CSC 138 Structured Programming CHAPTER 3: RECORDS (STRUCT) [PART 1]

Transcription

1 CSC 138 Structured Programming CHAPTER 3: RECORDS (STRUCT) [PART 1]

2 LEARNING OBJECTIVES Upon completion, you should be able to: o define records (structures) that contain array data type as one of the member. o declare and initialize C++ record (struct) as an array. o differentiate between array and record (struct). o write program using array and function in record (struct).

3 RECORD OVERVIEW 3.1 RECORDS DEFINITION 3.2 RECORDS DECLARATION 3.3 RECORDS INITIALIZATION 3.4 ACCESSING RECORDS MEMBERS 3.5 RECORDS ASSIGNMENT 3.6 AGGREGATE OPERATIONS WITH STRUCT

4 CHAPTER 3 RECORDS / STRUCT DEFINITION DECLARATION INITIALIZATION ACCESSING RECORD MEMBERS RECORD ASSIGNMENT AGGREGATE OPERATIONS WITH STRUCT

5 RECORD INTRODUCTION o Assume that we want to store the following students information: name Ic number Matric number Year entry o Below are the example of data representation that might be used: char Name[15], ICno[15], MatricNo[7]; int YearEntry; o We cannot use an array to group all of the items associated with a student. How to group elements of different types together?

6 RECORD INTRODUCTION o An array is a homogeneous data structure; a struct is typically a heterogeneous data structure. o A collection of fixed number of components in which components are accessed by name. The components may be of different types. o Allowed a group of related data being accessed and manipulate as a unit(record) struct o For example, data associated with student are: Name IcNo MatricNo YearEntry - string 15 characters - string 15 characters/array - string 7 characters - integer

7 RECORD DEFINITION o struct: collection of a fixed number of components (members), accessed by name Members may be of different types o Syntax:

8 RECORD DECLARATION o A struct is a definition, not a declaration OR struct studenttype { string firstname; string lastname; char coursegrade; int testscore; int programmingscore; double GPA; //variable declaration } newstudent, student; C++ Programming: From Problem Analysis to Program Design, Fourth Edition 8

9 RECORD DECLARATION C++ Programming: From Problem Analysis to Program Design, Fourth Edition 9

10 o Initialization can be done during declaration. o Example 1: o Example 2: RECORD INITIALIZATION struct student stud1 = { Nadiah, , A11000, 1998}; struct date { int day; int month; int year; } struct date Birth_Day = {1, 1, 1982}; Birth_Day 1 1 day month 1982 year Company Logo CSC 138 Structured programming

11 RECORD INITIALIZATION o Example 3: struct date Birth_Day = {31, 10, 1966}; day month year Birth_Day Company Logo CSC 138 Structured programming

12 ACCESSING RECORDS MEMBERS o The syntax for accessing a struct member is: o The dot (.) is an operator, called the member access operator C++ Programming: From Problem Analysis to Program Design, Fourth Edition 12

13 ACCESSING RECORDS MEMBERS o To initialize the members of newstudent: newstudent.gpa = 0.0; newstudent.firstname = "John"; newstudent.lastname = "Brown"; C++ Programming: From Problem Analysis to Program Design, Fourth Edition 13

14 ACCESSING RECORDS MEMBERS More examples: cin>>newstudent.firstname; cin>>newstudent.testscore>>newstudent.programmingscore; Score=(newStudent.testScore + newstudent.programmingscore) / 2; if (score >= 90) newstudent.coursegrade = 'A'; else if (score >= 80) newstudent.coursegrade = 'B'; else if (score >= 70) newstudent.coursegrade = 'C'; else if (score >= 60) newstudent.coursegrade = 'D'; else newstudent.coursegrade = 'F'; C++ Programming: From Problem Analysis to Program Design, Fourth Edition 14

15 ACCESSING RECORDS MEMBERS o Example 3: struct student S1; strcpy(s1.name, "C Bin D"); strcpy(s1.icno, " "); strcpy(s1.matricno, "A11122"); S1.YearEntry = 1990; S1 Name IcNo MatricNo YearEntry 'C'' ' 'B''i' 'n' ' ''D'\0' '6' '6''1''1''2''2' '-' '0''2' '-' '5' '5' '5''4''\0' 'A''1''1''1''2''2''\0' 1990 CSC 138 Structured programming

