ds & oops lab manual.doc

8/11/2019 DS & OOPS LAB MANUAL.doc

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EX.NO:1.A FUNCTION WITH DEFAULT ARGUMENTS

DATE :

AIM:

To write a C++ program to find the sum for the given variables using function withdefault arguments.

ALGORITHM:

1) Start the program.

2) Declare the variables and functions.3) ive the values for two arguments in the function declaration itself.

!) Call function sum") with three values such that it ta#es one default arguments.

$) Call function sum") with two values such that it ta#es two default arguments.%) Call function sum") with one values such that it ta#es three default arguments

$) &nside the function sum")' calculate the total.%) (eturn the value to the main") function.) Displa* the result.

) Stop the program.

PROGRAM:

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,include-iostream.h

,include-conio.h

void main")/

float sum"float a'int b01'int c01$'int d02)

int a02'b03'c0!'d0$clrscr")

cout--4nsum0--sum")

cout--4nsum0--sum"a'b'c'd)cout--4nsum0--sum"a'b'c)

cout--4nsum0--sum"a'b)

cout--4nsum0--sum"a)

cout--4nsum0--sum"b'c'd)getch")

5

float sum"float i' int 6' int #' int l)

/return"i+6+#+l)

5

OUTPUT:

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sum0!$

sum01!

sum027sum0!

sum0!

sum032

RESULT:

Thus' the given program for function with default arguments has been written and e8ecutedsuccessfull*.

EX.NO:1.B IMPLEMENTATION OF CALL BY VALUE

DATE :

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AIM:

To write a C++ program to find the value of a number raised to its power that

demonstrates a function using call b* value.

ALGORITHM:

1) Start the program.2) Declare the variables.

3) et two numbers as input

!) Call the function power to which a cop* of the two variables is passed .

$) &nside the function' calculate the value of 8 raised to power * and store it in p.%) (eturn the value of p to the main function.

) Displa* the result.

) Stop the program.

PROGRAM:

,include-iostream.h

,include-conio.h

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void main")

/

int 8'*double power"int' int)

clrscr")

cout--9nter 8'*:--endlcin8*

cout--8-- to the power --* -- is -- power"8'*)

getch")5

double power"int 8'int *)

/

double pp01.

if"*0)

while"*;;)

pT9( ? ' @:

2 32 TA TB9 A9( 3 &S

5


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RESULT:

Thus' the given program for implementation of call b* value has been written ande8ecuted successfull*.

EX. NO: 1.C IMPLEMENTATION OF CALL BY ADDRESS

DATE:

AIM:

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To write a C++ program and to implement the concept of Call b* Eddress.

ALGORITHM:

1. Start the program.

2. &nclude suitable header file.3. Declare a function swap with two pointes variables arguments.

!. Declare and initialiFe the value as two variable in main ").

$. rint the value of two variable before swapping.%. Call the swap function b* passing address of the two variable as arguments.

. rint the value of two variable after swapping.

. Stop the program.

PROGRAM:

,include-iostream.h

,include-conio.hvoid swap"int


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/

clrscr")

int i'6i01

602

cout--4n the value of i before swapping is:--icout--4n the value of i before swapping is:--6

swap "Gi'G6)

cout--4n the value of i after swapping is:--icout--4n the value of i after swapping is:--6

getch")

return")

5void swap"int


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RESULT:

Thus to write a c++ program and to implement the concept of Call b* Eddress wassuccessfull* completed.

EX. NO: 2 COPY CONSTRUCTORS

DATE:

AIM:

To write the c++ program and to implement the concept of cop* constructor

ALGORITHM:

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1. Start the program

2. &nclude suitable header file

3. Declare constructor with empt* arguments and another 'ob6ect as arguments!. &nitialiFe the value of a variable b* calling a constructor in main function

$. Essign the value of one ob6ect to another ob6ect

%. 98ecute the program. rint the statement

. Stop the program

PROGRAM:

,include-iostream.h,include-conio.hclass code

/

int id

public:code")/5

code"int a)

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/

5

code"codeG8)/

id08.id

5void displa*"void)

/

cout--id5

5

int main")

/clrscr")

code a"1)

code b"a)

code c0acode d

d0acout--4nid of a: a.displa*")

cout--4nid of b: b.displa*")

cout--4nid of c: c.displa*")cout--4nid of d: d.displa*")

getch")

return

5

OUTPUT:

id of E:1id of H:1

id of C:1id of D:1

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RESULT:

Thus to write the c++ program and implement the concept of cop* constructor was

successfull* completed.

EX. NO: 3.A OPERATOR OVERLOADING

DATE:

AIM:

To write a c++ program to concatenate two string using the concept of operator

Averloading.

