C++ Assignment

ASSIGNMENT 1



PROBLEM STATEMENT: STATEMENT :

Write a C++ program that models employee database using inheritance. The program should perform the following: a) Start with a base class employee. b) From this class derive the other classes-manager, scientist and   laborer. c) The manager and scientist classes contain additional information. d) Also write proper member functions to handle these operations .



ALGORITHM: ALGORITHM:

set constant LEN=80         /*Array size for name of employee*/ /*Algorithm for employee class*/ class employee begin private members: variable name[LEN] (character) variable number (unsigned long) public members: /*Algorithm for getdata()*/ function getdata() begin write "\n enter last name:" read name; write "\n enter number:" read number; end of function getdata() /*Algorithm for putdata()*/ function putdata() begin write "name:" write name write "\nnumber:" write number end of function putdata() end of class employee /*Algorithm for manager class which is derived publicly from employee class*/ class manager: public employee

1

begin private members: variable title[LEN] (character) variable dues (double) public members: /*Algorithm for getdata(). Overloading the getdata() of employee*/ function getdata() begin call employee::getdata() /*Calling employee’s getdata()*/ write "\n enter title:" read title write "\n enter golf club dues:" read dues end of function getdata() /*Algorithm for putdata(). Overloading the putdata() of employee*/ function putdata() begin call employee::putdata() /*Calling employee’s putdata()*/ write "\n title:" write title write "golf club dues:" write dues end of function putdata() end of class manager /*Algorithm for scientist class which is derived publicly from employee class*/ class scientist: public employee begin private members: variable pubs (integer) public members: /*Algorithm for getdata(). Overloading the getdata() of employee*/ function getdata() begin call employee:: employee::getda getdata() ta() /*Calling /*Calling employee’s employee’s getdata()*/ write "\n enter name of pubs:" read pubs end of function getdata() /*Algorithm for putdata(). Overloading the putdata() of employee*/ function putdata() begin call employee::putdata() /*Calling employee’s putdata()*/ write "\n name of pubications:" write pubs

2

end of function putdata() end of class scientist /*Algorithm for laborer class which is derived publicly from employee class*/ class laborer: public employee begin end of class laborer function main() begin object m1,m2 (manager type) object s1 (scientist type) object l1 (laborer type) write "\n" write "\n enter data for manager-1" call m1.getdata() write "\n enter data for manager-2" call m2.getdata() write "\n enter data for scientist-1" call s1.getdata(); write "\n enter data for laborer1-1" call l1.getdata() write "\n data for manager-1" call m1.putdata() write "\n data for manager-2" call m2.putdata() write "\n data for scientist-1" call s1.putdata(); write "\n data for laborer-1" call l1.putdata() end of function main()



PROGRAM CODE: CODE:

#include #include const int LEN=80; class employee { private: char name[LEN];               //employee name unsigned long number;         //employee number public: void getdata() { cout<<"\n enter last name:";

3

cin>>name; cout<<"\n enter number:"; cin>>number; } void putdata() { cout<<"name:"<>title; cout<<"\n enter golf club dues:"; cin>>dues; } void putdata() { employee::putdata(); cout<<"\n title:"<>pubs; } void putdata() { employee::putdata();

4

cout<<"\n name of pubications:"<
INPUT AND OUTPUT : enter data for manager-1 enter last name: palak enter number: 23 enter title: president enter golf club dues:10000 enter data for manager-2 enter last name: vijay enter number: 26 enter title: vice-president enter golf club dues:14867 enter data for scientist-1 enter last name: edward enter number:1231 enter name of pubs:125 enter data for laborer1-1 enter last name: shyamal

5

enter number: 456 data for manager-1 name: palak number: 23 title: president golf club dues:10000 data for manager-2 name: vijay number: 26 title: vice-president golf club dues:14867 data for scientist-1 name: edward number: 1231 name of pubications:125 data for laborer-1 name: shyamal number: 456



DISCUSSIONS:

1) In the program there is a base class called employee. From the base class we have derived three other classes namely- manager, scientist and laborer. Actually this program show the application of  inheritance. 2) This program also shows explicitly how the derived functions can use the member functions of  the base class.

6

ASSIGNMENT 2



PROBLEM STATEMENT:

Create a class that can store a matrix. Then overload the +,-,* operators such they can be applied with the objects of the matrix class . Also provide the member functions to get data for the matrix, to print the matrix and to transpose the matrix. A matrix object may or may not be initialized at the time of creation and if the proper dimensions is not given then the dimensions should be set to 0*0.



ALGORITHM:

/*Algorithm for class matrix*/ matrix class begin private members: enumerated data max, cmax set max=10, cmax=10 variable mat[max][cmax] (integer) variable row (integer) variable column (integer) public members: function getdata() /*function declaration*/ /*Algorithm for matrix()*/ matrix()    //no argument constructor begin set row=0 set column=0 end of maitrix() /*Algorithm for matrix(integer r, integer c)*/ matrix(integer r,integer c)    //2 argument constructor begin if(r>max OR c>cmax)then write”The given dimension is too large” write”Please specify the dimension within range” write”row=” read row write”column=” read column else set row=r set column=c endif  Call function getdata() end of matrix(integer r, integer c) /*function declrations*/ function putdata()

7

friend function operator +(matrix&,matrix&) friend function operator -(matrix&,matrix&) friend function operator *(matrix&,matrix&) function transpose(matrix&) end of class matrix /*Algorithm for getdata()*/ function matrix::getadata() begin if(row=0 AND column=0)then write ”The dimensions of the matrix is 0*0” write  “please enter the no. of rows of the matrix:” read row write “Enter the no. of columns of the matrix:” read column endif  Write “Enter the elements of the matrix” for i=1 to row do begin for j=1 to column do begin read mat[i][j] endfor endfor end of function getdata() /*Algorithm for putdata()*/ function matrix::putdata() begin Write “The matrix is:” for i=1to row do begin for j=1 to column do begin write mat[i][j] endfor endfor end of function putdata() /*Algorithm for overloading + operator*/ function operator +(matrix &m1,matrix &m2) begin object m (matrix type) if(m1.row!=m2.row OR m1.column!=m2.column) then write “Addition not possible” return m else set m.row=m1.row set m.column=m1.column for i=1 to m.row do begin for j=1 to m.column do

8

begin set m.mat[i][j]=m1.mat[i][j]+m2.mat[i][j] endfor endfor return m endif  end of function operator +(matrix &m1,matrix &m2) /*Algorithm for overloading - operator*/ function operator -(matrix &m1,matrix &m2) begin object m (matrix type) if(m1.row!=m2.row OR m1.column!=m2.column) then write “Subtraction not possible” return m else set m.row=m1.row set m.column=m1.column for i=1 to m.row do begin for j=1 to m.column do begin set m.mat[i][j]=m1.mat[i][j]-m2.mat[i][j] endfor endfor return m endif  end of function operator -(matrix &m1,matrix &m2) /*Algorithm for transpose(matrix &m1)*/ function matrix::transpose(matrix &m1) begin set row=m1.column set column=m1.row for i=1 to m1.row do begin for j=1 to m1.column do set mat[j][i]=m1.mat[i][j] endfor endfor end of function matrix::transpose(matrix &m1) /*Algorithm for overloading * operator*/ function operator *(matrix &m1,matrix &m2) begin object m (matrix type) if(m1.column!=m2.row) then write “Multiplication not possible” return m endif  set m.row=m1.row set m.column=m2.column

9

for i=1 to m1.row do begin for j=1 to m1.column do begin set m.mat[i][j]=0 for k=1 to m2.row do begin set m.mat[i][j]= m.mat[i][j]+ m1.mat[i][k]*m2.mat[k][j] endfor endfor endfor return m end of function operator *(matrix &m1,matrix &m2) function main() begin objects m1,m2,m3,m4,m5 (matrix type) call m1.getdata() call m1.putdata() call m2.getdata() call m2.putdata() set m3=m1+m2 write “After addition-“ call m3.putdata() set m4=m1-m2 write “After Subtraction-“ call m4.putdata() set m5=m1*m2 write “After Multiplication-“ call m5.putdata() variable m6(3,5),m7 (matrix type) call m6.putdata() call m7.transpose(m6) write “After Transpose-“ Call m7.putdata() End of function main()