16 ACCESSING RECORDS MEMBERS o Example 4: struct date Birth_Day; cout<< When is your birthday?(dd MM YYYY)\n"; cin>>birth_day.day>>birth_day.month>> Birth_Day.year; cout<< Your birthday is << Birth_Day.day<< / <<Birth_Day.month<< / << Birth_Day.year<<endl; day month year????????? Company Logo Birth_Day CSC 138 Structured programming

17 o Example ACCESSING RECORDS MEMBERS struct stud S1; cout<<"name, Ic no, Matric no, year?\n"); gets(s1.name); //also act as input data for char [cin] gets(s1.icno); //needs to include stdio.h in using gets()& puts() gets(s1.matricno); cin>>s1.yearentry; puts(s1.name); // also act as display data for char [cout] Company Logo S1 Name IcNo MatricNo YearEntry???????????????????????????????????????? CSC 138 Structured programming

18 RECORD ASSIGNMENT o Value of one struct variable can be assigned to another struct variable of the same type using an assignment statement o The statement: student = newstudent; copies the contents of newstudent into student C++ Programming: From Problem Analysis to Program Design, Fourth Edition 18

19 RECORD ASSIGNMENT o The assignment statement: student = newstudent; is equivalent to the following statements: student.firstname = newstudent.firstname; student.lastname = newstudent.lastname; student.coursegrade = newstudent.coursegrade; student.testscore = newstudent.testscore; student.programmingscore = newstudent.programmingscore; student.gpa = newstudent.gpa; C++ Programming: From Problem Analysis to Program Design, Fourth Edition 19

20 AGGREGATE OPERATIONS WITH STRUCTURES o Recall that arrays had none (except reference parameter) o Structures DO have aggregate operators assignment statement = parameter (value or reference) return a structure as a function type 20

21 AGGREGATE OPERATIONS WITH STRUCTURES o Limitations on aggregate operations no I/O no arithmetic operations cout << old_part; cin >> new_part; no comparisons old_part = new_part + old_part; if (old_part < new_part) cout <<...; 21

22 COMPARISON (RELATIONAL OPERATORS) o Compare struct variables member-wise No aggregate relational operations allowed o To compare the values of student and newstudent: C++ Programming: From Problem Analysis to Program Design, Fourth Edition 22

23 INPUT/OUTPUT o No aggregate input/output operations on a struct variable o Data in a struct variable must be read one member at a time o The contents of a struct variable must be written one member at a time C++ Programming: From Problem Analysis to Program Design, Fourth Edition 23

24 ARRAYS VERSUS STRUCTS C++ Programming: From Problem Analysis to Program Design, Fourth Edition 24

25 Exercise 1 1. Define a struct, checkingaccount, to store the following data about a checking account; account holder s name(string), account number (int), balance (double), and interest rate (double). 2. Define a struct, movietype, to store the following data about a movie: movie name(string), movie director(string), producer (string), the year movie was released (int),and number of copies in stock. C++ Programming: From Problem Analysis to Program Design, Fourth Edition 25

26 Exercise 2 Assume the definition of Exercise 1 1. Declare a checkingaccount variable and write C++ statements to store the following information: account holder s name Jason Miller, account number , balance , interest rate -2.5%. 2. Declare a variable of type movietype to store the following data: movie name Summer Vacation, director Tom Blair, producer Rajiv Merchant, year the movie released 2005, the number of copies in stock 34 C++ Programming: From Problem Analysis to Program Design, Fourth Edition 26

27 Exercise 3 a) Write a C++ statement for each of the following questions: i. Define a struct named Student_Information which has four members of different types. The members are student name, gender( F or M ), semester and CGPA. C++ Programming: From Problem Analysis to Program Design, Fourth Edition 27

28 Exercise 4 a) Consider the following statements: Struct nametype { }; string first; string last; double score; nametype name; nametype name2; Mark the following statements as valid or invalid. If statement is invalid explain why. i. name.first = Nafisah ; ii. iii. iv. cout<< nametype; name2= name; cin >> nametype; v. cout<< name2. last; vi. vii. nametype.last = Amin ; double total = score; viii. double average; average= name2.score /2; C++ Programming: From Problem Analysis to Program Design, Fourth Edition 28

29 Mark the following statements as valid or invalid based on the struct definition below: struct booktype { char authorname[30]; float price; char edition[15]; int yearpublished; char publisher[50]; }book1, book2; a) book1.price=50.00; b) book1.publisher[30]; c) book2==book1; Exercise 5 d) if (strcmp(book1.edition,book2.edition)==0) e) strcpy(book2.publisher, Pearson );

30 Exercise 6 a) Define a struct named Course with 3 members: coursename, coursecode and coursefee. Declare 2 struct variables namely course1 and course2. b) A company has six salespersons. Every month they go on road trips to sell the company s product. At the end of each month, the total sales for each salesperson, together with that salesperson s ID and month is recorded. Write an appropriate struct definition for this problem.

