavior. Q2. What is the significance of access labels in a class? Ans. A class provides three access labels namely Private, Protected and public. A member declared as private remains hidden form outside world and it can only be accessed by the member functions and friend functions of the class whereas protected members can be used by its child or derived class. A member declared as public is made available to the outside world. i.e. it can be accessed by any function, any expression in the program but only by using an object of the same class type. These access labels enforce data hiding and abstraction, OOP concepts. Q3. What do you mean data members and member functions in OOPS? Ans Data Members: Data members are the data type properties that describe the characteristics of a class. There may be zero or more data members of any type in a class. Member Functions: These are the set of operations that may be applied to objects of that class. There may be zero or more member functions for a class. They are referred to as the class interface. Q4. How is a member function differ from an ordinary function? Ans Member functions have full access privilege to both the public and private members of the class while, an ordinary functions have access only to the public members of the class. Member functions are defined within the class scope that is they are not visible outside the scope of class. Ordinary functions are also visible outside the scope of class. Q5. How is memory allocated to a class and its objects? Study Material Class XII (Comp.Sc.) :: 6::

17 Ans When a class is defined, memory is allocated for its member functions and they are stored in the memory. When an object is created, separate memory space is allocated for its data members. All objects work with the one copy of member function shared by all. Q6. What are the advantage and disadvantage of inline functions? Ans Advantage: The advantage of the inline function is that they run faster than the normal functions as there will be no function calling overheads. Disadvantage: The disadvantage of inline function is that it takes more memory space because the body of the function substituted in place of function call. 4 Mark Questions Q7. Suppose a class is defined as follows answer the following questions below: class animal Private: char name[5]; int legs; Public: Void input() cin>>name>>legs; Char *getname() return name; int num_of_legs() return legs; ; (i) How many data members does the class animal have? Ans: 2 (ii) How many methods does the class animal have? Ans: 3 (iii) Ans: (iii) Ans: How to declare an object of Class Animal? Animal mydog; Write statement to print the name and the number of legs of a dog object named mydog. Cout<<mydog.getname()<< has <<mydog.num_of_legs()<< legs <<endl; Q8. Define a class student for the following specifications. Private members are : rollno integer name array of characters of size 20 class_st array of characters of size 8 marks array of integers of size 5 percentage float Write a function to calculate that calculates overall percentage marks Public members are: Study Material Class XII (Comp.Sc.) :: 7::

18 Ans: Readmarks( ) reads marks and invokes the calculate function Displaymarks( ) prints all the members of the class. class student int rollno; char name[20]; char Class_st[8]; int marks[5]; float percentage; void calculate (); public: void readmarks(); void displaydata(); ; void student :: calculate() int tot=0; for(int i=0;i<5;i++) tot=tot + marks[i]; percentage=(tot/500)*00; void student :: readmarks() cout<< Enter the roll number: ; cin>>rollno; cout<< Enter name: ; gets(name); cout<< Enter class: ; gets(class_st); cout<< Enter marks in five subjects: ; for(int i=0;i<5;i++) cin>>marks[i]; calculate(); void student :: displaydata() cout<< Roll number is : <<rollno<<endl; cout<< Name is : <<name<<endl; cout<< class is : <<class_st<<endl; cout<< Marks in five subjects : ; for(int i=0;i<5;i++) cout<<marks[i]<<endl; cout<< Percentage is : <<percentage<< % ; Q9. Define a class serial in C++ with the following specifications: Private members of class serial Serialcode integer Title 20 character Duration float Noofepisodes integer Study Material Class XII (Comp.Sc.) :: 8::

19 Public member function of class serial A constructor to initialize duration as 30 and noofepisodes as 0. Newserial() function to accept values for Serialcode and Title. Otherentries () function to assign the values of Duration and Noofepisodes with the help of corresponding values passed as parameters to this function. Dispdata() function to display all the data members on the screen. Ans. class serial int serialcode; char Title[20]; float Duration; int Noofepisodes; Public: serial() Duration =30; Noofepisodes =0; void Newserial() cout<< \n Enter the code for serial : ; cin>>serialcode; cout<< \n Enter the title of a serial : ; gets(title); void otherentries(float dur, int noe) Duration=dur; Noofepisodes=noe; ; void Dispdata () cout<< \n Serial Code : <<serialcode; cout<< \n Title : <<Title; cout<< \n Duration : <<Duration; cout<< \n No. of Episodes : <<Noofepisodes; Q. 0. A class TRAVEL with the following descriptions: Private members are : Tcode, no_of_pass, no_of_buses integer (ranging from 0 to 55000) Place string Calculate function Calculates no_of_buses according to the following rules : Number of passengers Number of buses Less than 40 Equal to or more than 40 & less than 80 2 Equal to or more than 80 3 Public members are: A function INDETAILS( ) to input all information except no_of_buses and invoke calculate function A function OUTDETAILS( ) to display all the information. Ans. class Travel unsigned int tcode; unsigned int No_of_pass; Study Material Class XII (Comp.Sc.) :: 9::

