The document discusses structures in C++. It defines a structure called Employee that contains data members for an employee's first name, last name, and date of birth. It shows how to declare a structure variable Emp initialized with values, and print out the employee's details using dot operators to access individual data members like Emp.dob.month. Structures allow storing different data types together and can be nested, like including a date structure within the Employee structure.

1. STRUCTURE
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2.  At the end of this lecture you will able to
 Realize how to define a structure of your own data type
using a keyword <struct >
 Understand how Store large data of different type with
in a single structure name
 Identify different ways of structure instance creation
and initializing structure instance with data members
 Access individual structure members using dot( . )
access operator
 Use arrays of structure and able to store more records
 Use structure as a function argument
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3. Only one value will store at a time
It holds one type no data type diversity
No possibility to use as a new data type
 Class Activity
Why regular Variable?
Why Array?
discus by Their:
It stores large amount of data at a time
All data are of the same type
It is not possible to use as a new data type
 Storage capacity(amount of data store at a time)
 Diversity of data type
 Capacity to be new user defined data type
Why structure?
(5 Min)
It stores Large amount of data at a time
It is a collection of multi- diversified data types
We can have data type of our own
It is an instance of full record about one object
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4.  Arrays allow you to define variables that
combine several data items of the same type
 But structure is another user defined data type
which allows you to combine a group of data
items(members) of different data type and
length with in a single structure name.
 Structures are used to represent a record,
Suppose you want to keep track of your books in
a library. You might want to track the following
attributes about each book:
Title Author
ISBN Book_No
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5.  To declare structure of your own data type with
different data member we use keyword <truct>
 General syntax
struct Struct_Name {
type2 DataMember 1;
type2 DataMember 2;
type1 DataMember 3;
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};
New user Defined Data type
Keyword Struct
Data members with
different Data type

6.  Struct_Name :- we can use as a user defined
data type we can create a new object of
struct_name as following:
Syntax :-
We Use as our new data type
struct_name Obj_Name;
 We can also create an object of user defined
structure during structure declaration like:
struct Struct_Name {
type2 DataMember 1;
type2 DataMember 2;
type1 DataMember 3;
}stru1,stru2,stru3;
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7.  Create structure called Book which has
attributes Title, Author, ISBN, Book_No
Struct Book {
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
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};
Struct Book {
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
} bookObj;

8.  As with arrays and variables, structure members
can also be initialized. This is performed by
enclosing the values to be initialized inside the
braces { and } after the structure variable name
while it is defined.
Struct Book {
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
} Book1 = {“C++”, ” John”, “12-34-034-234”,12};
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9.  The Second way to initialize structure
Struct Book {
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
} ;
int main()
{
Book Book1 = {“C++”, ” John”, “12-34-034-234”,12};
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}

10.  Structure members can’t access directly as a
normal local or global variables …
 To access any member of a structure, we use the
member access operator either called dot (.)
operator for non pointer structure variable or
arrow operator () for pointer structure
variables. The member access operator is coded
as a dot between the structure variable name and
the structure member that we wish to access.
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11.  Syntax
struct_var.Member1=//some operation;
 Example
Struct Book
{
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
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}Book1;
Assign Individual Member
Book1 .Title= “C++”;
Book1.Author= ”John”;
Book1.ISBN= “12-34-034-234”
Book1.Book_No=12;

