12

INDEX

S.NO DATE TOPIC PAGE NO

STAFF SIGNATURE

1 DISTANCE CONVERSION PROBLEM

1

2 OVERLOADING OBJECTS 10

3 OVERLOADING CONVERSIONS

16

4 OVRELOADING MATRIX 22

5 AREA COMPUTATION USING DERIVED CLASS

32

6 VECTOR PROBLEM 37

7 INHERITANCE 43

EX.NO:1

DATE:

DISTANCE CONVERSION PROBLEM

AIM:

Create two classes DM and DB which store the value of distances. DM

store the value ofdistances. DM stores distances in meters and centimeters in

DB in feet and inches. Write a Program that can create the values of the class

objects and add one object DM with another object DB.

Use a friend function to carry out addition operation. The object that

stores the result may be DM object or DB object depending on the units in

which results are required.

The display should be in the order of meter and centimeter and feet or

inches depending on the order of display.

ALGORITHM:

STEP 1:

Start the program execution.STEP 2:

To create a class metric and british.

STEP 3:

To create a function input() and display().

STEP 4:

Accept the feet,inches value of british.

STEP 5: Accept the meter ,cm,value of metric.

STEP 6:

To convert metric value to british value.

STEP 7:

To convert british value to metric value.

STEP 8:

To display the metric values and british values.

STEP 9:

Stop the program execution.

CODING:

#include<iostream.h>

#include<conio.h>

#include<math.h>

class metric;

class british

{

public:

int feet,inches;

void input();

friend british addbritish(british & b,metric & m);

void display();

};

class metric

{

public:

int meter,cm;

void input();

friend metric addmetric(metric & m,british & b);

void display();

};

void british::input()

{

cout<<"\n\t enter the value of feet:";

cin>>feet;

cout<<"\n\t enter the value of inches:";

cin>>inches;

}

british addbritish(british & b,metric & m)

{

british b1;

int ftin,tmcm,tin,mtcm,tm,cm,tmcmtin,tot;

ftin=(b.feet)*12;

tin=ftin+b.inches;

mtcm=(m.meter)*100;

tmcm=mtcm+m.cm;

tmcmtin=tmcm/2.5;

tot=tin+tmcmtin;

b1.feet=tot/12;

b1.inches=tot%12;

return(b1);

}

void british::display()

{

cout<<"\n\t the feet value is:\t"<<feet;

cout<<"\n\t the inches value is:\t"<<inches;

}

void metric::input()

{

cout<<"\n\t enter the value of meter:\t";

cin>>meter;

cout<<"\n\t enter the value of cm:\t";

cin>>cm;

}

metric addmetric(metric & m,british & b)

{

metric m1;

int tin,ftin,mtcm,tincmt,tot;

ftin=(b.feet)*12;

tincmt=(ftin+b.inches)*2.5;

mtcm=(m.meter)*100;

tot=tincmt+mtcm+m.cm;

m1.meter=tot/100;

m1.cm=tot%100;

return(m1);

}

void metric::display()

{

cout<<"\n\t the meter value is \t"<<meter;

cout<<"\n\t the centimeter value is \t"<<cm;

}

void main()

{

int op;

clrscr();

british b1,b2;

metric m1,m2;

cout<<"\n Distance conversion";

b1.input();

m1.input();

do

{

clrscr();

cout<<"\n MENU";

cout<<"\n 1.Metric to British";

cout<<"\n 2.British to Metric";

cout<<"\n 3.Exit";

cout<<"\n Enter the choice:";

cin>>op;

switch(op)

{

case 1:

cout<<"\n\t metric to british";

cout<<"\n\t output";

b1.display();

m1.display();

b2=addbritish(b1,m1);


b2.display();

break;

case 2:

cout<<"\n\t british to metric";


b1.display();

m1.display();

m2=addmetric(m1,b1);

cout<<"\n output";

m2.display();

break;

case 3:

cout<<"\n exit";

break;

}

getch();

}

while (op!=3);