20 unsigned int no_of_buses; char Place[20]; void calculate(); public: void INDETAILS( ); void OUTDETAILS( ); ; void Travel :: calculate( ) if (no_of_pass <40) no_of_buses=; else if (no_of_pass<80) no_of_buses=2; else no_of_buses=3; void Travel :: INDETAILS( ) cout<< Enter the Travel Code <<endl; cin>>tcode; cout<< Enter the no. of passengers <<endl; cin>>no_of_pass; cout<< Enter place <<endl; gets(place); calculate(); Void Travel :: OUTDETAILS( ) cout<< Tcode <<tcode; cout<< Number of passengers <<no_of_pass; cout<< Place <<place; cout<< Number of buses required are <<no_of_buses; Q. Define a class Library with following specifications: Private Members Book_no, Book_Name, price, no_of_copies, no_of_copies_issued Public Members - Constructor() to initializes all the data members with values i.e book information - issue_book() to issue book from library by checking its availability. - return_book() to return book to library and update the data of the library. - display() to display book information. Ans: class Library int book_no; char book_name[20]; int no_of_copies; int no_of_copies_issued; public: //constructor to initialized all data members Library(int b_no, char b_name[], int nc, int ni) Study Material Class XII (Comp.Sc.) :: 20::

21 book_no=b_no; strcpy(book_name,b_name); no_of_copies=nc; no_of_copies_issued=ni; // function of issue book from library void book_issue() If no_of_copies>= no_of_copies--; no_of_copies_issued++; else cout<< \n Book not available in library ; //function to return book to the library void return_book() no_of_copies++; no_of_copies_issued--; void display() cout<< \n Book Number : <<book_no; cout<< \n Book Name : <<book_name; cout<< \n No of copies : <<no_of_copies; cout<< \n Book issued : <<no_of_copies_issued; ; //end of class Study Material Class XII (Comp.Sc.) :: 2::

22 CHAPTER- 5 : Constructor & Destructor Key Points: Constructor: Constructor is a special member function of class whose task is to initialize data members of the object. It is automatically invoked when object of class is created. Its name is same as class name. The constructor functions have certain special characteristics:- Constructor functions are invoked automatically when the objects are created. Constructor has no return type even void. They can not be inherited. A destructor can not be static. Example: class student private: int Rn, Marks; Public: Student( ) // Constructor Rn=0; Marks=0; ; Types of constructor: Constructor has three types: Default Constructor: A constructor takes no arguments is called default constructor. It is automatically invoked when an object is created without providing any initial values. In case, the programmer has not defined a constructor in the class, the compiler will automatically generate default constructor. Parameterized Constructor: A constructor can accept arguments is known as parameterized constructor. Parameterized constructor is used to initialize the data member of object with different values. Copy Constructor: A copy Constructor is used to initialize an object from another object. Copy constructor always takes argument as a reference object. Destructor: As the name implies it is used to destroy the object that has been created by the Constructor. It is used to clean up the storage that is no longer required. Like a Constructor the destructor is a member function whose name is same as the class name but it is preceded by (~ )tilde sign. It is good practice to declare destructor in a program since it release memory space for future use. When controlled exited from main function then Destructor destroy the object obj, because it is no longer accessible. Characteristics of Destructor: It is automatically invoked when control exit from program or block and destroy the object. It never takes any arguments. It has no return type It can not be inherited. A destructor can not be static. Study Material Class XII (Comp.Sc.) :: 22::