PROGRAM CODE:

#include #include #include #include class matrix { private: enum{max=10,cmax=10}; int mat[max][cmax]; int row;

10

int column; public: void getdata(); matrix() { row=0; column=0; } matrix(int r,int c) { if(r>max||c>cmax) { cout<>row; cout<<”column=”; cin>>column; } else { row=r; column=c; } getdata(); } void putdata(); friend matrix operator +(matrix&,matrix&); friend matrix operator -(matrix&,matrix&); friend matrix operator *(matrix&,matrix&); void transpose(matrix&); }; void matrix::getdata() { if(row==0&&column==0) { cout<>row; cout<>column; } cout<>mat[i][j]; } void matrix::putdata() { cout<
11

for(int i=1;i<=row;i++) { cout<
12

mat[j][i]=m1.mat[i][j]; } matrix operator *(matrix &m1,matrix &m2) { matrix m; if(m1.column!=m2.row) { cout<
13



INPUT AND OUTPUT:

The dimensions of the matrix is 0*0 please enter the no. of rows of the matrix: 2 Enter the no. of columns of the matrix: 2 Enter the elements of the matrix 13 45 The matrix is: 13 45 The dimensions of the matrix is 0*0 please enter the no. of rows of the matrix: 2 Enter the no. of columns of the matrix: 2 Enter the elements of the matrix 56 35 The matrix is: 56 35 After additionThe matrix is: 6  9 7 10 After SubtractionThe matrix is: -4 -3 1   0 After MultiplicationThe matrix is: 14  21 35  49



DISCUSSIONS:

The above matrix class can form a 10x10 matrix. It can be further modified to take larger matrix. Here few operations are done for matrix. It can be further modified to cover all the matrix operations.

14

ASSIGNMENT 3



PROBLEM STATEMENT: STATEMENT :

Write a C++ program that performs the concatenation of two strings using operator overloadingoverload the operator ‘+’ and ‘=’ in your program. 

ALGORITHM: ALGORITHM:

/*Algorithm for string class*/ class string begin private members: variable str[20] (character) public members: /*Algorithm for one argument constructor*/ string() begin call strcpy(str," "); end of string() /*function declarations*/ function operator +(string ss); function operator =(string ss); /*Algorithm for init()*/ function init() begin write "enter the string:" read str end of function init() /*Algorithm for show()*/ function show()

15

begin write "\nthe concatenated string is:" write str end of function show() end of class string

/*Algorithm for overloading = operator*/ function string::operator =(string ms) begin call strcpy(str,ms.str) return *this end of fuction operator =(string ms)

/*Algorithm for overloading + operator*/ function string:: operator +(string ss) begin if(call strlen(str) + call strlen(ss.s) <= 80) then object temp (string type) call strcpy(temp.str,str) call strcat(temp.str,ss.str) strcat(temp.str,ss.str) return temp endif  end of function operator +(string ss)

/*Algorithm for function main()*/ function main() begin object A,B,C (string type) call  A.init() call B.init() set C=A+B call C.show()

16

end of main()



PROGRAM CODE: CODE:

#include #include

class string { private: char str[20]; public: string() { strcpy(str," "); } string operator +(string ss); string operator =(string ss); void init() { cout<<"enter the string:"; cin>>str; } void show() { cout<<"\nthe concatenated string is:"<
string string::operator =(string ms) {

17

strcpy(str,ms.str); return *this; }

string string:: operator +(string ss) { if(strlen(str)+strlen(ss.s)<=80) { string temp; strcpy(temp.str,str); strcat(temp.str,ss.str); return temp; } }

void main() { string A,B,C; A.init(); B.init(); C=A+B ; C.show(); }



INPUT AND OUTPUT: enter the string:Hello enter the string:World the concatenated string is: HelloWorld



DISCUSSIONS:

18

The class string uses a precompiled function strcpy(). This function can be avoided by writing own function for copying a string. This above class usage few string oriented function. This class may be further modified for defining other string related functions.

19

ASSIGNMENT 4

 PROBLEM STATEMENT:

Write a C++ program to create a linked list using template.

 ALGORITHM: /*Defination of structure link*/ template structure link begin T data; pointer variable next (link type) end of structure link /*Algorithm for class linkedlist*/ template class linkedlist begin private members: pointer variable head (link) varibale p (integer) varibale num (T) public members: /*Algorithm for one argument constructor*/ linkedlist() begin set head=NULL end of linkedlist() /*Algorithm for getdata()*/ function getdata() begin write "enter number of items you want to insert:" read p for i=0 to p-1 do begin write "\n enter data:" read num call additem(num) endfor end of function getdata() /*function declaration*/

20

function additem(T item) function display() end of class linkedlist /*algorithm for aditem(T item)*/ template function linkedlist::additem(T item) begin pointer varibale temp (link) set temp=new link set temp->data=item set temp->next=NULL if(head==NULL) then set head=temp else pointer variable list (link) set list=head while(list->next!=NULL) do begin set list=list->next endwhile set list->next=temp endif  end of function additem(T item) /*Algorithm for display()*/ template function linkedlist::display() begin pointer varibale start (link) set start=head write "\n\n" while(start!=NULL) do begin write start->data write "-->" set start=start->next; endwhile write "END" end of function display() /*Algorithm for main()*/ function main() begin write "\n\t------For integer data type------\n" object A linkedlist) call A.getdata() call A.display() write "\n\t------For double data type------\n" object B (linkedlist) call B.getdata() call B.display()

21

write "\n\t------For character data type------\n" object C (linkedlist) call C.getdata() call C.display() call getch(); end of main()

 PROGRAM CODE:

#include #include #include template struct link { T data; link *next; }; template class linkedlist { private: link *head; int p; T num; public: linkedlist() { head=NULL; } void getdata() { cout<<"enter number of items you want to insert:"; cin>>p; for(int i=0;i>num; additem(num); } } void additem(T item); void display(); }; template void linkedlist::additem(T item)

22

{ link *temp; temp=new link; temp->data=item; temp->next=NULL; if(head==NULL) head=temp; else { link *list; list=head; while(list->next!=NULL) list=list->next; list->next=temp; } }

template void linkedlist::display() { link *start; start=head; cout<<"\n\n"; while(start!=NULL) { cout<data<<"-->"; start=start->next; } cout<<"END"; } void main() { cout<<"\n\t------For integer data type------\n"; linkedlist A; A.getdata(); A.display(); cout<<"\n\t------For double data type------\n"; linkedlist B; B.getdata(); B.display(); cout<<"\n\t------For character data type------\n"; linkedlist C; C.getdata(); C.display(); getch(); }

 INPUT AND OUTPUT:

23

------For integer data type-----enter number of items you want to insert:4 enter data:34 enter data:67 enter data:12 enter data:89 34-->67-->12-->89-->END ------For double data type-----enter number of items you want to insert:3 enter data:567.67 enter data:341.22 enter data:789.908 567.67-->341.22-->789.908-->END ------For character data type-----enter number of items you want to insert:5 enter data:H enter data:E enter data:L enter data:L enter data:O H-->E-->L--> L-->O-->END



DISCUSSIONS:

The advantage of this class is that it can of any datatype. The other advantage of this class is it can be of any size. Here only few operations of linked list are shown. The class can be further modified to perform other operations of linked list.

24

ASSIGNMENT 5

 PROBLEM STATEMENT:

Write a C++ program to implement generic stack using array. Also provide methods to handle exceptions that will arise during push or pop operations.

