Introduction to Object-Oriented Programming

1. Introduction to Object-Oriented Programming Prof. Shie-Jue Lee Dept. of Electrical Engineering National Sun Yat-sen University

2. MAINTAINING AN ADDRESS BOOK ---Problem Problem Write a program to maintain an address book. The program should repeatedly ask the user whether to display the names and addresses, to add a name and address, to delete a name and address, or to quit. When the names are displayed, they should be listed in alphabetical order. When the program terminates, the names and addresses should be written to a file. The first time the program is run, the program should create the address book; thereafter, each time the program is run, it should read the file and then present the options to the user.

3. MAINTAINING AN ADDRESS BOOK---Sample Input/Output

4. MAINTAINING AN ADDRESS BOOK---Sample Input/Output

5. MAINTAINING AN ADDRESS BOOK---Sample Input/Output

6. MAINTAINING AN ADDRESS BOOK---Sample Input/Output

7. MAINTAINING AN ADDRESS BOOK---Sample Input/Output

8. MAINTAINING AN ADDRESS BOOK---C++ Implementation C++ Implementation #include <iostream> #include <fstream> #include <iomanip> #include <cstdlib> #include <cstring> using namespace std; const int MaxNo = 30; // max numb of names and addresses const int MaxFld = 81; // max length of one record const char filename[ ] = "address.dat"; const int RespSize = 30; // max length of user response void show( char list[ ][ MaxFld ], int no ); bool add( char name[ ], char list[ ][ MaxFld ], int& no ); bool del( int i, char list[ ][ MaxFld ], int& no ); void init( char list[ ][ MaxFld ], int& no ); void quit( char list[ ][ MaxFld ], int no );

9. MAINTAINING AN ADDRESS BOOK---C++ Implementation int main() { char addr[ MaxNo ][ MaxFld ]; char name[ MaxFld ]; char resp[ RespSize ]; int number; int index; // read in names from a file if it exists // and set number to the number of records init( addr, number ); do { cout << "\n\n[S]how names and addresses\n" << "[A]dd a name and address\n" << "[D]elete a name and address\n" << "[Q]uit\n" << "Your choice? "; cin.getline( resp, RespSize );

10. MAINTAINING AN ADDRESS BOOK---C++ Implementation switch ( resp[ 0 ] ) { case 'S': case 's': show( addr, number ); break; case 'A': case 'a': cout << "Name and address to add\n"; cin.getline( name, MaxFld ); if ( !add( name, addr, number ) ) cout << "\a*** Out of room, unable to add: " << name << endl; break;

11. MAINTAINING AN ADDRESS BOOK case 'D': case 'd': cout << "Number of name to delete? "; cin.getline( resp, RespSize ); index = atoi( resp ); if ( !del( index, addr, number ) ) cout << "\a*** Unable to delete number: " << index << endl; break; case 'Q': case 'q': quit( addr, number ); break; default: cout << "\a*** Illegal choice; try again\n"; break; } } while ( resp[ 0 ] != 'Q' && resp[ 0 ] != 'q' ); return 0; }

12. MAINTAINING AN ADDRESS BOOK // show prints the records and numbers them starting with zero. The // numbers are printed in a field of width 3, which is why setw( 3 ) // is used. // show also prints PerScreen (20) records per screen. void show( char list[ ][ MaxFld ], int no ) { int i; char resp[ RespSize ]; const int PerScreen = 20; for ( i = 0; i < no; i++ ) { cout << setw( 3 ) << i << ' ' << list[ i ] << endl; if ( ( i + 1 ) % PerScreen == 0 ) { cout << "Hit RETURN to continue: "; cin.getline( resp, RespSize ); } } }

13. MAINTAINING AN ADDRESS BOOK // add adds a record if possible and updates the number of records bool add( char name[ ], char list[ ][ MaxFld ], int& no ) { int i; // out of room? if ( no >= MaxNo ) return false; // find correct position for name for ( i = no - 1; i >= 0 && strcmp( name, list[ i ] ) < 0; i-- ) strcpy( list[ i + 1 ], list[ i ] ); strcpy( list[ i + 1 ], name ); no++; return true; }

14. MAINTAINING AN ADDRESS BOOK---C++ Implementation // del deletes the record at index I and updates the number of // records bool del( int i, char list[ ][ MaxFld ], int& no ) { int j; // is i in bounds? if ( i < 0 || i >= no ) return false; // move names down to delete entry i for ( j = i; j < no - 1; j++ ) strcpy( list[ j ], list[ j + 1 ] ); no--; return true; }