23 Questions and Answers 2 Mark Questions Q. Answer the questions (i) and (ii) after going through the following class: class Exam int year; public: Exam (int y) year=y; //Constructor Exam (Exam & t); ///Constructor 2 ; Create an object, such that it invokes Constructor. Write complete definition for Constructor 2. Ans I. Exam obj (2009); II. Exam (Exam &t) year=t.year; Q2. Answer the questions (i) and (ii) after going through the following class: class Science char Topic [20]; int Weightage; public: Science ( ) //Function strcpy (Topic, Optics ); Weightage = 30; cout<< Topic Activated ; ~Science( ) //Function 2 cout<< Topic Deactivated ; ; (i) Name the specific features of class shown by Function and Function 2 in the above example. (ii) How would Function and Function 2 get executed? Ans: (i) Function : Constructor/ Default Constructor Function 2: Destructor (ii) Function is executed or invoked automatically when an object of class Science is created. Ex: Science s; Function 2 is invoked automatically when the scope of an object of class Science comes to an end. Q3. Answer the questions (i) and (ii) after going through the following class : class Computer char C_name[20]; char Config[00]; public: Computer(Computer &obj); // function Computer(); //function 2 Computer(char *n,char *C); // function3 ; Study Material Class XII (Comp.Sc.) :: 23::

24 i. Complete the definition of the function (ii) Name the specific feature of the OOPs shown in the above example Ans: (i) Computer (Computer &obj) strcpy (C_name, obj.c_name); strcpy(config, obj.config); (ii) Constructor overloading. Q4. Answer the questions (i) and (ii) after going through the following class: class Exam int Marks; char Subject[20]; public: Exam () //Function Marks = 0; strcpy (Subject, Computer ); Exam (char S []) //Function 2 Marks = 0; strcpy(subject,s); Exam (int M) //Function 3 Marks = M; strcpy (Subject, Computer ); Exam (char S[], int M) //Function 4 Marks = M; strcpy (Subject,S); ; (i) Write statements in C++ that would execute Function 3 and Function 4 of class Exam. (ii) Which feature of Object Oriented Programming is demonstrated using Function, Function 2, Function 3 and Function 4 in the above class Exam? Ans: (i) Exam obj (75); //To execute function 3: Exam obj2 ( computer, 85); // To execute function 3 (ii) Constructor Overloading Q5. Answer the question (i) and (ii) after going through the following class. class Bazar char Type[20]; char Product[20]; int Qty; float Price; Bazar( ) strcpy(type, Electronic ); strcpy(product, Calculation ); Qty = 0; Price = 225; Study Material Class XII (Comp.Sc.) :: 24::

25 ; public: void Disp( ) cout<<type<< - <<Product << : <<Qty <<price <<endl; void main( ) Bazar B; //statement B.Disp( ); //statement2 (i) Will statement initialize all the data member for object B with the values given in the Function? (Yes OR No). Justify your answer suggesting the correction(s) to be made in the above code. (ii)what shall be the possible output when the program gets executed? (Assuming, if required the suggested corrections are made in the program) Ans. (i) No, Since the default constructor Bazar( ) is declared inside private: section, it can not initialize the objects declared outside the class. Correction needed is the constructor Bazar( ) should be declared inside public: section. (ii) Electronic-calculator 225 Study Material Class XII (Comp.Sc.) :: 25::

26 CHAPTER- 6 : Inheritance Key Points: Inheritance is the capability of one class inherit properties from another class. There are two type of class one is base class and another is derived class. Derived class inherits from the base class. Inheritance may take place in many form. There are different type of inheritance:. Single inheritance:- one class inherited by one class. 2. Multiple inheritance : - One child inherited from two or more base classes 3. Multilevel inheritance: one class inherited by another class. Another class inherited by one another class. 4. Hierarchical inheritance:- two class inherit one class 5. Hybrid inheritance:- combines two or more forms of inheritance. Study Material Class XII (Comp.Sc.) :: 26::

27 Questions and Answers 4 Mark Questions Q. Answer the questions (i) to (iv) based on the following code : Class abc int a; char ch[0]; float cost; void ss(); public: int n; abc(); //Constructor void get_data(); // member function void put_data(); ; Class xyz : public abc int a[0]; char ch; void xx(); public: xyz(); void read_data(); void write_data(); ; //member function //Constructor //member function //member function (i) What is the size of object xyz in bytes (ii) Write name of all the member function accessible form the object of class xyz. (iii)which class constructor will be called first at the time of declaration of an object of class xyz. (iv) Name the data member(variable) which accessible from the object of class abc. Ans. (i) Calculate the size: =39 (ii) (iii) (iv) get_data() and put_data from abc class read_data() and write_data from xyz class. Abc() then xyz() n Study Material Class XII (Comp.Sc.) :: 27::

