Cs2312 OOPS LAB MANUAL

LAB MANUAL

MAILAMENGINEERING COLLEGE

CS2312 OBJECT ORIENTED PROGRAMMING LAB

LIST OF PROGRAMS USING C++ :

1. Inline Function Overloading Using Default Arguments And User Input.

2. Complex Number Using Operator Overloading.

3. To read a value of distance from one object and add with a value in another object using friend function

4. String concatenation using dynamic memory allocation concept

5. Type Conversion Using Complex Numbers

6. Managing bank account using inheritance concept

7. Stack Class With Exception Handling

8. Queue Class With Exception Handling Using Templates

LIST OF PROGRAMS USING JAVA :

9. Writing simple programs in Java.

10. Palindrome Using Strings In Java

11. Use of interfaces in Java

12. Using Inheritance In Java

13. Exception handling in Java

14. Design of multithreaded programs in Java.

1

LAB MANUAL
EXPT NO:1 Inline Function Overloading Using Default Arguments And User Input.

AIM:

To write a c++ program for Inline function overloading using default arguments and user input.

ALGORITHM:

Step 1: start the program

Step 2: declare three swap inline functions of two variables each in integer, float and character

Step 3: call the function 1 with defaults arguments.

Step 4: get user input for function 1

Step 5: repeat step 3 and 4 for function 2 and 3

Step 6: stop the program

PROGRAM:

#include<iostream.h>

#include<conio.h>

void swap(int x ,int y)

{

x=x+y;

y=x-y;

x=x-y;

cout<<endl<<"t"<<"CALLING FUNCTION 1 "<<endl;

cout<<endl<<"after swapvalue of x: "<<x<<endl;

cout<<endl<<"after swapvalue of y: "<<y<<endl;

}

void swap(float x ,float y)

{

x=x+y;

y=x-y;

x=x-y;
2

LAB MANUAL

cout<<endl<<"after swapvalue of x: "<<x<<endl;

cout<<endl<<"after swapvalue of y: "<<y<<endl;

}

void swap(char x1 ,char y1)

{

char t;

t=x1;

x1=y1;

y1=t;


cout<<endl<<"after swapvalue of x1: "<<x1<<endl;

cout<<endl<<"after swapvalue of y1: "<<y1<<endl;

}

void main()

{

int a,b;

float c,d;

char e,f,t;

clrscr();

a=10;

b=20;

cout<<endl<<"***FUNCTION OVERLOADING WITH DEFAULT ARGUMENTS***"<<endl;

cout<<endl<<"before swapvalue of a: "<<a<<endl;

cout<<endl<<"before swapvalue of b: "<<b<<endl;

swap(a,b); //swapping with default arguments

cout<<endl<<"enter two integer values"<<endl;
3

LAB MANUAL
cin>>a>>b;

swap(a,b);//swapping with user input arguments

c=92.7;

d=106.4;

cout<<endl<<"before swapvalue of c: "<<c<<endl;

cout<<endl<<"before swapvalue of d: "<<d<<endl;

swap(c,d);//swapping with default arguments

cout<<endl<<"enter two float values"<<endl;

cin>>c>>d;

swap(c,d);//swapping with user input arguments

e='A';

f='Z';

cout<<endl<<"before swapvalue of e: "<<e<<endl;

cout<<endl<<"before swapvalue of f: "<<f<<endl;

swap(e,f);//swapping with default arguments

cout<<endl<<"enter two characters"<<endl;

cin>>e>>f;

swap(e,f);//swapping with user input arguments

getch();

}

Output:

***FUNCTION OVERLOADING WITH DEFAULT ARGUMENTS***

before swapvalue of a: 10

before swapvalue of b: 20

CALLING FUNCTION 1

after swapvalue of x: 20
4

LAB MANUAL
after swapvalue of y: 10

enter two integer values

5

7

CALLING FUNCTION 1

after swapvalue of x: 7

after swapvalue of y: 5

before swapvalue of c: 92.7

before swapvalue of d: 106.4

CALLING FUNCTION 2

after swapvalue of x: 106.4

after swapvalue of y: 92.7


5.55

7.89

CALLING FUNCTION 2

after swapvalue of x: 7.89

after swapvalue of y: 5.55

before swapvalue of e: A

before swapvalue of f: Z

CALLING FUNCTION 3

after swapvalue of x: Z

after swapvalue of y: A


P

Q

CALLING FUNCTION 3
5

LAB MANUAL
after swapvalue of x: Q

after swapvalue of y: P

RESULT:

Thus the c++ program for inline function overloading with default arguments and user input has been
executed successfully

----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:2 Complex Number Using Operator Overloading.

AIM:

To write a c++ program for complex numbers using operator overloading

ALGORITHM:


Step 2: initialize the complex function of image and real in float

Step 3: declare the complex add for complex two

Step 4: return the complex temp of the real and image in float

Step 5: increment the real and image value

Step 6: print the output


PROGRAM:


#include<conio.h>

class complex

{

float real;

float imag;

public:
6

LAB MANUAL
complex(float tempreal=0,float tempimag=0)

{

real=tempreal;

imag=tempimag;

}

complex add(complex comp2)

{

float tempreal;

float tempimag;

tempreal=real+comp2.real;

tempimag=imag+comp2.imag;

return complex(tempreal,tempimag);

}

complex operator+(complex comp2)

{

float tempreal;

float tempimag;

tempreal=real+comp2.real;

tempimag=imag+comp2.imag;

return complex(tempreal,tempimag);

}

complex operator ++()

{

real++;

return complex(real,imag);

}

complex operator++(int dummy)
7

LAB MANUAL
{

imag++;

return complex(real,imag);

}

void display()

{

cout<<real<<"+"<<imag<<"in";

}

};

void main()

{

getch();

complex comp1(10,20);

complex comp2(20,30);

complex compresult1,compresult2;

compresult1=comp1.add(comp2);

compresult1.display();

compresult2=comp1+comp2;

compresult2.display();

++comp1;

comp1.display();

++comp2;

comp2.display();

}

OUTPUT:

30+50I
8

LAB MANUAL
11+20i

21+30i

RESULT:

Thus the c++ program for complex numbers using operator overloading has been executed successfully

----------------------------------------------------
XXX------------------------------------------------------------------------

To read a value of distance from one object and add with a value in another object
EXPT NO:3
using friend function

AIM:
To write a c++ program to read a value of distance from one object and add with a value in another
object using friend function.
ALGORITHM:

Step 2: Create two classes AB and AC and store the value of distances.
Step 3: Declare friend function.
Step 4: Read the value from the classes.
Step 5: Perform addition to add one object of AB with another object of AC.
Step 6: Display the result of addition.

PROGRAM:

#include<conio.h>
#include<math.h>
class AC;
class AB
{
float d;
9

LAB MANUAL
public:
AB()
{
cout<<"Enter the first distance(in feet):";
cin>>d;
}
friend void add(AB,AC);
};
class AC
{
float d;
public:
AC()
{
cout<<"Enter the second distance(in inches):";
cin>>d;
}
friend void add(AB a,AC b)
{
float total;
b.d=b.d/12;
total=a.d+b.d;
cout<<"Total Distance:"<<total<<"feet";
}
};
void main()
{
clrscr();
AB A;
AC B;
add(A,B);
getch();
}
OUTPUT:
Enter the first distance(in feet):12
Enter the second distance(in inches):30
Total Distance:14.5feet
RESULT:

10

LAB MANUAL
Thus the c++ program for reading a value of distance from one object and add with a value in another
object using friend function has been executed successfully
----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:4
String concatenation using dynamic memory allocation concept

AIM:

To write a c++ program for string concatenation function by using dynamic memory allocation
concept.
ALGORITHM:

Step 2: Create class STRING with two constructors.