12. ?
Example :1
Create structure called Books which have
four data members Title, Author, ISBN,
Book_No that used to store three different
Books and display all Book you initialize from
the initialization when you instantiate an
instance of structure Books
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13. #include<iostream>
using namespace std;
struct Books {
char Title[50];
char Author[25];
char ISBN[15];
int Book_No; };
int main() {
Books Book1 = {"Java","Johe","123-3243-2312",1};
Books Book2 = {"C++","Johe","123-3243-2313",3};
Books Book3 = {"C#","Johe","123-3243-2314",2};
cout<<"tTitle tAuthor tISBN ttBook Number"<<endl;
cout<<"t................................................................"<<endl;
cout<<"t"<<Book1.Title<<"t"<<Book1.Author<<"t“
Initializing 3 Books
With 4 data members
for each book
<<Book1.ISBN<<"t"<<Book1.Book_No<<endl;
cout<<"t"<<Book2.Title<<"t"<<Book2.Author<<"t“
<<Book2.ISBN<<"t"<<Book2.Book_No<<endl;
cout<<"t"<<Book3.Title<<"t"<<Book3.Author<<"t“
<<Book3.ISBN<<"t"<<Book3.Book_No<<endl;
system("pause");
return 0;
}
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•Printing 3 books
data by using dot
operator access
operator
Structure declaration and
member definition

14. Why Arrays of Structure?
 Arrays of a structure is a collection of structure
 A single structure instance will represent a single
record at a time. If we need to store large
amount of record in a single structure we need
to create many instances of a give structure and
this is inefficient to manage large record
 It is possible to work by creating individual
structure variable for small number of records
which is less than 10 record but …
 In order to manage large record of same
category we need to develop an arrays of
structure which is an array stores different types
of structure member variables
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15.  Image you have structure Books with attribute
Title, Author, ISBN, Book_No and you need to
store 1000 Books with their corresponding
attribute
Is it easy to create 1000 Books Variable?
 It is not easy! so what is the solution? …
1. We can solve this using arrays of structure …
2. Lets compare and contrast individual structure
variable creation and array usage
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16. Struct Books
{
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
}Book1, Book2, Book3, Book4 …Book1000;
Individual
structure variable
Using arrays of structure
Struct Books
{
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
} BookObj[1000];
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17.  To access individual elements of arrays of
structure we need to specify the index number
with in a square bracket of the structure
instance like simple array operation
 Assume that you have structure as follow
Struct Books
{
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
} BookObj[10];
How to access the first
three books from your book
storage structure?
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18. Book one information Book Two information Book Three information
BookObj[0].Title
BookObj[0].Author
BookObj[0].ISBN
BookObj[0].Book_No
BookObj[1].Title
BookObj[1].Author
BookObj[1].ISBN
BookObj[1].Book_No
BookObj[2].Title
BookObj[2].Author
BookObj[2].ISBN
BookObj[2].Book_No
Title
Author
ISBN
Book_No
BookObj[0]
Title
Author
ISBN
Book_No
BookObj[1]
Title
Author
ISBN
Book_No
BookObj[2]
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19.  Structure can pass into function as an argument
rather than passing individual elements it is
better to pass the whole of structure
 We can pass structure either by reference or by
value but it is recommended to pass structure by
their reference because to minimize the cost of
copying the value
 Its default value of passing structure is by nature
uses by reference
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20. #include<iostream>
using namespace std;
struct Time
{
int h, m, s;
};
int ToSecond(Time now){
return (now.h*60*60 + now.m*60 + now.s);
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}i
nt main(){
Time T;
cout<<"Enter Time In Hour, Min and Second:";
while( cin>>T.h>>T.m>>T.s)
{
cout<<"Total Second:"<<ToSecond(T);
}
system("pause");
return 0;
}
Example
Passing structure time T