28 Q2. Answer the questions (i) to (iv) based on the following code : class Employee int id; protected : char name[20]; char doj[20]; public : Employee(); ~Employee(); void get(); void show(); ; class Daily_wager : protected Employee int wphour; protected : int nofhworked; public : void getd(); void showd(); ; class Payment : private Daily_wager char date[0]; protected : int amount; public : Payment(); ~Payment(); void show(); ; (i) Name the type of Inheritance depicted in the above example. (ii) (iii) (iv) (v) Name the member functions, which are accessible by the objects of class Payment. From the following, Identify the member function(s) that can be called directly from the object of class Daily_wager class show(), getd(), get() Find the memory size of object of class Daily_wager. Is the constructors of class Employee will copied in class Payment? Due to inheritance. Ans: (i) Multilevel inheritance (ii) void show() (iii) getd() (iv) 46 Note: Employee= =42 daily_wager=2+2 =04 total =46 (v) not copied Q3. Consider the following code and answer the questions: class typea int x; protected: Study Material Class XII (Comp.Sc.) :: 28::

29 int k; public: typea(int m); void showtypea(); ; class typeb : public typea float p,q; protected: int m; void intitypeb(); public: typeb(float a, float b); void showtypeb(); ; class typec : private typeb int u,v; public: typec(int a, int b); void showtypec(); ; (i) How much byte does an object belonging to class typec require? (ii) Name the data member(s), which are accessible from the object(s) of class typec. (iii) Name the data members, which can be accessed from the member functions of class typec? (iv) Is data member k accessible to objects of class typeb? Ans:- (i) 8 Class a= 2+2 =4 Class b=4+4+2 =0 Class c= 2+2 =4 Total =8 (ii) None (iii) m,k,u,v (iv) No Q4. Answer the question (i) to (iv) based on the following code: Class Medicines char Category[0]; char Dateofmanufacture[0]; char Company[20]; public: Medicines(); void entermedicinedetails(); void showmedicinedetails(); ; class Capsules : public Medicines protected: char capsulename[30]; char volumelabel[20]; public: float Price; Capsules(); Study Material Class XII (Comp.Sc.) :: 29::

30 void entercapsuledetails(); void showcapsuledetails(); ; class Antibiotics : public Capsules int Dosageunits; char sideeffects[20]; int Usewithindays; public: Antibiotics(); void enterdetails(); void showdetails(); ; (i) How many bytes will be required by an object of class Medicines and an object of class Antibiotics respectively? (ii) Write names of all the member functions accessible from the object of class Antibiotics. (iii) Write names of all the members accessible from member functions of class Capsules. (iv) Write names of all the data members which are accessible from objects of class Antibiotics. Ans: (i) medicines=40 and antibiotics=8 Note: Medicines = =40 Capsules = =54 Antibiotics = =24 Total= 8 (ii)entermedicinedetails(), showmedicinedetails(), entercapsuledetails(), showcapsuledetails(), void enterdetails(); showdetails(); (iii) entermedicinedetails(),showmedicinedetails(), entercapsuledetails(), showcapsuledetails(), price, capsulename, volumelabel (iv) price Q5. Answer the following questions (i) to (iv) based on the following code : class DRUG char catg[0]; protected: char DOF[0], comp[20]; public: DRUG( ); void endrug( ); void showdrug( ); ; class TABLET : public DRUG protected: char tname[30],volabel[20]; public: TABLET( ); void entab( ); void showtab( ); ; class PAINKILLER : public TABLET int dose, usedays; char seffect[20]; public : Study Material Class XII (Comp.Sc.) :: 30::

31 (i) (ii) (iii) (iv) void entpain( ); void showpain( ); ; How many bytes will be required by an object of TABLET? Write names of all the data members accessible from member functions of class TABLET? Write names off all members functions accessible from object of class PAINKILLER. Write names of all data members accessible from member functions of class PAINKILLER. Ans: i) 90 drug=0+0+20=40 Tablet=30+20 =50 Total =90 ii) tname, volabel, DOF, Comp iii) endrug(), showdrug(),entab(),showtab(), entpain(), showpain() iv) does, usedays, tname, valabel, seffect, catg, comp, DOF Study Material Class XII (Comp.Sc.) :: 3::