}

OUTPUT

Distance conversion

enter the value of feet: 10

enter the value of inches: 6

enter the value of meter: 4

enter the value of cm: 2

MENU

1.Metric to British

2.British to Metric

3.Exit

Enter the choice:1

metric to british

output

the feet value is: 10

the inches value is: 6

the meter value is 4

the centimeter value is 2

output



MENU

1.Metric to British

2.British to Metric

3.Exit

Enter the choice:2

british to metric

output





output



MENU

1.Metric to British

2.British to Metric

3.Exit

Enter the choice:3

Exit

RESULT

EX.NO:2

DATE:

OVERLOADING OBJECTS

AIM:

Create a class FLOAT that contains one float data member overload all

the four arithmetic operators so that operate on the objects of FLOAT.

ALGORITHM :

STEP 1:

Start the program execution

STEP 2:

To create a class FLOAT

STEP 3:

Accept the input values

STEP 4:

Overload the arithmetic operator +

STEP 5:

Overload the arithmetic operetor –

STEP 6:

Overload the arithmetic operator *

STEP 7:

Overload the arithmetic operator /

STEP 8:

To display the output values.

STEP 9:


CODING:


#include<conio.h>

#include<math.h>

class FLOAT

{

float x,D;

public:

FLOAT()

{

}

FLOAT(float a)

{

x=a;

}

FLOAT operator+(FLOAT);

FLOAT operator-(FLOAT);

FLOAT operator*(FLOAT);

FLOAT operator/(FLOAT);

void outdata();

void indata()

{

cout<<"\n\n\t enter the value:";

cin>>x;

}

};

FLOAT FLOAT::operator+(FLOAT D)

{

FLOAT temp;

temp.x=x+D.x;

return(temp);

}

FLOAT FLOAT::operator-(FLOAT D)

{

FLOAT temp;

temp.x=x-D.x;

return(temp);

}

FLOAT FLOAT::operator*(FLOAT D)

{

FLOAT temp;

temp.x=x*D.x;

return(temp);

}

FLOAT FLOAT::operator/(FLOAT D)

{

FLOAT temp;

temp.x=x/D.x;

return(temp);

}

void FLOAT::outdata(void)

{

cout<<x<<endl;

}

int main()

{

FLOAT o1,o2,ADD,SUB,MUL,DIV;

clrscr();

o1.indata();

o2.indata();

ADD=o1+o2;

SUB=o1-o2;

MUL=o1*o2;

DIV=o1/o2;

cout<<"\n\n\t o1:";

o1.outdata();

cout<<"\n\n\t o2:";

o2.outdata();

cout<<"\n\n\t ADDITION VALUE IS (o1+o2)=";

ADD.outdata();

cout<<"\n\n\t SUBTRACTION VALUE IS (o1-o2)=";

SUB.outdata();

cout<<"\n\n\t MULTIPLICATION VALUE IS (o1*o2)=";

MUL.outdata();

cout<<"\n\n\t DIVISION VALUE IS (o1/o2)=";

DIV.outdata();

getch();

return(0);

}

2.OUTPUT:

enter the value:4

enter the value:4

o1:4

o2:4

ADDITION VALUE IS (o1+o2)=8

SUBTRACTION VALUE IS (o1-o2)=0

MULTIPLICATION VALUE IS (o1*o2)=16

DIVISION VALUE IS (o1/o2)=1

RESULT:

EX.NO:3

DATE:

OVERLOADING CONVERSIONS:AIM:

Design a class polar which describes a point in a plane using polar

Co-ordinates radius and angle. A point in polar Co-ordinates is as shown

below.

Use the overloader + operator to add two objects of polar. Note that we

cannot add polar values of two points directly. This requires first the

conversion.

Points into rectangular Co-ordinates and finally converting the result

into polar Co-ordinates.

You need to use following trigonometric formulas.

X= r * cos (a); Y= r * sin (a);

a= sqrt (X * X +Y * Y);

ALGORITHM :

STEP 1:

Start the program execution.

STEP 2:

To create class polar.

STEP 3:

Accept the radius and angle values.

STEP 4:

First we accept the radius and angle value is 0.