15. MAINTAINING AN ADDRESS BOOK---C++ Implementation // init reads names from the file address.dat and sets no to the // number of records read. If address.dat does not exist, init // simply sets no to zero and returns. void init( char list[ ][ MaxFld ], int& no ) { ifstream in; in.open( filename ); no = 0; // check if file exists; if not, return. if ( !in ) return; // read records until out of room or end-of-file while ( no < MaxNo ) { in.getline( list[ no ], MaxFld ); if ( !strlen( list[ no ] ) ) break; no++; } in.close(); }

16. MAINTAINING AN ADDRESS BOOK---C++ Implementation // quit writes the records to address.dat void quit( char list[ ][ MaxFld ], int no ) { ofstream out; out.open( filename ); int i; // write records to address.dat for ( i = 0; i < no; i++ ) out << list[ i ] << endl; out.close(); }

17. A TIME STAMP CLASS ---Problem 17

18. A TIME STAMP CLASS ---Problem 18

19. A TIME STAMP CLASS ---Sample Output 19

20. A TIME STAMP CLASS ---Sample Output 20

21. A TIME STAMP CLASS ---Sample Output 21

22. A TIME STAMP CLASS ---C++ Implementation 22 C++ Implementation

23. A TIME STAMP CLASS ---C++ Implementation 23

24. A TIME STAMP CLASS ---C++ Implementation 24

25. A TIME STAMP CLASS ---C++ Implementation 25 #include <iostream> #include <ctime> #include <cstring> using namespace std; class TimeStamp { public: void set( long s = 0 ); time_t get(); const char* getAsString(); const char* getYear(); const char* getMonth(); const char* getDay(); const char* getHour(); const char* getMinute(); const char* getSecond();

26. A TIME STAMP CLASS ---C++ Implementation 26 private: const char* extract( int, int ); time_t stamp; char string[ 30 ]; // holds ctime's return string }; void TimeStamp::set( long s ) { if ( s <= 0 ) stamp = time( 0 ); else stamp = s; } time_t TimeStamp::get() { return stamp; } const char* TimeStamp::getAsString() { return ctime( &stamp ); }

27. A TIME STAMP CLASS ---C++ Implementation 27 const char* TimeStamp::extract( int offset, int count ) { char temp[ 30 ]; strcpy( temp, ctime( &stamp ) ); strncpy( string, temp + offset, count ); string[ count ] = '\0'; // ensure a string return string; } const char* TimeStamp::getYear() { return extract( 20, 4 ); } const char* TimeStamp::getMonth() { return extract( 4, 3 ); }

28. A TIME STAMP CLASS ---C++ Implementation 28 const char* TimeStamp::getDay() { return extract( 0, 3 ); } const char* TimeStamp::getHour() { return extract( 11, 2 ); } const char* TimeStamp::getMinute() { return extract( 14, 2 ); } const char* TimeStamp::getSecond() { return extract( 17, 2 ); }

29. A SEQUENCE HIERARCHY ---Problem 29 Problem Develop an inheritance hierarchy to handle sequences of strings and sorted sequences of strings. A sequence is a list in which there is a first element, a second element, and so on. For example, in the sequence Abby George Ben Abby is the first member, George is the second member, and Ben is the third member. This sequence is considered distinct from the sequence George Ben Abby

30. A SEQUENCE HIERARCHY ---Problem 30 because, for example, the first member, George, is different from the first member, Abby, of the first sequence. A sorted sequence is a sequence in which the elements are in sorted(ascending)order. For example, the sequence Abby Ben George Is a sorted sequence because the elements are in sorted order. The sequence Abby George Ben Is not a sorted sequence because Ben should precedeGeorge in sorted order.

31. A SEQUENCE HIERARCHY ---Problem 31 Class Sequence has data members • To hold strings. • To hold a file name. • To hold the index of the last string. • To handle input and output files. These members are protected so that they are visible throughout the class hierarchy. Class Sequence has public methods to • Add a string at a designated position. • Delete a string at a designated position. • To output the sequence.

32. A SEQUENCE HIERARCHY ---Problem 32 Class Sequence also has a default constructor, a one-parameter char [ ] constructor, and a destructor. The default constructor • Sets the index of the last string to -1 to indicate that no strings are in the sequence. • Sets the file name to the null string to indicate that no file name has been given.