32 CHAPTER-7 : Arrays Key Points: An array is a finite, ordered set of homogeneous elements that stored in contiguous memory locations, elements can be accessed by the index. There are two type of array. Single Dimension Array (-D) Two Dimension Array (2-D) Single Dimension Array:- One dimensional array is simply a named group of finite number of similar data elements under one index. e.g elements Index Two Dimension Array:- Two dimensional array is simply a named group of finite number of similar data elements under two index. e.g. A[0,0] A[0,] A[0,2] A[0,3] A[,0] A[,] A[,2] A[,3] A[2,0] A[2,] A[2,2] A[2,3] Address Calculation in Arrays- We may compute the address of any subscript of array in memory. Since, array is stored in contiguous memory locations. The facts like base address and data type of array elements should be consider in address calculation.. Address Computation of Single dimensional array- Since the elements of one dimensional array are stored in the memory location by sequential allocation techniques, the address of I th element of the array can be obtained, if we know- (a) The Base address (the address of first element) of the array and denoted by B. (b) The size of the element in the array denoted by S and LB is Lower bound. In C++ language LB is 0. Address of I th Element = B+(I LB) * s 2. Address Computation of Two -dimensional array: Since elements of the two-dimensional array can be stored in the computer memory in contiguous manner in row wise order. In general, the address of an array element may be computed as- Address of an element = Base address +(no of elements before this element) * size Let B is base address, S is size of each element, M and N is the number of row and column respectively. We obtain the formula as follows- Address of a[i,j]=b+s [(i*n)+j] Operations on Array Traversing:- Accessing (or Printing ) the elements of array. Insertion: Insertion of an element in the array at specified position or at end. Deletion: Removing an element from an array from specific location. Searching: Finding a given value in the array. Sorting: Arranging elements of the array in Ascending or descending order. Study Material Class XII (Comp.Sc.) :: 32::

33 Q. write a program to show a linear search. Answer:- #include<iostream.h> #include<conio.h> int search(int A[], int size,int N) clrscr(); int i; for(i=0;i<size;i++) if(a[i]==n) return(i); return(-); Q.2 Suppose A,B,C are array of integer of size M,N and M+N respectively. The numbers in array A appears in ascending order while the numbers in array B is in descending order. Write a user defined function in c++ to produce III rd array.(by merging solution) Answer:- #include<iostream.h> #include<conio.h> void merge(int A[],int M,int B[],int N,int C[]) int a,b,c; for(a=0,b=n-,c=0;a<m&&b>0;) if(a[a]<=b[b]) C[c]=A[a]; c++; a++; else C[c]=B[b]; c++; b--; if(a<m) while(a<=m) c[c++]=a[a++]; else while(b>=0) c[c++]=b[b--]; Q.3 Suppose a one dimensional array ARR containing integers arranged in ascending order. Write a user defined function in C++ to search for one integer from ARR with the help of binary search method. The function should return an integer 0 to show absence of the number & the integer to show the presence of the number in the array. The function should have three parameters as () an array ARR (2) the number DATA to be searched (3) number of element N. Answer:- #include<iostream.h> #include<conio.h> Study Material Class XII (Comp.Sc.) :: 33::

34 int Bsearch(int ARR[],int n, int Data) int beg,last,mid; beg=0;last=n-; while(beg<=last) mid = (beg+last)/2; if(data = =ARR[mid]) return ; else if (Data>Arr[mid]) beg = mid+; else last = mid-; return 0; Q.4 write a function to sort an array using selection sort method. Answer:- void insert(int a[], int size) int i, j, n, t; clrscr(); for (i=0;i<n-;i++) for (j=i+;j<n;j++) if (a[i]> a[j]) t = a[i]; a[i]=a[j]; a[j]=t; Q.5 Write a function to implement bubble sort to sort an array. Answer:- void bubble(int a[], int size) int i, j, n, temp; clrscr(); for (i=0;i<n-;i++) for (j=0;j<n--i; j++) If( a[j]>a[j+]) temp = a[j]; a[j] = a[j+]; a[j+] = temp; Q6. An array MAT [20][0] is stored in the memory along the row with each element occupying 4 bytes of memory. Find out the base address & the address of elements MAT [0][5], if the location of MAT [3][7] is stored at address 000. Answer:- Let Base address is B No. of column (n) = 0 Element size (W) = 4 Study Material Class XII (Comp.Sc.) :: 34::