Step 3: The first is an empty constructor, which allows declaring an array of strings.

Step 4: The second constructor initializes the length of the strings, and allocates necessary space for
the string to be stored and creates the string itself.

Step 5: Create a member function to concatenate two strings.

Step 6: Estimate the combined length of the strings to be joined and allocates memory for the
combined string using new operator and then creates the same using the string functions
strcpy() and strcat().

Step 7: Display the concatenated string.


PROGRAM:
#include<string.h>
#include<conio.h>

11

LAB MANUAL
class string
{
char *name;
int length;
public:
string()
{
length=0;
name=new char[length+1];
}
string(char*s)
{
length=strlen(s);
strcpy(name,s);
}
void display(void)
{
cout<<name<<"n";
}
void join(string &a,string &b);
};
void string::join(string &a,string &b)
{
length=a.length+b.length;
delete name;
strcpy(name,a.name);
strcat(name,b.name);

12

LAB MANUAL
};
void main()
{
clrscr();
string name1 ("Object ");
string name2 ("Oriented ");
string name3 ("Programming Lab");
string s1,s2;
s1.join(name1,name2);
s2.join(s1,name3);
name1.display();
name2.display();
name3.display();
s1.display();
s2.display();
getch();
}
OUTPUT:
Object
Oriented
Programming Lab
Object Oriented
Object Oriented Programming Lab

RESULT:

Thus the c++ program for String concatenation using dynamic memory allocation concept

has been executed successfully
----------------------------------------------------
XXX------------------------------------------------------------------------

13

LAB MANUAL

EXPT NO:5 Type Conversion Using Complex Numbers

AIM:
To write a c++ program for type conversion using complex numbers.

ALGORITHM:

Step 1: start the program.

Step 2: create the class complex

Step 3: declare the variables x and y.

Step 4: construct the default constructor.

Step 5: define the variables

Step 6: create the object c1, c2, c3 and initialize it.

Step 7: display the output.

Step 8: stop the program.

PROGRAM:

using namespace std;
class complex
{

int x, y;

public:

complex(){}

complex(float r, float i)

{

x=int(r);

y= int (y);

}

14

LAB MANUAL
complex (float r, int i)

{

x=int (r);

y=i;

}

complex(int r, int i)

{

x=r;

y=int(i);

}

complex operator +(complex);

void display(void);

};

complex complex :: operator+(complex c)

{

complex t;

t.x=x+c.x;

t.y=y+c.y;

return (t);

}

void complex :: display (void)

{

cout<<x<<”+i”<<y<<”n”;

}

int main()

{

complex c1,c2,c3;
15

LAB MANUAL
c1=complex (2,3.5);

c2=complex(1,2.7);

c3=c1+c2;

cout<<”c1=”;c1.display();



c1=complex(6,2.4);

c2=complex(4,3.5);

c3=c1+c2;




c1=complex(6,6.2);

c2=complex(3,2.1);

c3=c1+c2;




return 0;

}

OUTPUT:

C1=2+i3

C2=1+i2

C3=3+i5

C1=6+i2

C2=4+i3

C3=10+i5
16

LAB MANUAL
C1=6+i6

C2=3+i2

C3=9+i8

RESULT:

Thus the C++ program for type conversion in complex number has been executed successfully.

----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:6 Managing bank account using inheritance concept

AIM:
To write a c++ program to manage the account information of the customer using inheritance concept.

ALGORITHM:

Step 1: start the program.
Step 2: Create a class with the following member variables. Customer name, account number and account
type.
Step 3: Create the derived classes with following member variables.
• for current account information Balance, Deposit and withdrawal amount
• for savings account information Balance and Deposit amount
Step 4: Write a member function to get the Deposit and withdrawal amount and to update the balance
information for current account.
Step 5: Write a member function to get the Deposit amount and to update the balance information for saving
account.
Step 6: Write a member function to Display the balance information for respective account type.
Step 7: Stop the program.