STEP 5:

To create arguments in polar.

STEP 6:

To overload the + operator to add two object of polar.

STEP 7:


STEP 8:


CODING:


#include<conio.h>

#include<math.h>

class polar

{

double radius,angle;

public:

polar()

{

radius=0;

angle=0;

}

polar(double r,double a)

{

radius=r;

angle=a;

}

void disp();

polar operator+(polar);

};

void polar::disp()

{

cout<<"\n\n\t radius value is:";

cout<<radius;

cout<<"\n\n\t angle value is:";

cout<<angle;

}

polar polar::operator+(polar p1)

{

polar p2;

double x1,x2,y1,y2,x,y;

x1=radius*cos(angle*3.14/180);

x2=p1.radius*cos(p1.angle*3.14/180);

y1=radius*sin(angle*3.14/180);

y2=p1.radius*sin(p1.angle*3.14/180);

cout<<"\n output\n";

cout<<"*****\n";

cout<<"\n\n radius 1="<<x1<<"\t angle 1="<<y1;

cout<<"\n\n radius 2="<<x2<<"\t angle 2="<<y2;

x=x1+x2;

y=y1=y2;

cout<<"\n\n sum of the radius in rectangular coordinates is:";

cout<<x;

cout<<"\n\n sum of the angle in rectangular coordinates is:";

cout<<y;

p2.radius=sqrt(x*x+y*y);

p2.angle=atan(y/x)*(180/3.14);

return(p2);

}

void main()

{

clrscr();

cout.precision(2);

polar o1,o2,o3;

double r1,r2,a1,a2;

char c;

do

{

cout<<"\n enter the radius 1 and angle 1:";

cin>>r1>>a1;

cout<<"\n enter the radius 2 and angle 2:";

cin>>r2>>a2;

o1=polar(r1,a1);

o2=polar(r2,a2);

o3=o1+o2;

o3.disp();

cout<<"\n\n Do you want to continue(y/n):";

cin>>c;

}

while(c=='y'||c!='n');

getch();

}

3.OUTPUT

enter the radius 1 and angle 1:

11

13

enter the radius 2 and angle 2:

13

15

output

*****

radius 1=10.72 angle 1=2.47

radius 2=12.56 angle 2=3.36

sum of the radius in rectangular coordinates is:23.28

sum of the angle in rectangular coordinates is:3.36

radius value is:23.52

angle value is:8.23

Do you want to continue(y/n):n

RESULT:

EX.NO:4

DATE:

OVRELOADING MATRIX:

AIM:

Create a class MAT of size M*N. Define all possible matrix operations

for MAT type objects. Verify the identity.

(A-B)^2 = A^2+B^2 –2*A*B

ALGORITHM :

STEP 1:


STEP 2:

To create a class MAT.

STEP 3:

To accept the element of matrices.

STEP 4:

To create a function inp(int) and disp().

STEP 5:

To overload the operator the operators +,-,*,/,^.

STEP 6:

To using for condition.

STEP 7:

To accept the element of first and second matrix.

STEP 8:

To calculate (A-B)^2.

STEP 9:

To calculate A^2+B^2-2*A*B.

STEP 10:

To verify (A-B)^2=A^2+B^2-2AB the identity is true using it condition.

STEP 11:

To verify (A-B)^2!=A^2+B^2-2AB the given identity is false.

STEP 12:

To create a output values.

STEP 13:


CODING:


#include<conio.h>

#include<math.h>

const size=3;

class mat

{

int nrc;

int a[size][size];

public:

void inp(int);

void disp();

mat operator+(mat);

mat operator-(mat);

mat operator*(mat);

mat operator*(int);

mat operator^(int);

int operator==(mat);

};

void mat::inp(int n)

{

nrc=n;

for(int i=0;i<nrc;i++)

for(int j=0;j<nrc;j++)

cin>>a[i][j];

}

void mat::disp()

{


{


cout<<a[i][j]<<" ";

cout<<endl;

}

}

mat mat::operator+(mat m)