31 Exercise 7 Given the following struct definition: struct booktype { char authorname[30]; float price; char edition[15]; int yearpublished; char publisher[50]; }book1, book2; write a C++ statement for each of the following: a) Set the value of yearpublished for book2 to b) Set the price of book2 equal to the price of book1. c) Set the value of authorname for book2 as D.S. Malik.

32 CSC 138 Structured Programming CHAPTER 3: RECORDS (STRUCT) [PART 2]

33 RECORD OVERVIEW 3.7 RECORDS AND ARRAY 3.8 RECORDS AND FUNCTION

34 CHAPTER 3 RECORDS AND ARRAY ARRAY IN RECORD RECORDS IN ARRAY

35 ARRAY IN RECORD (STRUCT) o Example 1: const int arraysize = 500; struct listtype { //array in struct int elements[arraysize]; int listlength; }; Memory: listone elements elements[0] elements[1]... elements[499] listlength listtype listone; //declared a struct variable

36 ARRAY IN RECORD (STRUCT) o Example 1: listone.listlength = 0; listone.elements[0] = 11; listone.listlength++; Memory: listone elements elements[0] elements[1]... elements[499] listone.elements[1] = 22; listlength 0 listone.listlength++;

37 ARRAY IN RECORD (STRUCT) o Example 1: listone.listlength = 0; listone.elements[0] = 11; listone.listlength++; Memory: listone elements elements[0] elements[1]... elements[499] 11 listone.elements[1] = 22; listlength 0 listone.listlength++;

38 ARRAY IN RECORD (STRUCT) o Example 1: listone.listlength = 0; listone.elements[0] = 11; listone.listlength++; Memory: listone elements elements[0] elements[1]... elements[499] 11 listone.elements[1] = 22; listlength 0 1 listone.listlength++;

39 ARRAY IN RECORD (STRUCT) o Example 1: listone.listlength = 0; listone.elements[0] = 11; listone.listlength++; Memory: listone elements elements[0] elements[1]... elements[499] listone.elements[1] = 22; listlength 1 listone.listlength++;

40 ARRAY IN RECORD (STRUCT) o Example 1: listone.listlength = 0; listone.elements[0] = 11; listone.listlength++; Memory: listone elements elements[0] elements[1]... elements[499] listone.elements[1] = 22; listlength 1 2 listone.listlength++;

41 ARRAY IN RECORD (STRUCT) o Example 1 (Complete Program): const int arraysize = 500; struct listtype { int elements[arraysize]; //array in struct int listlength; }; int main(){ listtype listone; //Declares a struct variable cout << "Enter number of element(s): "; cin >> listone.listlength; for(int i = 0; i < listone.listlength; i++) listone.elements[i] = (i+1); } for(int i = 0; i < listone.listlength; i++) cout << listone.elements[i] << endl;

42 ARRAY IN RECORD (STRUCT) o Example 2 (struct definition & variable declaration): struct studenttype { string studname; char coursegrade; int testscore[2]; int asgnscore; double GPA; }; studenttype student; Memory: student studname coursegrade testscore [0] testscore [1] asgnscore GPA

43 ARRAY IN RECORD (STRUCT) o Example 2 (variable declaration & initialization): studenttype student = { Jamaluddin, A, 90, 87, 85, 3.85 }; Memory: student studname coursegrade testscore [1] asgnscore GPA Jamaluddin A testscore [0]

44 ARRAY IN RECORD (STRUCT) o Example 2 (main function): int main() { studenttype student = { Jamaluddin, A, 90, 87, 85, 3.85 }; //Assignment statement: Method 1 studenttype newstudent; //Assignment statement: Method 2 for(int i = 0; i < 2; i++) newstudent.studname = student.studname; newstudent.coursegrade = student.coursegrade; newstudent.testscore[i] = student.testscore[i]; newstudent.asgnscore = student.asgnscore; newstudent.gpa = student.gpa;