PROGRAM:

17

LAB MANUAL
#include<conio.h>
class bank
{
public:
char*cname;
long int acno;
void display()
{
cout<<cname;
cout<<acno;

18

LAB MANUAL
}
};

class savings : public bank
{
public:
int wdraw, dep, bal;
void saves()
{
cout<<"Enter the amount to be deposited=Rs.";
cin>>dep;
bal+=dep;
cout<<"Enter the amount to be withdrawn=Rs.";
cin>>wdraw;
bal-=wdraw;
cout<<"Your A/c Balance is=Rs."<<bal<<endl;
}
};

class current : public savings
{
public:
void info()
{
cout<<"Last amount witdrawn=Rs."<<wdraw<<endl;
cout<<"Last amount deposited=Rs."<<dep<<endl;
cout<<"Your A/c Balance is=Rs."<<bal<<endl;
}
};

void main()
{
int ch;
current ac;
clrscr();
cout<<"Enter customer name:";
cin>>ac.cname;
cout<<endl<<"Enter your A/c no.:";
cin>>ac.acno;
cout<<endl;
while(1)
{
cout<<"Enter the A/c type,1.Savings 2.Current 3.Exitn";
cin>>ch;
switch(ch)
{
case 1:
ac.saves();
break;
case 2:
19

LAB MANUAL
ac.info();
break;
case 3:
break;
default:
cout<<"Invalid Choice";
break;
}
if (ch==3)
break;
}
getch();
}

OUTPUT:

Enter customer name: XXX

Enter your A/c no.:666

Enter the A/c type: 1.Savings 2.Current 3.Exit
1
Enter the amount to be deposited=Rs.5000
Enter the amount to be withdrawn=Rs.500
Your A/c Balance is=Rs.4500
2
Last amount withdrawn=Rs.500
Last amount deposited=Rs.5000
Your A/c Balance is=Rs.4500
3

RESULT:

Thus the C++ program for managing bank account has been executed successfully.

----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:7 Stack Class With Exception Handling

20

LAB MANUAL
AIM:
To write a c++ program for stack class with exception handling
ALGORITHM:
Step 2: declare the two function (i) PUSH (ii) POP
Step 3: initialize the stack () constructor top element is to be NULL
Step 4: check for the condition if(top>=9) throw to exception 8 assign the value
Step 5: in POP function check for the condition if (top<0)
PROGRAM:
#include<conio.h>
#include<process.h>
class Stack
{
int s[10],top;
public:
class Size{};
Stack()
{
top=-1;
}
~Stack()
{
}
void push(int i)
{
if(top<=9)
{
s[++top]=i;
}
else
21

LAB MANUAL
cout<<"OverFlow occurs......Size is"<<top<<"> than stack size";
}
int pop()
{
if(top<0)
{
cout<<"UnderFlow occurs......Size is";
return top+1;
}
else
return s[top--];
}
};
void main()
{
Stack a;
int i,ae,size;
clrscr();
cout<<"Enter the Satck size:";
cin>>size;
cout<<"nPush Operation.....n";
for(i=0;i<size;i++)
{
cout<<"Enter the Value:";
cin>>ae;
a.push(ae);
}
cout<<endl<<"Pop Operation.....n";
for(i=0;i<=size;i++)
{
cout<<a.pop()<<endl;
}
getch();
}
OUTPUT:
22

LAB MANUAL
Enter the stack size:3
Push operation ….
Enter the value : 10
Enter the value: 20
Enter the value: 30
Pop operation …
30
20
10
underflow occurs…size is 0
RESULT:
Thus the c++ program for stack class with exception handling has been executed successfully.
----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:8 QUEUE Class With Exception Handling and Templates

AIM:
To write a c++ program for queue class with exception handling using templates
ALGORITHM:
Step 2: create the class queue and decleare the front and rear values
Step 4: initialize the front and rear == -1
Step 4: then the condition is true incremented the front value
Step 5: then the condition is flase then display the output the queue is empty
Step 6: after that incremented the rear value
Step 7: the elements are inserted and display the queue
PROGRAM:
#include <iostream>
# define MAX 4
class queue