{

mat temp;

temp.nrc=nrc;



temp.a[i][j]=a[i][j]+m.a[i][j];

return temp;

}

mat mat::operator-(mat m)

{

mat temp;

temp.nrc=nrc;



temp.a[i][j]=a[i][j]-m.a[i][j];

return temp;

}

mat mat::operator*(mat m)

{

mat temp;

temp.nrc=nrc;



{

temp.a[i][j]=0;

for(int k=0;k<nrc;k++)

temp.a[i][j]+=a[i][j]*m.a[i][j];

}

return temp;

}

mat mat::operator*(int x)

{

mat temp;

temp.nrc=nrc;



temp.a[i][j]=x*a[i][j];

return temp;

}

mat mat::operator^(int)

{

mat temp;

temp.nrc=nrc;



{

temp.a[i][j]=0;

for(int k=0;k<nrc;k++)

temp.a[i][j]+=a[i][j]*a[i][j];

}

return temp;

}

int mat::operator==(mat m)

{

int t=1;

int f=0;



if(a[i][j]!=m.a[i][j])

return f;

else

return t;

return 0;

}

void main()

{

clrscr();

mat m1,m2,mat1,mat2;

int n;

cout<<"\n enter the order of matrices:";

cin>>n;

cout<<"\n enter the element of first matrix:";

m1.inp(n);

cout<<"\n enter the element of second matrix:";

m2.inp(n);

cout<<"\n first matrix is:";

m1.disp();

cout<<"\n second matrix is:";

m2.disp();

mat1=(m1-m2)^2;

cout<<"\n result of (A-B)^2 is:\n";

mat1.disp();

mat2=(m1^2)+(m2^2)-((m1*m2)*2);

cout<<"\n result of (A^2+B^2-2AB) is:\n";

mat2.disp();

if(mat1==mat2)

cout<<"\n the given identity is true:\n";

else

cout<<"\n the given identity is false:\n";

getch();

}

4.OUTPUT:

enter the order of matrices:2

enter the element of first matrix:

1

1

1

1

enter the element of second matrix:

1

1

1

1

first matrix is:

1 1

1 1

second matrix is:

1 1

1 1

result of (A-B)^2 is:

0 0

0 0

result of (A^2+B^2-2AB) is:

0 0

0 0

the given identity is true:

RESULT:

EX.NO:5

DATE:

AREA COMPUTATION USING DERIVED CLASS

AIM:

Area of rectangle = X*YArea of triangle = ½ * X * Y

ALGORITHM :

STEP 1:


STEP 2:

To create a class rectangle and triangle.

STEP 3:

Create a function get(),cal() and disp().

STEP 4:

Accept the two sides of rectangles and triangle.

STEP 5:

Calculate the area of rectangle are=X*Y.

STEP 5:

Calculate the area of triangle are =1/2*X*Y.

STEP 6:

Display the outputs.

STEP 7:


CODING


#include<conio.h>

#include<math.h>

class rectangle

{

float area;

protected:

float x,y;

public:

void get();

void calc()

{

area=x*y;

}

void disp();

};

void rectangle::get()

{

cout<<"\n enter the sides of rectangle and triangle:";

cin>>x>>y;

}

void rectangle::disp()

{

cout<<"\n area of the rectangle is:"<<area<<endl;

}

class triangle:public rectangle

{

float area;

public:

void calc()

{

area=(x*y)/2;

}

void disp()

{

cout<<"\n area of the triangle is:"<<area;

}

};

void main()

{

clrscr();

triangle tr;

tr.get();

tr.rectangle::calc();

tr.rectangle::disp();

tr.calc();

tr.disp();

getch();

}

5.OUTPUT:

enter the sides of rectangle and triangle:

4

5

area of the rectangle is:20

area of the triangle is:10

RESULT:

EX.NO:6

DATE:

VECTOR PROBLEM:

AIM:

Define a class for vector containing scalar values. Apply overloading

concepts for vector addition, Multiplication of a vector by a scalar quantity,

replace the values in a position vector.

ALGORITHM

STEP 1:


STEP 2:

To create a class vector.