45 ARRAY IN RECORD (STRUCT) o Example 2 (main function cont.):... cout << " STUDENT'S INFORMATION" << endl; } cout << " NAME\t: " << student.studname[0] << " " << student.studname[1]; cout << "\n GPA\t: " << student.gpa; cout << "\n COURSE GRADE : " << student.coursegrade; cout << "\n\n ASSIGNMENT SCORE\t: " << student.asgnscore; cout << "\n TEST SCORE\t\t: [1] " << student.testscore[0] << " [2] " << student.testscore[1];

46 o Example 2 (Output): ARRAY IN RECORD (STRUCT)

47 EXERCISE (3.2.1) Extends the codes from previous program in Example 1 to: 1. display the elements of the array in reverse order. 2. display the odd elements of the array.

48 ANSWER (3.2.1) const int arraysize = 500; struct listtype { }; int elements[arraysize]; //array in struct int listlength; main(){ : //Display elements in reverse order for(int i = (listone.listlength - 1); i >= 0; i--) cout << listone.listelement[i] << endl; } //Display odd elements for(int i = 0; i < listone.listlength; i++) if(listone.listelement[i] % 2!= 0) cout << listone.listelement[i] << endl;

49 RECORDS (STRUCT) IN ARRAY o Suppose a company has 50 employees. The employer need to print the employees monthly paychecks and keep track of how much money has been paid to each employee. o Step 1 : Define struct const int noofemployee = 50; struct employeetype { char name[50], deptid[15], employeeid[15]; int hourworked; double basicsalary, monthlysalary; }; name deptid employeeid hourworked basicsalary monthlysalary

50 RECORDS (STRUCT) IN ARRAY o Step 2 : Declare an array of struct variable employeetype employees[noofemployee]; Declares an array employees with 50 components of type employeetype. Every elements is a struct. Memory: employees[0] name deptid employeeid hourworked basicsalary monthlysalary Memory: employees[1] name deptid employeeid hourworked basicsalary monthlysalary Memory: employees[49] name deptid employeeid hourworked basicsalary monthlysalary

51 RECORDS (STRUCT) IN ARRAY o Step 3 : Input / Read data for each of the employees for(int x = 0; x < noofemployee; x++) { //Input statement //Calculate the monthly salary } Output screen:

52 RECORDS (STRUCT) IN ARRAY //Input statement cout << " Employee " << x+1 << endl; cout << " \n"; cout << " Name : "; cin.getline(employees[x].name, 50); cout << " Department : "; cin.getline(employees[x].deptid,15); cout << " Employee ID : "; cin.getline(employees[x].employeeid,15); cout << " Hours Worked : "; cin >> employees[x].hourworked; cout << " Basic Salary : RM "; cin >> employees[x].basicsalary; //Calculate the monthly salary employees[x].monthlysalary = employees[x].hourworked * employees[x].basicsalary; cout << endl;

53 RECORDS (STRUCT) IN ARRAY o Step 4 : Access elements of the array, elements of the struct //print all the name for each employees for(int x = 0; x < noofemployee; x++) cout << employees[x].name; //print all the employeeid for(int x = 0; x < noofemployee; x++) cout << employees[x].employeeid;

54 RECORDS (STRUCT) IN ARRAY //Display the record for each employee cout << setw(10) << "EMPLOYEE ID" << setw(20) << "BASIC SALARY (RM)" << setw(25) << "MONTHLY SALARY (RM)" << endl; cout << setfill('-') << setw(55) << endl; cout << setfill(' '); } for(int i = 0; i < noofemployee; i++) { cout << setw(8) << employees[i].employeeid; cout << setw(16) << employees[i].basicsalary; cout << setw(16) << employees[i].monthlysalary; cout << endl; } system("pause");

55 EXERCISE (3.2.2) Extends the codes from previous program to: 1. search employee record by their employee ID. 2. display the particular details as the following output screen:

56 ANSWER (3.2.2) main() { employeetype employees[noofemployee]; char searchid[15]; cout << "\n Enter Employee ID to be searched: "; cin >> searchid; for(int i = 0; i < noofemployee; i++) { if(strcmp(employees[i].employeeid, searchid) == 0) { //Display the record for the employee cout << endl << setw(10) << " EMPLOYEE ID" << setw(15) << "EMPLOYEE NAME" << setw(25) << "DEPARTMENT" << endl; cout << " " << endl; cout << setfill(' '); cout << setw(10) << employees[i].employeeid; cout << setw(15) << employees[i].name; cout << setw(25) << employees[i].deptid; cout << endl; } else cout << " Record doesn't exist!";} }

57 EXERCISE (3.2.3) Suppose FSKM has 5 students. We need to calculate the total score obtained by each student. The total score is by adding the mark of test score and programming score. From the total score, find the course grade of each student. Lastly, display all information of all students. struct studenttype { string firstname; string lastname; char coursegrade; int testscore; int programmingscore; double GPA; int score; } ; studenttype student[5];

58 EXERCISE (3.2.4) Suppose a company has 10 full-time employed. We need to print their monthly paychecks and keep track of how much money has been paid to each employee in the year-to-date. First, let s define an employee s record: Input : firstname, lastname, personid, deptid, yearlysalary, monthlybonus struct employeetype { string firstname, lastname; string deptid; int personid; double yearlysalary; double monthlysalary; double yeartodatepaid; double monthlybonus; } ; employeetype employee[10];

59 ANSWER (3.2.3) int main() { for(int i=0; i<5; i++) { cout<<"insert first name : "; cin>>student[i].firstname; cout<<"insert last name : "; cin>>student[i].lastname; cout<<"insert test score : "; cin>>student[i].testscore; cout<<"insert programming score : "; cin>>student[i].programmingscore; cout<<"insert GPA : "; cin>>student[i].gpa; } } for(int i=0; i<5; i++) { student[i].score = student[i].testscore+student[i].programmingscore; if(student[i].score >=90) student[i].coursegrade='a'; else if(student[i].score >=80 && student[i].score<90) student[i].coursegrade='b'; else if(student[i].score >=70 && student[i].score<80) student[i].coursegrade='c'; else if(student[i].score >=60 && student[i].score<70) student[i].coursegrade='d'; else student[i].coursegrade='f'; } for(int i=0; i<5; i++) { cout<<"first Name : <<student[i].firstname<<endl; cout<<"last Name : <<student[i].lastname<<endl; cout<<"course Grade : <<student[i].coursegrade<<endl; cout<<"test Score : <<student[i].testscore<<endl; cout<<"programming Score : <<student[i].programmingscore<<endl; cout<<" GPA : <<student[i].gpa<<endl; }

60 ANSWER (3.2.4) int main() { for(int i=0; i<10; i++) { cout << \ninsert first name : "; cin >> employee[i].firstname; cout << \ninsert last name : "; cin >> employee[i].lastname; cout << \ninsert id : "; cin >> employee[i].personid; cout << \ninsert dept id : "; cin >> employee[i].deptid; cout << \ninsert yearly salary: "; cin >> employee[i].yearlysalary; cout << \ninsert monthly bonus: "; cin >> employee[i].monthlybonus; } for(int i=0; i<10; i++) { employee[i].monthlysalary = employee[i].yearlysalary / 12; double paycheck = employee[i].monthlysalary + employee[i].monthlybonus; employee[i].yeartodatepaid = paycheck; cout << "Year To Date Paid : <<employee[i].yeartodatepaid << endl; } }

61 RECORD OVERVIEW 3.7 RECORDS AND ARRAY 3.8 RECORDS AND FUNCTION

62 CHAPTER 3 RECORDS AND FUNCTION PASS RECORD VARIABLE AS PARAMETER PASS RECORD MEMBER AS PARAMETER PASS ARRAY OF RECORDS AS PARAMETER RETURN RECORDS

63 PASS RECORD VARIABLE AS PARAMETER o Recall the difference between array and struct involving aggregate operation: AGGREGATE OPERATION ARRAY STRUCT Arithmetic No No Assignment No Yes Input / Output No (except strings) No Comparison No No Parameter passing By reference only By value or reference Function returning a value No Yes

64 PASS RECORD VARIABLE AS PARAMETER o The following struct definition were given: struct studenttype { char name[25]; char matrixno[15]; int icno; }; name matrixno icno

65 PASS RECORD VARIABLE AS o Construct main function: PARAMETER main() { studenttype s;//struct variable declaration cout << Enter Matrix No.: ; cin.getline(s.matrixno, 15); cout << Enter Name: cin.getline(s.name, 25); cout << Enter IC No.: ; cin >> s.icno; } display(s) //function call

66 PASS RECORD VARIABLE AS PARAMETER (Pass entire struct) o Construct function prototype: void display(studenttype); o Construct function definition: Pass entire struct at once void display(studenttype x) { cout << x.name << << x.matrixno << << x.icno; } Member s name CANNOT BE RENAMED! But the arrangement of variable CAN BE MODIFIED.