23

LAB MANUAL
{
private :
int arr [ MAX] ;
int front, rear ;
public :
class q full
{
};
class q empty
{
};
queue ()
{
front=-1;
rear=-1;
}
void add q(int item)
{
if (rear==MAX-1)
throw q full () ;
rear ++ ;
arr [0]=11
arr[1]=12
arr[2]=13
front=0
arr[3]=14
arr[rear]=item;
if(front==-1)
front=0;
count<<"n Item added into queue : <<arr[real];
}int del q()
{
24

LAB MANUAL
int data;
if(front==-1)
throw q emty ();
data=arr[front] ;
if(front==rear=-1;
else
front++;
return data ;
}
};
void Main()
{
queue a;
try
{
a.add q (11);
a.add q (12);
a.add q(13);
a.add q (14);
a.add q (15); queue is full
}
catch (queue ::qfull)
{
cout<<endl<<"QUEUE IS FULL";
}
int i;
try
{ i=a.delq();
cout<<endl<<"Item deleted=<<i;
i=a.del q();
cout<<eendl<<"Item deleted =" <<i;
i=a.del q ();
25

LAB MANUAL
cout <<endl<<"Item deleted=" <<i;
i=a,del q ();
cout<<endl<<"Item deleted="<<i;
i=a.del()
cout <<endl<<"Item deleted=<<i;
i=a.delq());
}
OUTPUT:
item added into queue:11
queue is full
item deleted:11
item deleted:12
item deleted:13
item deleted:14
queue is empty
RESULT:
Thus the c++ program for queue class with exception handling has been executed successfully.
----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:9 To Print Multiplication Table Of A Given Number

AIM:

To write a java program to print multiplication table of a given number

Algorithm:

Step 1: Start the Program

Step 2: create a class with the name mtable.

26

LAB MANUAL
Step 3: Declare two integer variables i and k.

Step 4: Get input for variable k from the user.

Step 5: Use for loop to perform multiplication table operation

Step 6: Print the result.

Step 7: Stop the program

PROGRAM:

import java.io.*;

class mtable

{

public static void main(String ar[])throws IOException

{

int i,k;

DataInputStream di= new DataInputStream(System.in);

System.out.println("enter the number ");

k=Integer.parseInt(di.readLine());

System.out.println("***multiplication table *****");

for(i=1;i<=15;i++)

{

System.out.println("[ "+i+" * "+k+" ] "+" [ "+i*k+" ]");

}

}

}

COMPILATION AND EXECUTION:

D:javaprg>javac mtable.java
D:javaprg>java mtable

OUTPUT:

27

LAB MANUAL
enter the number

9

***multiplication table *****

[ 1 * 9 ][ 9 ]

[ 2 * 9 ][ 18 ]

[ 3 * 9 ][ 27 ]

[ 4 * 9 ][ 36 ]

[ 5 * 9 ][ 45 ]

[ 6 * 9 ][ 54 ]

[ 7 * 9 ][ 63 ]

[ 8 * 9 ][ 72 ]

[ 9 * 9 ][ 81 ]

[ 10 * 9 ][ 90 ]

[ 11 * 9 ][ 99 ]

[ 12 * 9 ][ 108 ]

[ 13 * 9 ][ 117 ]

[ 14 * 9 ][ 126 ]

[ 15 * 9 ][ 135 ]

RESULT:
Thus the program to perform multiplication table is done using java.

----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:10 Palindrome using strings in java

28

LAB MANUAL
AIM:

To write a java program to print multiplication table of a given number

Algorithm:

Step 1: Start the Program

Step 2: create a class with the name palin.

Step 3: Declare a string and two integer variable.

Step 4: Assign value for the string.

Step 5: Store the length of the string in a integer variable.

Step 6: Reverse the string with charAt() method.

Step 7: check if the reversed string and input string are equal.