 ALGORITHM:

/*Algorithm for class stack*/ template class stack begin private members: variable st[4] (T) variable top (integer) public members: /*for exception handling*/ class Full{ } class Empty{ } /*function and constructor declaration*/ stack() function push(T var) function pop() function display() end of class stack /*Algorithm for no argument constructor*/ template stack :: stack() begin set top=-1 end of stack() /*Algorithm for push*/ template function stack :: push(T var) begin if(top>=3) then throw Full() endif  set st[++top]=var end of function push(T var)

25

/*Algorithm for pop*/ template function stack :: pop() begin if(top<0) then throw Empty() endif  return st[top--] end function pop() /*Algorithm for display()*/ template function stack :: display() begin write "\n\t The stack is:" for i=0 to top do begin write st[i] write " " endfor write "\n\n" end of function display() /*Algorithm for main()*/ function main() begin object A (stack) variables item,n : integer while(true) do begin try begin write "\n\tEnter" write "\n\t1.for push" write "\n\t2.for pop" write "\n\t3.for display" write "\n\t4.for exit" write "\n\n enter your choice:" read n; switch(n) begin case 1: write "\n enter item:" read item call A.push(item) break case 2: write "\n popped element:" write A.pop() break case 3:

26

call A.display() break case 4: call exit(0) break default: write "Wrong choice" break endswitch endtry catch(stack::Full) begin write "\nstack is full\n" endcatch catch(stack::Empty) begin write "\nstack is empty\n" end of catch(stack::Empty) endwhile call getch() end of function main()

 PROGRAM CODE:

#include #include #include #include template class stack { private: T st[4]; int top; public: class Full{ }; class Empty{ }; stack(); void push(T var); T pop(); void display(); }; template stack :: stack() { top=-1; }

27

template void stack :: push(T var) { if(top>=3) throw Full(); st[++top]=var; } template T stack :: pop() { if(top<0) throw Empty(); return st[top--]; } template void stack :: display() { cout<<"\n\t The stack is:"; for(int i=0;i<=top;i++) cout< A; int item,n; while(1) { try { cout<<"\n\tEnter"; cout<<"\n\t1.for push"; cout<<"\n\t2.for pop"; cout<<"\n\t3.for display"; cout<<"\n\t4.for exit"; cout<<"\n\n enter your choice:"; cin>>n; switch(n) { case 1: cout<<"\n enter item:"; cin>>item; A.push(item); break; case 2: cout<<"\n popped element:"<
28

break; case 4: exit(0); default: cout<<"Wrong choice"; break; } } catch(stack::Full) { cout<<"\nstack is full\n"; } catch(stack::Empty) { cout<<"\nstack is empty\n"; } } getch(); }

 INPUT AND OUTPUT: Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:1 enter item:55 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:1 enter item:45 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:1 enter item:78 Enter 1.for push 2.for pop 3.for display

29

4.for exit enter your choice:1 enter item:47 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:3 The stack is: 55 45 78 47 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:1 enter item:101 stack is full Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:2 popped element:47 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:2 popped element:78 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:2 popped element:45 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:2

30

popped element:55 Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:2 stack is empty Enter 1.for push 2.for pop 3.for display 4.for exit enter your choice:4

 DISCUSSION

To show the exception handling we use here a array of size 4. To handle large amount of data the size can be extended. The advantage of this class is that it can work with any datatype.

31

ASSIGNMENT 6

 PROBLEM STATEMENT:

Write a C++ program that performs the addition of two polynomials using linked list.

 ALGORITHM:

/*definition of structure poly*/ structure poly begin variable coeff (integer) variable pow (integer) pointer variable next (ploy type) end of structure poly

/*Algorithm for add2poly class*/ class add2poly begin private members: pointer variables poly1, poly2, poly3 (poly type) public members: /*Algorithm for no argument constructor*/ add2poly() begin set poly1=poly2=poly3=NULL end of add2poly() /*function declaration*/ function addpoly()

32

function display() end of class add2poly

/*Algorithm for addpoly()*/ function add2poly :: addpoly() begin variable i,p (integer) painter variables newl, end (poly type) set new1=end=NULL write "Enter highest power for x\n" read p /*Creation of 1st polynomial*/ write "\nFirst Polynomial\n" for i=p to 0 do begin set newl=new poly set newl->pow=p write "Enter Co-efficient for degree" write i write "::  " read newl->coeff  set newl->next=NULL if(poly1==NULL) then set poly1=newl else set end->next=newl end if  set end=newl endfor /* Creation of 2 nd polynomial */ write "\n\nSecond Polynomial\n" set end=NULL

33

for i=p to 0 do begin set newl=new poly set newl->pow=p write "Enter Co-efficient for degree" write i write "::  " read newl->coeff  set newl->next=NULL if(poly2==NULL) then set poly2=newl else set end->next=newl endif  set end=newl endfor

/*Addition Logic*/ pointer variable p1, p2 (poly type) set p1 = poly1 set p2 = poly2 set end=NULL while(p1 !=NULL AND p2!=NULL) do begin if(p1->pow == p2->pow) than set newl=new poly set newl->pow=p-set newl->coeff=p1->coeff + p2->coeff  set newl->next=NULL if(poly3==NULL) then set poly3=newl else

34

set end->next=newl endif  set end=newl endif  set p1=p1->next set p2=p2->next endwhile end of function addpoly()

/*Algorithm for display*/ function add2poly :: display() begin pointer variable t (poly type) set t=poly3 write "\n\nAnswer after addition is : " while(t!=NULL) do begin write t->coeff  write " X" write t->pow write "  " set t=t->next endwhile end of function display()

/*Algorithm for main function*/ function main() begin call clrscr(); object obj (add2poly type) call obj.addpoly() call obj.display()

35

call getch() end of function main()

 PROGRAM CODE:

#include #include #include

struct poly{ int coeff; int pow; poly *next; };

class add2poly { poly *poly1, *poly2, *poly3; public: add2poly(){poly1=poly2=poly3=NULL;} void addpoly(); void display(); };

void add2poly :: addpoly() { int i,p; poly *newl=NULL,*end=NULL; cout<<"Enter highest power for x\n"; cin>>p;

36

//Read first poly cout<<"\nFirst Polynomial\n"; for(i=p;i>=0;i--) { newl=new poly; newl->pow=p; cout<<"Enter Co-efficient for degree"<>newl->coeff; newl->next=NULL; if(poly1==NULL) poly1=newl; else end->next=newl; end=newl; }

//Read Second poly cout<<"\n\nSecond Polynomial\n"; end=NULL; for(i=p;i>=0;i--) { newl=new poly; newl->pow=p; cout<<"Enter Co-efficient for degree"<>newl->coeff; newl->next=NULL; if(poly2==NULL) poly2=newl; else end->next=newl; end=newl; }

37

//Addition Logic poly *p1=poly1,*p2=poly2; end=NULL; while(p1 !=NULL && p2!=NULL) { if(p1->pow == p2->pow) { newl=new poly; newl->pow=p--; newl->coeff=p1->coeff + p2->coeff; newl->next=NULL; if(poly3==NULL) poly3=newl; else end->next=newl; end=newl; } p1=p1->next; p2=p2->next; } }

void add2poly :: display() { poly *t=poly3; cout<<"\n\nAnswer after addition is : "; while(t!=NULL) { cout<coeff; cout<<" X"<pow; cout<<"  ";

38

t=t->next; } }

void main() { clrscr(); add2poly obj; obj.addpoly(); obj.display(); getch(); }

 INPUT AND OUTPUT: Enter highest power for x 4 First Polynomial Enter Co-efficient Enter Co-efficient Enter Co-efficient Enter Co-efficient Enter Co-efficient

for for for for for

Second Polynomial Enter Co-efficient for Enter Co-efficient for Enter Co-efficient for Enter Co-efficient for Enter Co-efficient for

degree4:12 degree3: 9 degree2: 0 degree1: 7 degree0: 23 degree4: degree3: degree2: degree1: degree0:

0 4 9 9 4

Answer after addition is : 12 X4  13 X3  9 X2  16 X1  27 X0

 DISCUSSIONS: The above program only performs addition of two polynomial. It can be further modified to permor  other operations of polynomial.

39

ASSIGNMENT 7



PROBLEM STATEMENT:

Write a C++ program that overloads stream operators-‘>>’ and ‘<<’.