33. A SEQUENCE HIERARCHY ---Problem 33 The one-parameter char [ ] constructor • Sets the index of the last string to -1 to indicate that no strings are in the sequence. • Copies the file name passed into the data member that holds a file name. • Attempts to open the file for input. If the file cannot be opened, the constructor simply returns. • Reads the sequence from the file until end-of-file or until storage is exhausted, whichever occurs first. • Closes the file.

34. A SEQUENCE HIERARCHY ---Problem 34 The destructor • Returns if the file name is the null string. • Open the file for output. • Writes the sequence to the file. • Closes the file.

35. A SEQUENCE HIERARCHY ---Sample Input/Output 35

36. A SEQUENCE HIERARCHY ---Sample Input/Output 36

37. A SEQUENCE HIERARCHY ---C++ Implementation 37

38. A SEQUENCE HIERARCHY ---C++ Implementation 38

39. A SEQUENCE HIERARCHY ---C++ Implementation 39

40. A SEQUENCE HIERARCHY ---C++ Implementation 40 #include <iostream> #include <fstream> #include <cstdlib> #include <cstring> using namespace std; const int MaxStr = 50; const int MaxSize = 80; class Sequence { public: bool addS( int, char [ ] ); bool del( int ); void output(); Sequence(); Sequence( char [ ] ); ~Sequence();

41. A SEQUENCE HIERARCHY ---C++ Implementation 41 protected: char s[ MaxStr ][ MaxSize ]; char filename[ MaxSize ]; int last; ifstream in; ofstream out; };

42. A SEQUENCE HIERARCHY ---C++ Implementation 42 bool Sequence::addS( int pos, char entry[ ] ) { if ( last == MaxStr - 1 || pos < 0 || pos > last + 1 ) return false; for ( int i = last; i >= pos; i-- ) strcpy( s[ i + 1 ], s[ i ] ); strcpy( s[ pos ], entry ); last++; return true; }

43. A SEQUENCE HIERARCHY ---C++ Implementation 43 bool Sequence::del( int pos ) { if ( pos < 0 || pos > last ) return false; for ( int i = pos; i < last; i++ ) strcpy( s[ i ], s[ i + 1 ] ); last--; return true; } void Sequence::output() { for ( int i = 0; i <= last; i++ ) cout << i << " " << s[ i ] << endl; }

44. A SEQUENCE HIERARCHY ---C++ Implementation 44 Sequence::Sequence() { last = -1; filename[ 0 ] = '\0'; } Sequence::Sequence( char fname[ ] ) { last = -1; strcpy( filename, fname ); in.open( filename ); if ( !in ) return; while ( last < MaxStr - 1 ) { in.getline( s[ last + 1 ], MaxSize ); if ( !strlen( s[ last + 1 ] ) ) break; last++; } in.close(); }

45. A SEQUENCE HIERARCHY ---C++ Implementation 45 Sequence::~Sequence() { if ( filename[ 0 ] == '\0' ) return; out.open( filename ); for ( int i = 0; i <= last; i++ ) out << s[ i ] << endl; out.close(); } class SortedSeq : public Sequence { public: bool addSS( char [ ] ); SortedSeq(); SortedSeq( char [ ] ); protected: void sort(); };

46. A SEQUENCE HIERARCHY ---C++ Implementation 46 void SortedSeq::sort() { char temp[ MaxSize ]; int j; for ( int i = 0; i <= last - 1; i++ ) { strcpy( temp, s[ i + 1 ] ); for ( j = i; j >= 0; j-- ) if ( strcmp( temp, s[ j ] ) < 0 ) strcpy( s[ j + 1 ], s[ j ] ); else break; strcpy( s[ j + 1 ], temp ); } }

47. A SEQUENCE HIERARCHY ---C++ Implementation 47 bool SortedSeq::addSS( char entry[ ] ) { int i; for ( i = 0; i <= last; i++ ) if ( strcmp( entry, s[ i ] ) <= 0 ) break; return addS( i, entry ); } SortedSeq::SortedSeq() : Sequence() { } SortedSeq::SortedSeq( char fname[ ] ) : Sequence( fname ) { sort(); }

48. A COMPLEX NUMBER CLASS ---Problem 48

49. A COMPLEX NUMBER CLASS ---Problem 49

50. A COMPLEX NUMBER CLASS ---Sample Output 50