67 PASS RECORD VARIABLE AS PARAMETER (Pass address of struct) o Construct function prototype: void display(studenttype&); o Construct function definition: Pass only the address of struct void display(studenttype& x) { cout << x.name << << x.matrixno << << x.icno; } Member s name CANNOT BE RENAMED! But the arrangement of variable CAN BE MODIFIED.

68 PASS RECORD MEMBERS AS PARAMETER o The following struct definition were given: struct studenttype { char name[25]; char matrixno[15]; int icno; }; name matrixno icno

69 o Construct main function: PASS RECORD MEMBERS AS PARAMETER main() { studenttype s; //struct variable declaration cout << Enter Matrix No.: ; cin.getline(s.matrixno, 15); cout << Enter Name: cin.getline(s.name, 25); cout << Enter IC No.: ; cin >> s.icno; } //function call display( s.matrixno, s.name, s.icno ) Pass members of struct individually

70 PASS RECORD MEMBERS AS PARAMETER o Construct function prototype: void display(char[], char[], int); o Construct function definition: Pass only the members of STRUCT individually void display(char matrix[], char name[], int ic) { cout << matrix << << name << } << ic; Member s name CAN BE RENAMED! But the arrangement of variables CANNOT BE MODIFIED

71 PASS ARRAY OF RECORDS AS PARAMETER o The following struct definition were given: struct studenttype { char name[25]; char matrixno[15]; int icno; }; name matrixno icno

72 o Main function: PASS ARRAY OF RECORDS AS PARAMETER main() { int arraysize = 10; studenttype s[arraysize]; getdata(s, arraysize); //function call } for(int i = 0; i < arraysize; i++) { cout << s[i].name << << s[i].matrixno << << s[i].icno; } Member s name CANNOT BE RENAMED! But the arrangement of variables CAN BE MODIFIED

73 o Function prototype: PASS ARRAY OF RECORDS AS void getdata(studenttype[]); PARAMETER Pass the array of struct o Function definition: void getdata(studenttype s[], int size) { for(int i = 0; i < size; i++) { cout << Enter Matrix No. and Name ; cin.getline(s[i].matrixno, 15); cin.getline(s[i].name, 25); cout << Enter IC No.: ; cin >> s[i].icno; } }

74 RETURN RECORD (By parameter) o The following struct definition were given: struct studenttype { char name[25]; char matrixno[15]; int icno; }; name matrixno icno

75 RETURN RECORD (By parameter) o Function prototype: void getdata(studenttype&); Pass the address of struct, by reference o Function definition: void getdata(studenttype& x) { cout << Enter Matrix No.: ; cin.getline(x.matrixno, 15); cout << Enter Name: cin.getline(x.name, 25); cout << Enter IC No.: ; cin >> x.icno; }

76 o Main function: RETURN RECORD (By parameter) main() { studenttype s;//struct variable declaration getdata(s) //function call } cout << s.name << << s.matrixno << << s.icno; Member s name CANNOT BE RENAMED! But the arrangement of variables CAN BE MODIFIED

77 RETURN RECORD (By return value) o The following struct definition were given: struct studenttype { char name[25]; char matrixno[15]; int icno; }; name matrixno icno

78 RETURN RECORD (By return value) o Function prototype: studenttype getdata(); o Function definition: studenttype getdata() { studenttype s; Pass the type of struct cout << Enter Matrix No. and Name ; cin.getline(s.matrixno, 15); cin.getline(s.name, 25); cout << Enter IC No.: ; cin >> s.icno; } return s;

79 RETURN RECORD (By return value) o Main function: main() { studenttype s;//struct variable declaration s = getdata() //function call } cout << s.name << << s.matrixno << << s.icno; Member s name CANNOT BE RENAMED! But the arrangement of variables CAN BE MODIFIED

80 EXERCISE (3.2.5) o Suppose that you have the following definitions: struct tourtype { char cityname[20]; int distance, hour, sec; double min; }; i. Declare the variable named destination of type tourtype. ii. Write C++ statements to store the following data in destination: cityname - chicago, distance miles, traveltime - 9 hours and 30 minutes. iii. iv. Write the definition of a function to output the data stored in a variable of type tourtype. Write the definition of value returning function that input data into a variable of type tourtype. v. Write the definition of void function with reference parameter of type tourtype to input data in a variable of type tourtype.