35 Lowest Row & Column indices major order is : address of I th, J th element of array in row Address of MAT [I][J] = B+uI- +J- MAT [3][7] = = B+4(0(3-6)+(7-0)) 000 = B+4(30+7) 000 = B+4(37) 000 = B+48 B = B = 852 MAT [0][5] = 852+4(0(0-0)+(5-0)) = 852+4(00+5) = 852+4(05) = = 272 Q7. Write a function in C++ which accepts an integer array & its size as arguments/parameters & assign the elements into a two dimensional array of integers in the following format:- If the array is,2,3,4,5,6 then resultant 2D array = And if the array is, 2, 3 then resultant 2D array = Answer:- void func(int arr[], int size) int a2[20][20]; int i,j; for (i=0;i<size;i++) for (j=0;j<size;j++) if(i>=j) a2[i][j] = arr[j]; else a2[i][j] = 0; cout<< a2[i][j]<< ; cout<<endl; Q8: Write a function in C++ which accepts an integer array & its size as arguments/parameters and assign the elements into two dimensional arrays of integers in the following format If the array is, 2, 3,4,5,6 then resultant 2D array = Study Material Class XII (Comp.Sc.) :: 35::

36 And if the array is, 2, 3 then resultant 2D array = Answer:- void func(int arr[], int size) int a2[20][20]; int i,j; for (i=0;i<size;i++) for (j=0;j<size;j++) if ((size-i)<j) a2[i][j] = 0; else a2[i][j] = arr[j]; cout<< a2[i][j]<< ; cout<<endl; Q.9: An array MAT [30][0] is stored in the memory column wise with each element occupying & bytes of memory. Find out the base address & the address of elements MAT [20][5], if the location of MAT [5][7]is stored at the address 000. Answer:- Base address B No. of rows (m) = 30 Element size (W) = 8 Lowest Row & Column indices column major order is : Address of MAT [I][J] = B+W(m(J- )+ (I- )) MAT [5][7] = = B+8(30(7-0)+(5-0)) 000 = B+8(20+5) 000 = B+8(25) 000 = B+720 B = B = -720 Hence Base address is Now address of MAT [20][5] is computed as: MAT [20][5] = (30(5-0)+(20-0)) = (50+20) = (70) = = 640 Hence address of MAT [20][5] is 640. address of I th, J th element of array in Study Material Class XII (Comp.Sc.) :: 36::

37 CHAPTER- 8 : Data Base Concept & SQL Key Points: Database is a collection of different kinds of data which are connecting with some relation. In different way, it is a computer based record keeping system. The collection of data referred to as a database. A database management system is answer to all these problems as it provides a centralized control of the data. Advantages of Database:. It reduce the data redundancy 2. Control data inconsistency 3. Facilitate sharing of data 4. Databases enforce standards 5. Ensure data security 6. Integrity can be maintained Data: It is row facts, figures, characters, etc with which we have to start. Information: The processed data or meaningful data is known as information. Database: An organized collection of relational data is called as database. DBMS: It is a Database Management System. It is a system used for create database, store the data, secure it and fetch data whenever required. MS Access, Oracle are the example of DBMS. Data Models : The external level and conceptual level user certain data structures to help utilize the database efficiently. There are three data models that are used for database management system are:. Relational Data Model 2. Hierarchical Data Model 3. Network Data Model Commonly used Terms related to Data Base: Tuple : The rows of tables are generally referred to as Tuples. Attributes : The columns of tables are generally referred to as attributes. Degree : The number of attributes in a relation determine the degree of a relation. A Table having five columns is said to be a relation of degree five. Cardinality : The number of tuples (rows) in relation (table) is called the Cardinality of the relation. Primary Key : It is a set of one or more attributes that can uniquely identify tuples within the relation. For e.g. EmpNo. is the primary key for the table Employee. Candidate Key : All attributes combinations inside a relation that can serve as primary key are Candidate keys as they are having the candidature to work as the Primary Key. Alternate Key : A candidate key that is not the primary key but can be used in place of primary key is called an alternate key. SQL Structured Query Language: SQL is the set of commands that is recognized by nearly all the RDBMS. It is a language that enables us to create and operate on relational databases, which are sets of related information stored in tables. Processing Capabilities of SQL: Study Material Class XII (Comp.Sc.) :: 37::