STEP 3:

To accept the input values.

STEP 4:

Overloading the *,+,<<,>> operators.

STEP 5:

To multiply the vector values.

STEP 6:

To calculate the sum of two vectors.

STEP 7:

To change the position of elements to be modified vector.

STEP 8:


STEP 9:


STEP 10:


CODING:


#include<conio.h>

#include<math.h>

const size=3;

class vector

{

public:

int *v;

vector()

{

v=new int[size];

}

int & operator[](int);

friend vector operator*(int,vector);

friend vector operator+(vector,vector);

friend istream & operator>>(istream &,vector&);

friend ostream & operator>>(ostream &,vector&);

};

int &vector::operator[](int pos)

{

return v[pos];

}

vector operator * (int a,vector b)

{

vector c;

for(int i=0;i<size;i++)

c.v[i]=a*b.v[i];

return c;

}

vector operator+(vector a,vector b)

{

vector c;


c.v[i]=a.v[i]+b.v[i];

return c;

}

istream & operator>>(istream &cin,vector &b)

{


cin>>b.v[i];

return cin;

}

ostream &operator<<(ostream &cout,vector &b)

{

cout<<","<<b.v[0];


{

cout<<","<<b.v[i];

}

return cout;

}

void main()

{

clrscr();

vector m,n;

int sc;

cout<<"\n enter the element of vector 1:";

cin>>m;

cout<<"\n enter the element of vector 2:";

cin>>n;

cout<<"\n output";

cout<<"\n ****************";

cout<<"\n vector 1_"<<m;

cout<<"\n vector 2_"<<n;

vector sum,mul;

cout<<"\n scalar multiplication value:";

cin>>sc;

mul=sc * m;

sum=m+n;

cout<<"\n scalar multiplication of vector 1 is:"<<mul;

cout<<"\n sum of the two vector is :"<<sum;

int pos,nval;

cout<<"\n enter the position of the element to be modified (vector):";

cin>>pos;

cout<<"\n\n enter the value:";

cin>>nval;

m[pos-1]=nval;

cout<<"\n\n modified value is:"<<m;

getch();

}

6.OUTPUT

enter the element of vector 1:

5

6

4

enter the element of vector 2:

4

8

6

output

****************

vector 1:5,6,4

vector 2:4,8,6

scalar multiplication value:3

scalar multiplication of vector 1 is: 15,18,12

sum of the two vector is : 9,14,10

enter the position of the element to be modified (vector):2

enter the value:5

modified value is:5,5,4

RESULT:

EX.NO:7

DATE:

INHERITANCE

AIM:

Create three classes alpha, beta and gamma, each containing one data member. The class gamma should be inherited from both alpha and beta. Use a constructor function in the class gamma to assign values to the data members of all the classes.

ALGORITHM:

STEP 1:


STEP 2:

To create a three classes alpha, beta and gamma

STEP 3:


STEP 4:

TO use constructor function

STEP 5:


STEP 6:


STEP 7:


CODING:


#include<conio.h>

#include<math.h>

class alpha

{

int x;

public:

alpha(int i)

{

x=i;

cout<<"alpha intialised\n";

}

void show_x(void)

{

cout<<"x="<<x<<"\n";

}

};

class beta

{

float y;

public:

beta(float j)

{

y=j;

cout<<"beta initialized\n";

}

void show_y(void)

{

cout<<"y="<<y<<"\n";

}

};

class gamma:public beta,public alpha

{

int m,n;

public:

gamma(int a,float b,int c,int d):alpha(a),beta(b)

{

m=c;

n=d;

cout<<"gamma initialized\n";

}

void show_mn(void)

{

cout<<"m="<<m<<"\n";

cout<<"n="<<n<<"\n";

}

};

void main()

{

clrscr();

gamma g(5,10.75,20,30);

g.show_x();

g.show_y();

g.show_mn();

getch();

}

7.OUTPUT

beta initialized

alpha intialised

gamma initialized

x=5

y=10.75

m=20

n=30

RESULT:

12

Engineering

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