81 EXERCISE (3.2.6) o The following struct definition consists of four data members: book s title, serial number, price and shelf location. struct Books { char title[50]; int serialno; float price; char location[20]; }; a) Create an array of 100 malay books b) Write the function definition for each of the following i. Function inputdata(): This function receives two parameters; an array of malay books and the size of the array. It then reads data from the user and stores them into the array. ii. Function preciousbook(): this function receives two parameters; an array of malay books and the size of the array. It determines the most expensive book and displays all the information about the book.

82 EXERCISE (3.2.7) A bookshop wants to manage the information of the books sold in their shop. You have been assigned to develop the program to update the books information. The information stored consists of the titles, quantities and prices of books. The title, quantity and price of book are stored in a structure named bookstock. Assuming that there are FIVE HUNDRED (500) books managed by the bookshop. Write definition for the following functions. Note : function parameter must at least have an array of type bookstock. o o o o InputData() : this function prompts user to enter book information and stores them in array of type bookstock. totalprice(): this function calculates total price by summing up multiplication result of quantities and prices. It will then return the result. displayrestock(): this function displays the details of restocked book by finding its quantity which is less than FIVE (5) books. searchtitle(): this function searches book title and display the book details. If the title is not found, display appropriate message. Searched book title is received through this function parameter.

83 RECORD OVERVIEW 3.7 RECORDS AND ARRAY 3.8 RECORDS AND FUNCTION

84 #include <iostream> #include <cstring> #include <conio.h> EXAMPLE 1 OF STRUCT WITH FUNCTION void printbook( struct Books book ); struct Books { char title[50]; char author[50]; char subject[100]; int book_id; }; int main( ) { struct Books Book1; struct Books Book2; // Declare Book1 of type Book // Declare Book2 of type Book strcpy( Book1.title, "Learn C++ Programming"); // book 1 specification strcpy( Book1.author, "Chand Miyan"); strcpy( Book1.subject, "C++ Programming"); Book1.book_id = ;

85 EXAMPLE 1 OF STRUCT WITH FUNCTION strcpy( Book2.title, "Telecom Billing"); // book 2 specification strcpy( Book2.author, "Yakit Singha"); strcpy( Book2.subject, "Telecom"); Book2.book_id = ; printbook( Book1 ); // Print Book1 info printbook( Book2 ); // Print Book2 info getch(); } void printbook( struct Books book ) { cout << "Book title : " << book.title <<endl; cout << "Book author : " << book.author <<endl; cout << "Book subject : " << book.subject <<endl; cout << "Book id : " << book.book_id <<endl; }

86 EXAMPLE 2 OF STRUCT WITH FUNCTION #include <iostream.h> #include <conio.h> #include <stdio.h> #include <cstring.h> void display_details (struct person Someone_I_know); //function prototype //declaraton of struct struct person { char name [30]; int eye_colour; float height; }; int main() { struct person sibling, mother; struct person spouse; //initialization for person : sibling, mother and spouse strcpy(sibling.name, "Diane"); sibling.eye_colour = 1; sibling.height = 1.61;

87 EXAMPLE 2 OF STRUCT WITH FUNCTION strcpy(mother.name,"mary"); mother.eye_colour = 2; mother.height = 1.44; strcpy(spouse.name,"helen"); spouse.eye_colour = 1; spouse.height = 1.7; display_details(sibling); //function call display_details(mother); display_details(spouse); getch(); } void display_details (struct person someone_i_know) { cout << "Name : " << someone_i_know.name <<endl; cout << "Eye colour : "; switch (someone_i_know.eye_colour) { case 1 : cout << "Brown"; break; case 2 : cout << "Blue"; break; case 3 : cout <<"Green"; break; default : cout << "Unknown"; } cout << endl; cout << "Height : " << someone_i_know.height << " metres" << endl; }

88 CSC 138 Structured Programming THE END..