ALGORITHM:

/*Algorithm for number class*/ class number begin private members: variable data_1 (integer) variable data_2 (integer) public members: /*Algorithm for no argument constructor*/ number() begin set data_1=0 set data_2=0 end of number() /*Algorithm for two argument constructor*/ number(integer num_1,integer num_2) begin set data_1=num_1 set data_2=num_2; end of number(integer num_1, integer num_2) /*declaration of functions*/ friend function operator<<(ostream &s, number &obj); friend function &operator>>(istream &s,number &obj); end of class number /*Algorithm for overloading << operator*/ function operator<<(ostream &s,number &obj) begin s<<"("<> operator*/ function operator>>(istream &s,number &obj) begin s>>obj.data_1>>obj.data_2

40

return s end of function operator>>(istream &s,number &obj) /*Algorithm for main function*/ function main() begin call clrscr() objects obj_1(5,6), obj_2(7,8), obj_3 (number type) write "\n Value of obj_1 = "<
41

istream &operator>>(istream &s,number &obj) { s>>obj.data_1>>obj.data_2; return s; } void main() { clrscr(); number obj_1(5,6); number obj_2(7,8); number obj_3; cout<<"\n Value of obj_1 = "<>obj_3; cout<<"\n Value of obj_3 = "<
INPUT AND OUTPUT: Value of obj_1 = (5,6) Value of obj_2 = (7,8) Enter the value of obj_3 : 25 32 Value of obj_3 = (25,32)



DISCUSSIONS:

Normally we can not read or write directly object of class using cin and cout respectively. But by overloading << and >> operators we can do that.

42

ASSIGNMENT 8



PROBLEM STATEMENT:

Write a c++ program to create a class fraction and which represents a fractional number and perform addition and subtraction of this fractional numbers.



ALGORITHM:

/*Algorithm for class fraction*/ class fraction begin private members: variable numerator,denominator (integer) public members: /*Algorithm for no argument constructor*/ fraction() begin set numerator=0 set denominator=1 end of fraction() /*Algorithm for two argument constructor*/ fraction(integer i, integer j) begin set numerator=i set denominator=j end of fraction(integer i, integer j) /*Algorithm for overload + operator*/ function operator +(fraction op) begin variable LCM (integer) set LCM=lcm(denominator,op.denominator) variable a (integer) set a=LCM/denominator variable b (integer) set b=LCM/op.denominator variable num (integer) set num=numerator*a+op.numerator*b variable deno (integer) set deno=LCM return(fraction(num,deno)) end of function operator +(fraction op) /*Algorithm for overloading * operator*/ function operator *(fraction op)

43

begin variable num (integer) set num=numerator*op.numerator variable deno (integer) set deno=op.denominator*denominator return(fraction(num,deno)) end of function operator *(fraction op) /*Algorithm for overloading = operator*/ function operator =(fraction op) begin set numerator=op.numerator set denominator=op.denominator end of function operator =(fraction op) /*Algorithm for overloading – operator*/ function operator -() begin set numerator=-numerator set denominator=denominator return(fraction(numerator,denominator)) end of function operator -() /*Algorithm for display()*/ function display() begin write "\ndenom=" write denominator write "\t numerator=" write numerator end of function display() /*function declaration*/ funtion lcm(integer a,integer b); end of class fraction /*Algorithm for lcm*/ function fraction::lcm(integer a,integer b) begin for i=1to infinity do begin if(i%a==0 AND i%b==0) than return i else continue endif  endfor end of function lcm(integer a,integer b)

function main() begin objects A(3,5),B(4,6),C(2,3),D(2,5),E (fraction type) set E=(-A+B)*C+D write "\n\t-----data for fraction A-----\n" call A.display()

44

write "\n\t-----data for fraction B-----\n" call B.display() write "\n\t-----data for fraction C-----\n" call C.display() write "\n\t-----data for fraction D-----\n" call D.display() write "\n\t-----Final Result-----\n" call E.display() call getch() end of function main()



PROGRAM CODE:

#include #include #include

class fraction { private: int numerator,denominator; public: fraction() { numerator=0; denominator=1; } fraction(int i,int j) { numerator=i; denominator=j; } fraction operator +(fraction op) { int LCM=lcm(denominator,op.denominator); int a=LCM/denominator; int b=LCM/op.denominator; int num=numerator*a+op.numerator*b; int deno=LCM; return(fraction(num,deno)); }

45

fraction operator *(fraction op) { int num=numerator*op.numerator; int deno=op.denominator*denominator; return(fraction(num,deno)); } void operator =(fraction op) { numerator=op.numerator; denominator=op.denominator; } fraction operator -() { numerator=-numerator; denominator=denominator; return(fraction(numerator,denominator)); } void display() { cout<<"\ndenom="<
46

D.display(); cout<<"\n\t-----Final Result-----\n"; E.display(); getch(); }



INPUT AND OUTPUT:

-----data for fraction A----denom=5 numerator=-3 -----data for fraction B----denom=6 numerator=4 -----data for fraction C----denom=3 numerator=2 -----data for fraction D----denom=5 numerator=2 -----Final Result----denom=90 numerator=40



DISCUSSIONS:

Only few operations for fractional numbers are shown here. The class can be further modified to include other operations of fractional numbers.

47

ASSIGNMENT 9

 PROBLEM STATEMENT:

Write a c++ program to overload pre-increment and post-increment operators.

 ALGORITHM:

/*Algorithm for index class*/ class index begin private members: variable count (integer)

public members: /*Algorithm for no argument constructor*/ index() begin set count=0 end of index() /*Algorithm for one argument constructor*/ index(integer i) begin write "\n one argument constructor is called:" set count=i end of index(integer i) /*Algorithm for display()*/ function display() begin write "\ndata=" write count end of function display()

48

/*Algorithm for copy constructor*/ index(index &a) begin set count=a.count end of index(index &a) /*Algorithm for overloading pre-increment operator*/ function operator ++() begin return(++count) end of function operator ++() /*Algorithm for post-increment operator*/ function operator ++(integer) begin return(count++) end of function operator ++(integer) /*Algorithm for overloading = operator*/ function operator =(index &b) begin set count=b.count return index(count) end of function operator =(index &b) /*Algorithm for destructor*/ ~index() begin write "\n\n destructor is called" end of ~index() end of class index

/*Algorithm for main()*/ function main() begin object A(34), B, C, E, F, D (index type)

49

set D=A

call A.display() call D.display() set C=D set E=F=D call C.display() call F.display() call E.display() set F=C++ call F.display() set E=++F call E.display() end of function main()

 PROGRAM CODE:

#include #include class index { private: int count; public: index() { count=0; } index(int i) { cout<<"\n one argument constructor is called:";

50

count=i; } void display() { cout<<"\ndata="<
void main() { index A(34),B,C,E,F;

51

index D=A; A.display(); D.display(); C=D; E=F=D; C.display(); F.display(); E.display(); F=C++; F.display(); E=++F; E.display(); }

 INPUT AND OUTPUT: one argument constructor is called: data=34 data=34 one argument constructor is called: destructor is called one argument constructor is called: one argument constructor is called: destructor is called destructor is called data=34 data=34 data=34 one argument constructor is called: one argument constructor is called: destructor is called destructor is called data=34 one argument constructor is called: one argument constructor is called: destructor is called destructor is called data=35

52

destructor is called destructor is called destructor is called destructor is called destructor is called destructor is called

 DISCUSSIONS:

The above class shows how destructors are called. This class’s objects may be use for counting purpose.

53

ASSIGNMENT 10

 PROBLEM STATEMENT:

Write a C++ program to implement Generic Bubble sort.