Step 8: Print the result.

Step 9: Stop the program.

PROGRAM:

import java.io.*;
class palin
{
public static void main(String args[])
{
String s="malayalam";
int i;
int n=s.length();
String str="";
for(i=n-1;i>=0;i--)
str=str+s.charAt(i);
System.out.println("n t GIVEN STRING:: "+s);
System.out.println("n t REVERSED STRING:: "+str);
if(str.equals(s))
System.out.println(s+ " :: IS A PALINDROME");
else
System.out.println(s+ " : :IS NOT A PALINDROME");
}
}
29

LAB MANUAL
D:javaprg>javac mtable.java
D:javaprg>java mtable

OUTPUT:

GIVEN STRING: malayalam

REVERSED STRING:: malayalam

Malayalam :: IS A PALINDROME

RESULT:
Thus the program to perform palindrome using java is done.
----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:11 To calculate area of rectangle and circle using interfaces

AIM:
To calculate area of rectangle and circle using interfaces.

ALGORITHM:

Step 1: Start the Program.

Step 2: Create an interface with name area.

Step 3: Create classes rectangle and circle which implements area.

Step 4: Create a class interface test.

Step5: Create object for rectangle, circle class and interface area.

Step6: Print the area of circle and triangle

Step 7: Stop the Program.

PROGRAM:

import java .io.*;

interface Area

{
30

LAB MANUAL
final static float pi=3.14F;

float compute(float x,float y);

}

class Rectangle implements Area

{

public float compute(float x,float y)

{

return(x*y);

}

}

class Circle implements Area

{

public float compute(float x, float y)

{

return(pi*x*x);

}

}

31

LAB MANUAL
class InterfaceTest
{
public static void main(String args[])
{
Rectangle rect=new Rectangle();
Circle cir=new Circle();
Area ar;
ar=rect;
System.out.println("Area of Rectangle="+ar.compute(10,20));
ar=cir;
System.out.println("Area of Circle="+ar.compute(10,0));
}
}

D:javaprg>javac InterfaceTest.java
D:javaprg>java InterfaceTest

OUTPUT:

Area of Rectangle=200.0
Area of Circle=314.0

RESULT:

Thus the programs for calculating area for circle and rectangle are done using java.

----------------------------------------------------
XXX------------------------------------------------------------------------

EXPT NO:12 To perform multilevel inheritance

AIM:
To perform multiple inheritance in java.

32

LAB MANUAL
ALGORITHM:


Step 2: Create class a1 and declare 3 variables.

Step 3: Create class b1 which extends a1.

Step 4: Class b1 has 3 variables and also derived from base class a1

Step 5: Create class c1 which is derived from base class b1

Step 6: Access variables from a1 and b1 and process the data in c1.


PROGRAM:

import java.io.*;

class a1

{

int s1,s2,s3;

}

class b1 extends a1 //simple inheritance

{

int s4=70, s5=90;

int total;

}

class c1 extends b1//multilevel inheritance

{

public static void main(String ar[])

{

int avg;

b1 b=new b1();

b.s1=70;
33

LAB MANUAL
b.s2 = 50;

b.s3 = 60;

b.total = b.s1+b.s2+b.s3+b.s4+b.s5;

avg = b.total/5;

System.out.println("***** multilevel inheritance*******");

System.out.println("values from class a1: " +b.s1 +"n");

System.out.println("values from class a1: " +b.s2+"n" );

System.out.println("values from class a1: " +b.s3 +"n");

System.out.println("values from class b1: " +b.s4 +"n");

System.out.println("values from class b1: " +b.s5 +"n");

System.out.println(" total value from class bt: " +b.total+"n" );

System.out.println("average of 5 numbers: "+avg+"n");

if ((avg>=91)&&(avg<=100))

{

System.out.println("GRADE IS (91-100) : S ");

}

else if ((avg>=81)&&(avg<=90))

{

System.out.println("GRADE IS (81-90) : A ");