21. ? Class activity
 Create a structure called Books to hold records
of 100 Books with Book detail of Title, Author,
ISBN and Book_No. Your program will accept an
input from the user and display all Book
information in table format
 Assume you will have a basic about
 Display():- Function to give output
 Input():- Function to accept input from keyboard
 Main():- To call an input() function
 Structure definition:-To declare all members
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22. #include<iostream>
using namespace std;
struct Books
{
char Title[50];
char Author[20];
char ISBN[15];
int Book_No;
};
const int size = 100;
void display(Books Book[size], int n)
{
cout<<"Title"<<"t"<<"Author"<<"t“
<<"ISBN"<<"t"<<"Book No"<<endl;
for(int i = 0; i < n ; i++) {
cout<<Book[i].Title<<"t"<<Book[i].Author<<"t"
<<Book[i].ISBN<<"t"<<Book[i].Book_No<<"t"
<<endl;
}
}
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void input()
{
Books Book[size];
cout<<"nEnter No of Books:";
int n; cin>>n;
for(int i = 0;i < n ; i++) {
cout<<"nEnter Title:“; cin>>Book[i].Title;
cout<<"nEnter Author:“; cin>>Book[i].Author;
cout<<"nEnter ISBN:“; cin>>Book[i].ISBN;
cout<<"nEnter Book No:"; cin>>Book[i].Book_No;
} display(Book,n);
}
int main()
{
input();
system("pause");;
return 0;
}
Input function
(initialize struct)
Output function
Structure declaration
Main function

23.  Structures can be implemented as the member of
other structure.
 This is termed as structure within structure we
call it nested structure
 Why we use nested structure?
 To organize multi valid attribute as a single value
 To eliminate data redundancy in different structures
We collect similar data members of two or more
structure with in a single structure name
 To minimize coding time
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24. struct date{
int day;
int month;
int year;
};
struct Employee {
char name[40];
long int Emp_ID;
char sex[5];
date dob;
Example: Using nested structure
to manipulate multi valid attribute
Example: Date of Birth has 3
integer values to manipulate it
we use Date as nested struct in
Employee struct
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};
Using date structure as Member
of Employee structure

25.  Assume you have struct Employee and Customer
with their data members as following
struct Employee{
string Name;
string EID;
int age;
String Contact_Name;
string Kebele;
string Woreda;
string PhoneMobil;
string PhonHome;
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};
struct Customer{
string Name;
string CustID;
int age;
string Kebele;
string Woreda;
string PhoneMobil;
string PhonHome;
};

26.  From the above structure definition address
information is redundant both for Employee and
customer structure
 So to eliminate this data redundancy we
separate address information in different
structure name called address and both customer
and employee will inherit from address structure
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27. struct Employee
{
string Name;
string EID;
int Empage;
Address EmpAddress;
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};
struct Customer
{
string Name;
string CustID;
int Custage;
Address CustAddress;
};
Struct Address
{
String Contact_Name;
string Kebele;
string Woreda;
string PhoneMobil;
string PhonHome;
}

28.  We can initialize a nested structure either by
putting all elements in order of one structure or
as block of data members
Example:-
Initialize all member of the above nested structure
definition with Data : Name =Dagim, ID =EMP009, Sex = M, Date of
Birth = 10-2-1989
Option 1:-
Employee Emp ={“Dagim”, “EMP009”,’M’,10,2,1989};
Option 2:-
Employee Emp ={“Dagim”, “EMP009”,’M’,{10,20,1989}};
Option 3:-
Emp.name=“Dagim”;
Emp.EMP_ID=“EMP009”;
Emp.sex=‘M’;
Emp.dob.day=10;
Emp.dob.month=2;
Emp.dob.dat=1989;
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29.  To access a nested structure data members we
will use to dot access operator
 Example:-
To access a particular month of Employee birth
date we can access as follow
Employee Emp;
Emp.dob.month ;
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30. #include<iostream>
using namespace std;
struct date {
int day;
int month;
int year;
};
struct Employee {
char First_Name[23];
char Last_Name[24];
date dob;
} Emp= {"DEMEKE","BIRHANU",{16 , 2, 2012}};
int main()
{
cout<<"ntFirst NametLast NametDate of Birth"<<endl;
cout<<"nt"<<Emp.First_Name<<"tt"<<Emp.Last_Name<<"tt"
<<Emp.dob.day<<"-"<<Emp.dob.month<<"-"<<Emp.dob.year<<endl;
system("pause");
return 0;
}
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31. Thanks !
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