 ALGORITHM:

/*Algorithm for bubble class*/ template class bubble begin private members: varibale a[25] (t) public members: /*function declarations*/ function get(integer) function sort(integer) function display(integer) end of class bubble

/*Algorithm for get() which collect data from user*/ template function bubble ::get(int n) begin write "\nEnter the array elements:" for i=0 to n-1 do begin write a[i] endfor end of function get(int n)

54

/*Algorithm for display()*/ template function bubble ::display(integer n) begin write "\n The sorted array is:\n" for i=0 to n-1 do begin write a[i] write " " endfor end of function display(integer n)

/*Algorithm for bubble sort*/ template function bubble ::sort(integer n) begin variable i,j (integer) variable temp (t) for i=0 to n do begin for j=i+1 to n do begin if(a[i]>a[j]) then temp=a[i]; a[i]=a[j]; a[j]=temp; endif  endfor endfor end of function sort(integer n)

55

/*Algorithm for main()*/ function main() begin variable n (integer) object b1 (bubble) object b2 (bubble) call clrscr() write "\n Bubble Sort on Integer Values" write "\n Enter the size of array:" read n call b1.get(n) call b1.sort(n) call b1.display(n) write "\n Bubble Sort on Float Values" write "\n Enter the size of array:\n" read n call b2.get(n) call b2.sort(n) call b2.display(n) call getch() end of function main()

 PROGRAM CODE:

#include #include

template class bubble { private:

56

t a[25]; public: void get(int); void sort(int); void display(int); };

template void bubble ::get(int n) { int i; cout<<"\nEnter the array elements:"; for(i=0; i>a[i]; }

template void bubble ::display(int n) { int i; cout<<"\n The sorted array is:\n"; for(i=0;i
template void bubble ::sort(int n) { int i,j; t temp; for(i=0;i
57

for(j=i+1;ja[j]) { temp=a[i]; a[i]=a[j]; a[j]=temp; } } } }

void main() { int n; bubble b1; bubble b2; clrscr(); cout<<"\n Bubble Sort on Integer Values"; cout<<"\n Enter the size of array:"; cin>>n; b1.get(n); b1.sort(n); b1.display(n); cout<<"\n Bubble Sort on Float Values"; cout<<"\n Enter the size of array:\n"; cin>>n; b2.get(n); b2.sort(n); b2.display(n); getch(); }

58

 INPUT AND OUTPUT: Bubble Sort on Integer Values Enter the size of array:8 Enter the array elements:12 34 3 56 33 67 7 66 The sorted array is: 3  7  12  33  34  56  66  67 Bubble Sort on Float Values Enter the size of array: 6 Enter the array elements:4.78 3.67 6.7 2.68 8.97 7.89 The sorted array is: 2.68   3.67   4.78   6.7   7.89   8.97

59

 DISCUSSIONS:

The bubble sort is not efficient in comparison to other sorting techniques. But for smaller lists this sorting technique works fine. The bubble sort is data sensitive that is if the list is sorted it will stop sorting. To take more elements program can be modified. The Complexity of bubble sort is

60

( ).

O n

2

ASSIGNMENT 11

 PROBLEM STATEMENT:

Write a C++ program to show the usefulness of virtual function.

 ALGORITHM:

/*Algorithm for Base class*/ class Base begin public members: /*Algorithm for display()*/ function display () begin write "\n Display base" end of function display() /*Algorithm for show() which is virtual*/ function virtual show() begin write "\n show base" end of function virtual show() end of class Base /*Algorithm for Derived class which derived publicly from Base class*/ class Derived:public Base begin public members: /*Algorithm for overloaded display()*/ function display() begin write "\n Display derived" end of function display() /*Algorithm for overloaded show()*/ function show() begin write "\n show derived" end of function show() end of class Derived /*Algorithm for main()*/ function main () begin object B (Base) object D (Derived) object pointer bptr (Base)

61

Write "\n bptr points to Base\n" set bptr=&B call bptr->display() call bptr->show() write "\n\n bptr points to Derived\n" set bptr=&D call bptr->display() call bptr->show() end of function main()

 PROGRAM CODE:

#include class Base { public: void display () { cout<<"\n Display base"; } virtual void show() { cout<<"\n show base"; } }; class Derived:public Base { public: void display () { cout<<"\n Display derived"; } void show () { cout<<"\n show derived"; } }; int main () { Base B; Derived D; Base *bptr; cout<<"\n bptr points to Base\n"; bptr=&B; bptr->display();

62

bptr->show() ; cout<<"\n\n bptr points to Derived\n"; bptr=&D; bptr->display() ; bptr->show() ; return 0; }

 INPUT AND OUTPUT: bptr points to Base Display base Show base bptr points to Derived Display base Show derived

 DISCUSSION:

Polymorphism refers to the property by which objects belonging to different classes are able to respond to the same message, but in different forms. When we use the same function name in both the base and derived classes,the function in base class is declared as virtual using the keyword virtual preceding its normal declaration. When bptr is made to point to the object D,the statement bptr-+displayO; calls only the function associated with the Base (i.eBase::displayO), whereas the statement bptr-+showO; calls the Derived version of showO.This is because the function display() has not been made virtual in the Base class.

63

ASSIGNMENT 12

 PROBLEM STATEMENT:

Write a C++ program that implements linear search using FRIEND functions & INLINE functions.

 ALGORITHM: /*Algorithm for array class*/ class array begin private members: pointer varibale p (integer) variable size (integer) public members: /*function declarations*/ function read(); function create(); function display(); friend function search(array); end of class array /*Algorithm for read()*/ inline function array::read() begin write "\n\nEnter the size of the array:" read size set p = new integer[size] /*allocate memory dynamically*/ end of inline function read() /*Algorithm for create()*/ function array::create() begin write "\n\nEnter the array elements\n\n" for i=0 to size-1 do begin write "element" write i+1 write ":" read *(p+i) endfor write "\n\nThe array is:" call display() end of function create() /*Algorithm for display()*/ function array::display() begin

64

for i=0 to size-1 do begin write " " write *(p+i) endfor /*Algorithm for search*/ finction search(array ob) begin variable item (integer) write "\n\nEnter the item to be searched:" read item for i=0 to ob.size-1 do begin if(*(ob.p+i)==item) then write "\n\nItem found at location:" write i+1 return endif  endfor write "\n\nItem not found" end of function search(array ob) /*Algorithm for main()*/ function main() begin object obj (array type) call clrscr() call obj.read() call obj.create() call search(obj) call getch() return(0) end of function main()

 PROGRAM CODE:

#include #include class array { int *p,size; public: void read(); void create(); void display(); friend void search(array); }; inline void array::read() { cout<<"\n\nEnter the size of the array:"; cin>>size;

65

p = new int[size]; //allocate memory dynamically } void array::create() { cout<<"\n\nEnter the array elements\n\n"; for(int i=0;i>*(p+i); } cout<<"\n\nThe array is:"; display(); } void array::display() { for(int i=0;i>item; for(int i=0;i
 INPUT AND OUTPUT:

Enter the size of the array: 5 Enter the array elements element1:12 element2:13

66

element3:24 element4:35 element5:46 The array is: 12 13 24 35 46 Enter the item to be searched: 24 Item found at location: 3

 DISCUSSION:

1. The complexity of BINARY SEARCH is measured by the number f(n) of comparison where n is the number of data elements & is given by O(n). 2. A solution to this complexity problem is to use Binary search technique having a complexity O(log 2(n)).