}


{

System.out.println("GRADE IS (71-80) : B ");

}


{

34

LAB MANUAL
System.out.println("GRADE IS (61-70) : c ");

}


{

System.out.println("GRADE IS (50-60) : D ");

}


{

System.out.println("GRADE IS (0-49) : F ");

} } }

D:javaprg>javac InterfaceTest.java
D:javaprg>java InterfaceTest
OUTPUT:

***** multilevel inheritance*******

values from class a1: 70



values from class b1: 70

values from class b1: 90

total value from class b1: 340

average of 5 numbers: 68

GRADE IS (61-70): C

RESULT:
Thus the programs for are done using java.

EXPT NO:13 Exception handling using java

35

LAB MANUAL
AIM:
To perform exception handling using java.

ALGORITHM:


Step 2: Create class exph1 and declare 3 variables.

Step 3: Use Try and Catch block for arithmetic exception.

Step 4: Use Try and Catch block for number format exception.

Step 5 Stop the Program.

PROGRAM:

class exph1

{

public static void main(String ar[])

{

int a = 10;

int b = 0;

int c;

int i,n=ar.length,sum=0,invalid=0;

for(i=0;i<n;i++)

{

try

{

sum+=Integer.parseInt(ar[i]);

}

catch(NumberFormatException e)

{

invalid++;
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LAB MANUAL
System.out.println("n"+"---------NUMBERFORMAT EXCEPTION-----");

System.out.println(ar[i]+" :: is not an integer"+"n" );

}

}

System.out.print("n"+"TOTAL NO OF ARGUMENTS : "+n);

System.out.println("n"+"invalid data:: "+invalid);

System.out.println("n"+"sumof valid integers:: "+sum);

try

{

c=a/b;

}

catch(ArithmeticException e)

{

System.out.println("n"+"--------ARITHMETIC EXCEPTION------ ");

System.out.println("n"+"dividend is :: "+a+"t"+"divisor is :: "+b);

System.out.println("Dvision by zero");

}

}

}

OUTPUT:

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LAB MANUAL

RESULT:

Thus the program for exception handling is done using java.

EXPT NO:14 Multithreading using java

AIM:
To perform multithreading using java.

ALGORITHM:


Step 2: Create a class sqr1 which extends Thread.

Step 3: Create a class cub1 which extends Thread.

Step 4: Create a class thrd1.

Step 5: Create objects for a class sqr1 and cub1.

Step 6: Execute the program


PROGRAM:

public class sqr1 extends Thread

{

public void run()
38

LAB MANUAL
{

for(int i=1;i<=30;i++)

System.out.println("n"+i+" square is:: "+i*i);

}

}

public class cub1 extends Thread

{

public void run()

{

for(int i=1;i<=10;i++)

System.out.println("n"+i+" cube is:: "+i*i*i);

}

}

public class thrd1

{

public static void main(String ar[])throws Exception

{

sqr1 s;

cub1 c;

s = new sqr1();

c = new cub1();

s.sleep(6000);

s.start();

c.start();

}

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LAB MANUAL
}

OUTPUT:

E:D.PRABHUthrd>javac sqr1.java
E:D.PRABHUthrd>javac cub1.java
E:D.PRABHUthrd>javac thrd1.java
E:D.PRABHUthrd>java thrd1

1 square is:: 1

2 square is:: 4

3 square is:: 9

4 square is:: 16

5 square is:: 25

6 square is:: 36

7 square is:: 49

8 square is:: 64

1 cube is:: 1

2 cube is:: 8

3 cube is:: 27

4 cube is:: 64

5 cube is:: 125

40

LAB MANUAL
6 cube is:: 216

7 cube is:: 343

8 cube is:: 512

9 cube is:: 729

10 cube is:: 1000

9 square is:: 81

10 square is:: 100
RESULT:

Thus the program for multithreading using java is done.

41

Cs2312 OOPS LAB MANUAL

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Cs2312 OOPS LAB MANUAL