67

ASSIGNMENT 13

 PROBLEM STATEMENT:

Write a program that implements a binary search tree(BST) and performs inorder traversal and searching operations on the tree.  ALGORITHM: /*Algorithm for BST class*/ class BST bagin private members: /*definition of tree structure*/ structure tree begin variable data (integer) pointer variables rchild, lchild (tree type) end of structure tree pointer variable root (tree type) public members: /*function declaration*/ function inorder(structure tree *root) function createBST() function Return() function search(structure tree *root) end of class BST /*Algorithm for Return()*/ function BST::Return() begin return(root); end of function Return() /*Algorithm for BST creation*/ Function CreateBST() begin pointer variable p,q,r (tree type) variable i,num (integer) set root=NULL write "\n\nenter the no.of nodes:" read num; for i=0 to num-1 do begin set p = new tree; /*create new node*/ write "\nenter value of node" write i+1 write ":" read p->data set p->lchild=p->rchild=NULL if(root==NULL) then set root=p

68

else set q=root set r=NULL while(q!=NULL) do begin if(p->datadata) then set r=q set q=q->lchild else set r=q set q=q->rchild endif  endwhile if(p->datadata) then set r->lchild=p else set r->rchild=p endif  endif  endfor end of function createBST() /*Algorithm for inorder traversal of BST*/ function BST::inorder(structure tree *root) begin if(root!=NULL) then call inorder(root->lchild) write root->data call inorder(root->rchild) endif  end of function inorder(structure tree *root) /*Algorithm for searching in BST*/ function BST::search(structure tree *root) begin varibale item,i=1 (integer) write "\n\nenter the value to be searched:" read item while(root!=NULL) do begin if(item==root->data) then write "\n\nitem found at location:" write i return endif  if(root->datarchild else set root=root->lchild endif  set i = i + 1 endwhile write "\n\nitem referred not found" end of function search(structure tree *root) /*Algorithm for main()*/ function main() begin variable ch (integer)

69

object ob (BST type) call clrscr() while(true) do begin write "\n\n\nwhat do you wish\n\n0.exit\n1.createBST\n2.inorder\n3.seacrh\n?" read ch if(ch==0) then return(0) endif  if(ch==1) then call ob.createBST() endif  if(ch==2) then write "\n\nthe inorder teaversal is as follows:" call ob.inorder(ob.Return()) write "\n\nassure that the list is sorted in ascending order" endif  if(ch==3) then call ob.search(ob.Return()) endif  endwhile end of function main()

 PROGRAM CODE: #include #include /* class declaration */ class BST { struct tree { int data; struct tree *rchild,*lchild; /*define node structure*/ }*root; public: void inorder(struct tree *root); void createBST(); tree* Return(); /*member function declarations / void search(struct tree *root); }; /* member function definition */ tree* BST::Return() { return(root); } //create BST void BST::createBST() { struct tree *p,*q,*r; int i,num; root=NULL; cout<<"\n\nenter the no.of nodes:"; cin>>num; for(i=0;i
70

{ p = new tree; /*create node*/ cout<<"\nenter value of node"<>p->data; p->lchild=p->rchild=NULL; if(root==NULL) { root=p; } else { q=root; r=NULL; while(q!=NULL) { if(p->datadata) { r=q; q=q->lchild; } else { r=q; q=q->rchild; } } if(p->datadata) r->lchild=p; else r->rchild=p; } } //end for } //end createBST //inorder void BST::inorder(struct tree *root) { if(root!=NULL) { inorder(root->lchild); cout<<" "<data; inorder(root->rchild); } } //end inorder //search void BST::search(struct tree *root) { int item,i=1; cout<<"\n\nenter the value to be searched:"; cin>>item; while(root!=NULL) { if(item==root->data) { cout<<"\n\nitem found at location:"<
71

if(root->datarchild; else root=root->lchild; i++; } cout<<"\n\nitem referred not found"; } //end search int main() { int ch; BST ob; clrscr(); while(1) { cout<<"\n\n\nwhat do you wish\n\n0.exit\n1.createBST\n2.inorder\n3.seacrh\n?"; cin>>ch; if(ch==0) return(0); if(ch==1) ob.createBST(); if(ch==2) { cout<<"\n\nthe inorder teaversal is as follows:"; ob.inorder(ob.Return()); cout<<"\nassure that the list is sorted in ascending order"; } if(ch==3) ob.search(ob.Return()); } } //end main()

72

 INPUT AND OUTPUT: what do you wish? 0.exit 1.createBST 2.inorder 3.seacrh ?1 enter enter enter enter

the no.of nodes:3 value of node1:10 value of node2:11 value of node3:9

what do you wish? 0.exit 1.createBST 2.inorder 3.seacrh ?2 The inorder traversal is as follows: 9 10 11 Assure that the list is sorted in ascending order 62 what do you wish? 0.exit 1.createBST 2.inorder 3.seacrh ?3 enter the value to be searched:9 item found at location: what do you wish 0.exit 1.createBST 2.inorder 3.seacrh ?0

 DISCUSSION:

1. Suppose we are searching for an item in a BST. The number of comparison is bounded by the depth of the tree. This comes from the fact that we proceed down single path of the tree as we search the item. Accordingly the running time of the search will be proportional to the depth of the tree . thus the time complexity of BST searching algorithm is o(log2n). 2. The inorder traversal gives a sorted list of elements.

73

ASSIGNMENT 14

 PROBLEM STATEMENT:

Write a C++ program that performs addition, subtraction, multiplication and division of two complex numbers. Overload operators ‘+’, ‘-‘, ‘*’,’/’ to perform the above operations.

 ALGORITHM:

/*Algorithm for complex class*/ class complex begin private members: variables real,img (float) public members: /*Algorithm for display()*/ function display() begin write "\n\n\t"; if(img>=0) then write real write "+" write img write "i" else write real write img write "i" endif  end of function display() /*Algorithm for no argument constructor*/ complex() begin set real=0 set img=0 end of complex() /*Algorithm for two argument constructor*/ complex(float r,float i) begin set real=r set img=i end of complex(float r,float i) /*function declaration*/ function operator +(complex &b)

74

function operator -(complex &b) function operator *(complex &b) function operator /(complex &b) end of class complex /*Algorithm for overloading + operator*/ function complex::operator +(complex &b) begin variable e,f (float) set e=real+b.real set f=img+b.img return(complex(e,f)) end of function operator +(complex &b) /*Algorithm for overloading - operator*/ function complex::operator -(complex &b) begin variable e,f : float set e=real-b.real set f=img-b.img return(complex(e,f)) end of function operator -(complex &b) /*Algorithm for overloading * operator*/ function complex::operator *(complex &b) begin varibale e,f : float set e=real*b.real-img*b.img set f=img*b.real+real*b.img return(complex(e,f)); end of function operator *(complex &b) /*Algorithm for overloading / operator*/ function complex::operator /(complex &b) begin variable num1,num2,deno : float variable e,f : float set num1=real*b.real+img*b.img set num2=img*b.real-real*b.img set deno=b.real*b.real+b.img*b.img set e=num1/deno set f=num2/deno return(complex(e,f)) end of function operator /(complex &b) /*Algorithm for main()*/ function main() begin objects x(3,2),y(2,5),z,a(8,8),b(3,2),e (complex type) set z=x+y write "\n complex number x(3,2)" call x.display() write "\n complex number y(2,5)" call y.display()

75

write "\n complex call z.display() set z=a-b; write "\n complex call a.display() write "\n complex call b.display() write "\n complex call z.display() set z=x*y write "\n complex call x.display() write "\n complex call y.display() write "\n complex call z.display() set e=a/b write "\n complex call a.display() write "\n complex call b.display() write "\n complex call e.display() end of function main()

number z=x+y"

number a(8,8)" number b(3,2)" number z=a-b"

number x(3,2)" number y(2,5)" number z=x*y"

number a(8,8)" number b(3,2)" number e=a/b"

 PROGRAM CODE: #include #include class complex { private: float real,img; public: void display() { cout<<"\n\n\t"; if(img>=0) cout<
76

{ real=r; img=i; } complex complex complex complex

operator operator operator operator

+(complex &b); -(complex &b); *(complex &b); /(complex &b);

}; complex complex::operator +(complex &b) { float e,f; e=real+b.real; f=img+b.img; return(complex(e,f)); } complex complex::operator -(complex &b) { float e,f; e=real-b.real; f=img-b.img; return(complex(e,f)); } complex complex::operator *(complex &b) { float e,f; e=real*b.real-img*b.img; f=img*b.real+real*b.img; return(complex(e,f)); } complex complex::operator /(complex &b) { float num1,num2,deno; float e,f; num1=real*b.real+img*b.img; num2=img*b.real-real*b.img; deno=b.real*b.real+b.img*b.img; e=num1/deno; f=num2/deno; return(complex(e,f)); } void main() { complex x(3,2),y(2,5),z,a(8,8),b(3,2),e; z=x+y; cout<<"\n complex number x(3,2)"; x.display();

77

cout<<"\n complex number y(2,5)"; y.display(); cout<<"\n complex number z=x+y"; z.display(); z=a-b; cout<<"\n complex number a(8,8)"; a.display(); cout<<"\n complex number b(3,2)"; b.display(); cout<<"\n complex number z=a-b"; z.display(); z=x*y; cout<<"\n complex number x(3,2)"; x.display(); cout<<"\n complex number y(2,5)"; y.display(); cout<<"\n complex number z=x*y"; z.display(); e=a/b; cout<<"\n complex number a(8,8)"; a.display(); cout<<"\n complex number b(3,2)"; b.display(); cout<<"\n complex number e=a/b"; e.display(); }

 INPUT AND OUTPUT: complex number x(3,2) 3+2i complex number y(2,5) 2+5i complex number z=x+y 5+7i complex number a(8,8) 8+8i complex number b(3,2) 3+2i complex number z=a-b 5+6i complex number x(3,2) 3+2i complex number y(2,5) 2+5i complex number z=x*y -4+19i complex number a(8,8) 8+8i complex number b(3,2) 3+2i complex number e=a/b 3.07692+0.615385i

78

 DISCUSSION:

The above class can be further modified to include other operations of complex numbers.

79

ASSIGNMENT 15

 PROBLEM STATEMENT:

Create a class called Time that has separate integer member data for hours,minutes and second. One constructor should initialize this data to zero and another initialize it to fix values. A member  function should display it in hh:mm:ss format. The final member function should add two objects of type Time passed as arguments. A main program should create to initialized the time objects,and one that isn't initialized. Then it should add two initialized values together, the result in the third time variable. Finally it should display the value of this third variable.

 ALGORITHM:

/*Algorithm for time class*/ class time begin private members: variable hour (integer) variable min (integer) variable sec (integer) public members: /*Algorithm for no argument constructor*/ time() baegin set hour=0 set min=0 set sec=0 end of time() /*Algorithm for three argument constructor*/ time(integre h, integer m, integer s) begin set hour=h set min=m set sec=s end of time(integre h, integer m, integer s) /*Algorithm for getdata()*/

80

function getdata() begin write "\nEnter the Hour:" read hour write "\nEnter the Minute:" read min write "\nEnter the Second:" read sec end of getdata() /*Algorithm for display()*/ function display() begin if (hour<10 AND min<10 AND sec<10) then write "0" write hour write ":0" write min write ":0" write sec write " Hr.\n" else if(hour<10 AND min<10) then write "0" write hour write ":0" write min write ":" write sec write " Hr.\n" else if (hour<10 AND sec<10) then write "0" write hour wrute ":" write min write ":" write "0" write sec write " Hr.\n" else if(sec<10 AND min<10) then write hour

81

write ":" write "0" write min write ":" write "0" write sec write " Hr.\n" else if(sec<10) than write hour write ":" write min write ":" write "0" write sec write " Hr.\n" else if(min<10) then write hour write ":" write "0" write min write ":" write sec write " Hr.\n" else if(hour<10) write "0" write hour write ":" write min write ":" write sec write " Hr.\n" else write hour write ":" write min write ":" write sec write " Hr.\n" endif 

82

end of function display() /*Algorithm for overloading + operator*/ function operator+(time t1) begin variable hr, mn, sc (integer) set hr=hour+t1.hour set mn=min+t1.min set sc=sec+t1.sec if(sc>=60) then set sc= sc – 60 set mn = mn + 1 endif  if(mn>=60) then set mn=mn-60 hr = hr + 1 endif  return time(hr,mn,sc); end of function operator+(time t1) end of class time /*Algorithm for main()*/ function main() begin objects t1,t2,t3 (time type) call clrscr() write "\nEnter the first time:" call t1.getdata() write "\nEnter the second time:" call t2.getdata() set t3=t1+t2 call clrscr() write "\n\n\n" write "\t\tThe entered Time::   " call t1.display() write "\n\n\n" write "\t\tThe entered Time::   " call t2.display() write "\n\n\t\t\t The Total Time  \n\n" call gotoxy(27,14) call t3.display()

83

call getch(); end of function main()

 PROGRAM CODE:

#include #include #include

class time { private: int hour; int min; int sec; public: time() {  hour=0;  min=0;  sec=0;} time(int h,int m,int s) { hour=h; min=m; sec=s; } void getdata() { cout<<"\nEnter the Hour:"; cin>>hour; cout<<"\nEnter the Minute:"; cin>>min; cout<<"\nEnter the Second:" ; cin>>sec; } void display() { if (hour<10 && min<10 && sec<10) { cout<<"0"<
84

} else if(hour<10 && min<10) cout<<"0"<=60) { sc-=60;  mn++; } if(mn>=60) { mn-=60; hr++; } return time(hr,mn,sc); } };

void main() { time t1,t2,t3; clrscr();

85

cout<<"\nEnter the first time:"; t1.getdata(); cout<<"\nEnter the second time:"; t2.getdata(); t3=t1+t2; clrscr(); cout<<"\n\n\n"; cout<<"\t\tThe entered Time::   "; t1.display(); cout<
OUTPUT:  OUTPUT:

Output Sample - I Enter the first time: Enter the Hour:2 Enter the Minute:52 Enter the Second:38

Enter the second time: Enter the Hour:6 Enter the Minute:25 Enter the Second:26 The entered Time::   02:52:38 Hr. The entered Time::   06:25:26 Hr. The Total Time 09:18:04 Hr. Output Sample -II

Enter the first time:

86

Enter the Hour:2 Enter the Minute:56 Enter the Second:48

Enter the second time: Enter the Hour:4 Enter the Minute:8 Enter the Second:15 The entered Time::   02:56:48 Hr. The entered Time::   4:08:15 Hr. The Total Time 07:05:03 Hr.

DISCUSSIONS:  DISCUSSIONS:

On the above program we take two input of time in different frame, hour, minute & second each object of time class taken user input data and the overloaded binary operator '+' user to add two time object and assigning it into the third object. The display function will call to display the time in [hh:mm:ss] format. That is if the hour, minute and second is less than 10 then '0' will be padded to its left.

87

ASSIGNMENT 16

 PROBLEM STATEMENT:

Write a C++ program to overload subscript operator.

 ALGORITHM: /*Algorithm for atype class*/ class atype begin private members: variable a[5] : integer public members: /*Algorithm for one argument constructor*/ atype(integer i, integer j, integer k) begin set a[0]=i set a[1]=j set a[2]=k end of atype(integer i, integer j, integer k) /*Algorithm for overloading [] operator*/ function operator [](integer i) begin return a[i] end of function operator [](integer i) /*Algorithm for showdata()*/ function showdata() begin write "\n data's are:" for i=0 to 2 do begin write "\n" write a[i] endfor end of function showdata()

88

end of class atype /*Algorithm for main()*/ function main() begin object A(12,34,45) (atype type) variable c (integer) for j=0 to 2 write "\ndata:" write (j+1) write "\t" write A[j] endfor set c=A[1]+A[0]*A[2] write "\nfinal answer is:" write c call A.showdata() call getch(); end of function main()

 PROGRAM CODE:

#include #include #include

class atype { private: int a[5]; public: atype(int i,int j,int k) { a[0]=i;a[1]=j;a[2]=k; } int operator [](int i) { return a[i];

89

} void showdata() { cout<<"\n data's are:"; for(int i=0;i<3;i++) cout<<"\n"<
 INPUT AND OUTPUT:

data:1 12 data:2 34 data:3 45 final answer is:574 data's are: 12 34 45

 DISCUSSION:

He above program shows a demo for overloading subscript ([]) operator. It can be used to other  programs to perform a specific operation.

90

ASSIGNMENT 17

 PROBLEM STATEMENT:

Write a C++ program to overload these operators- ‘+=’, ‘+’, ‘>=’, ‘<=’, ‘!=’, ‘==’.

 ALGORITHM: constant true = 1 constant false = 0 /*Algorithm for class sample*/ class sample begin private members: variable data (integer) public members: /*Algorith for one argument constructor*/ sample(integer i) begin set data=i end of sample(integer i) /*Algorithm for no argument constructor*/ sample() begin end of sample() /*Algorithm for overloading + operator*/ function operator +(sample s) begin return sample(data+s.data) end of function operator +(sample s) /*Algorithm for overloading +=*/ function operator +=(sample s) begin return sample(data+=s.data) end of function operator +=(sample s) /*Algorithm for overloading >=*/ function operator >=(sample s) begin if(data>=s.data) then return true else return false endif  end of function operator >=(sample s) /*Algorithm for overloading <= operator*/ function operator <=(sample s) begin if(data<=s.data) then return true else return false endif 

91

end of function operator <=(sample s) /*Algorithm for overloading != operator*/ function operator !=(sample s) begin if(data!=s.data) then return true else return false endif  end of function operator !=(sample s) /*Algorithm for overloading == operator*/ function operator ==(sample s) begin if(data==s.data) then return true else return false endif  end of function operator ==(sample s) /*Algorithm for showdata()*/ function showdata() begin write "\n the value is:" write data end of function showdata() end of class sample /*Algorithm for main()*/ function main() begin variable x,y (integer) write "\n enter two values for a and b:" read x read y objects a(x),b(y),c (sample) if(a>=b) then write "\n a is greater than or equal to b" endif  if(a<=b) then write "\n b is greater than or equal to a" end if  if(a==b) then write "\n a is equal to b" endif  if(a!=b) then write "\n a is not equal to b" endif  set c=a+b call c.showdata() set a+=b call a.showdata() end of function main()

 PROGRAM CODE:

#include

92

#define true 1 #define false 0 class sample { private: int data; public: sample(int i) { data=i; } sample() {} sample operator +(sample s) { return sample(data+s.data); } sample operator +=(sample s) { return sample(data+=s.data); } int operator >=(sample s) { if(data>=s.data) return true; else return false; } int operator <=(sample s) { if(data<=s.data) return true; else return false; } int operator !=(sample s) { if(data!=s.data) return true; else return false; } int operator ==(sample s) { if(data==s.data) return true; else return false; } void showdata() { cout<<"\n the value is:"<
93

int x,y; cout<<"\n enter two values for a and b:"; cin>>x>>y; sample a(x),b(y),c; if(a>=b) cout<<"\n a is greater than or equal to b"; if(a<=b) cout<<"\n b is greater than or equal to a"; if(a==b) cout<<"\n a is equal to b"; if(a!=b) cout<<"\n a is not equal to b"; c=a+b; c.showdata(); a+=b; a.showdata(); }

 INPUT AND OUTPUT: enter two values for a and b:2 3 b is greater than or equal to a a is not equal to b the value is:5 the value is:5

 DISCUSSIONS:

The above program shows a demonstration for overload logical operators. It can be used with some specific programs to perform a specific task.

94

ASSIGNMENT 18



PROBLEM STATEMENT:

Write a C++ program to create a singly linked list and perform the following functions: a)copy the list into another list b)remove duplicate elements from the list. c)display the list.

 ALGORITHM: /*Definition of structure link*/ structure link begin variable data : integer pointer variable next (link type) end of structure link /*Algorithm for linkedlist class*/ class linkedlist begin private members: pointer variable head (link type) variable p,num (integer) public members: /*Algorithm for no argument constructor*/ linkedlist() begin set head=NULL end of linkedlist() /*function declarations*/ function getdata() function additem(integer item) function copylist(linkedlist &A) function remDup() function display() end of class linkedlist /*Algorithm for getdata()*/ function linkedlist::getdata() begin write "enter number of items you want to insert:"; read p for i=0 to p-1 do begin

95

write "\n enter data:" read num call  additem(num) endfor end of function getdata() /*Algorithm for copy liked list*/ function linkedlist::copylist(linkedlist &A) begin structure link *temp,*q1 set q1=A.head while(q1!=NULL) do begin set temp=new (link type) set temp->data=q1->data set temp->next=NULL if(head==NULL) then set head=temp else pointer variable list (link type) set list=head while(list->next!=NULL) do set list=list->next endwhile set list->next=temp endif  set q1=q1->next endwhile end of function copylist(linkedlist &A) /*Algorithm for adding a item to the linked list*/ function linkedlist::additem(int item) begin pointer variable temp (link type) set temp=new link set temp->data=item set temp->next=NULL if(head==NULL) then set head=temp else pointer variable list (link type) set list=head while(list->next!=NULL) do begin set list=list->next endwhile set list->next=temp endif  end of function additem(int item) /*Algorithm for removing duplicates from linked list*/ function linkedlist::remDup() begin

96

pointer variables ptr1, ptr2, dup (link type) set ptr1 = head /* Pick elements one by one */ while(ptr1 != NULL AND ptr1->next != NULL) do begin set ptr2 = ptr1 /* Compare the picked element with rest of the elements */ while(ptr2->next != NULL) do begin /* If duplicate then delete it */ if(ptr1->data == ptr2->next->data) then /* sequence of steps is important here */ set dup = ptr2->next set ptr2->next = ptr2->next->next delete dup else set ptr2 = ptr2->next endif  endwhile set ptr1 = ptr1->next endwhile end of function remDup() /*Algorithm for displaying the linked list*/ function linkedlist::display() begin pointer variable start (link type) set start=head write "\n\n"; while(start!=NULL) do begin write start->data write "-->" set start=start->next endwhile write "END" end of function display() /*Algorithm for main()*/ function main() begin objects A,B (linkedlist type) call A.getdata(); write "\n 1st list:" call A.display() call B.copylist(A) write "\n copied list:" call B.display() call A.remDup() write "\n 1st list after removal of duplicate elements:" call A.display() call getch();

97

end of function main()



PROGRAM CODE:

#include #include #include struct link { int data; link *next; }; class linkedlist { private: link *head; int p,num; public: linkedlist() { head=NULL; } void getdata(); void additem(int item); void copylist(linkedlist &A); void remDup(); void display(); }; void linkedlist::getdata() { cout<<"enter number of items you want to insert:"; cin>>p; for(int i=0;i>num; additem(num); } } void linkedlist::copylist(linkedlist &A) { link *temp,*q1; q1=A.head; while(q1!=NULL)

98

{ temp=new link; temp->data=q1->data; temp->next=NULL; if(head==NULL) head=temp; else { link *list; list=head; while(list->next!=NULL) list=list->next; list->next=temp; } q1=q1->next; } } void linkedlist::additem(int item) { link *temp; temp=new link; temp->data=item; temp->next=NULL; if(head==NULL) head=temp; else { link *list; list=head; while(list->next!=NULL) list=list->next; list->next=temp; } }

void linkedlist::remDup() { link *ptr1, *ptr2, *dup; ptr1 = head; /* Pick elements one by one */ while(ptr1 != NULL && ptr1->next != NULL) { ptr2 = ptr1; /* Compare the picked element with rest of the elements */ while(ptr2->next != NULL) { /* If duplicate then delete it */ if(ptr1->data == ptr2->next->data) { /* sequence of steps is important here */

99

dup = ptr2->next; ptr2->next = ptr2->next->next; delete dup; } else ptr2 = ptr2->next; } ptr1 = ptr1->next; } }

void linkedlist::display() { link *start; start=head; cout<<"\n\n"; while(start!=NULL) { cout<data<<"-->"; start=start->next; } cout<<"END"; } void main() { linkedlist A,B; A.getdata(); cout<<"\n 1st list:"; A.display(); B.copylist(A); cout<<"\n copied list:"; B.display(); A.remDup(); cout<<"\n 1st list after removal of duplicate elements:"; A.display(); getch(); }



INPUT AND OUTPUT:

enter number of items you want to insert:7 enter data:12 enter data:56 enter data:89 enter data:45 enter data:89 enter data:12 enter data:22

100