ALGORITHM:

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2. &nclude suitable header file

3. Create a class with necessar* variables and function!. et the value as two variables as concatenating

$. Averload an operator to concatenate two string

%. 98ecute the program. Stop the program

PROGRAM:

,include-iostream.h

,include-string.h,include-conio.hconst int bufsiFe0$

class string

/

private:char strIbufsiFeJ

public:

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string")

/

strcp*"str' )5

string"char


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OUTPUT:

Hefore str30str10str2

str10stud

str20entstr30

Efterstr30str1+str2:

str10stud

str20ent

str30student

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RESULT:

Thus to write a c++ program to concatenate two string using the concept of operator

overloading was successfull* completed.EX.NO: 3.B UNARY OPERATOR OVERLOADING

DATE :

AIM:

To perform increment and decrement operations using unar* operator overloading.

ALGORITHM:

1) Start the program.

2) Create Class unar* with one constructors.3) The constructor ta#es no arguments and is used to create ob6ects that are not initialised.

!) Declare a function operator K++L and K;; inside the class unar* which are the function

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where overloading is done.

$) Create an ob6ect for the class.

%) Call the functions with declared ob6ect.) Stop the program.

PROGRAM:

,include-iostream.h,include-conio.h

class unar*

/

private:int 8'*'F

public:unar*"void)

/

cout-- 9nter En* Three &nteger >os. :

cin8*F5

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void displa*"void)

/

cout-- endl-- The Three >os. Ere : -- 8-- ' -- *-- ' -- F5

void operator ;;")

/8 0 ;;8

* 0 ;;*

F 0 ;;F5

void operator ++")

/

8 0 ++8* 0 ++*

F 0 ++F

5

5void main")

/clrscr")

unar* s

s.displa*");;s

cout-- endl-- endl-- endl--


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%. 98ecute the program

. Stop the program

PROGRAM:

,include -iostream.h

,include-conio.h

class comple8/

float 8

float *public:

comple8")/5

comple8"float real'float image)

/80real

*0image

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5

comple8 operator + "comple8)

void displa*"void)5

comple8 comple8::operator+"comple8 c)

/comple8 temp

temp.808+c.8

temp.*0*+c.*return"temp)

5

void comple8::displa*"void)

/cout--8--+6--*

5

int main")

/clrscr")

comple8 c1'c2'c3c10comple8"2.$'3.$)

c20comple8"1.%'2.)

c30c1+c2cout--c1 0

c1.displa*")

cout--c2 0

c2.displa*")cout--c3 0

c3.displa*")

getch")return

5

OUTPUT:

c102.$+63.$

c201.%+62.c30!.1+6%.2

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RESULT:

Thus to write a c++ program to concatenate two string using the concept Hinar* operator

overloading was successfull* completed.EX. NO: 3.D FUNCTION OVERLOADING

DATE:

AIM:To write a c++ program to find the volume using function overloading concept

ALGORITHM:


2. &nclude suitable header file3. Declare the function of same name with different parameter

!. Define the function for calculating the volume

$. Call the respective functions%. rint the volume

7. Stop the program

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PROGRAM:

,include-iostream.h

,include-conio.hint volume"int)

double volume"double'int)

long volume"long'int'int)int main")

/

clrscr")cout--Nvolume of cube0N

cout--volume"1)--4n

cout--Nvolume of c*linder0N

cout--volume"2.$')--4ncout--Nvolume of cubiod0N

cout--volume"1'$'1$)--4n

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return

getch")

5int volume"int s)

/

return"s


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RESULT:

Thus to write a c++ program to find the volume using function overloading concept was

successfull* completed.EX. NO: 4.A MULTIPLE INHERITANCE

DATE:

AIM:

To write a c++ program to implement the concept of multiple inheritance.

ALGORITHM:

1. Start the program.2. &nclude the suitable header file.

3. create a base class O and >.!. Declare a variable m and void getm function in class O.

$. Declare a variable n and void getn function in class >.

%. Derive a class from the base class OG> and declare a function void displa*.. Define the function void getm 'void getn and void displa* outside their respective class.

. et the value for the variable in getm and getn function.

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7. Call the displa* function for displa*ing the output.

1. Stop the program..

PROGRAM:

,include-iostream.h

class O/

protected:

int m

public:void getm"int)

5

class >

/protected:

int npublic:

void getn"int)

5

class :public O'public >/

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public:

void displa*"void)

5void O::getm"int 8)

/

m085

void >::getn"int *)

/n0*

5

void ::displa*"void)

/cout--m0--m--4n

cout--n0--n--4n

cout--m


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RESULT:

Thus to write a c++ program for multiple inheritance was successfull* completed.

EX. NO:4.B HYBRID INHERITANCE

DATE:

AIM:

To write a c++ program to implement the concept of h*brid inheritance.

ALGORITHM:

1. Start the program.

2. &nclude suitable header files.

3. Create a base class student and sports.

!. &n the base class student define the function void get number and put number.$. &n the base class sports define the function void get score and void put score.

%. Derive a class test form base student and define the function get mar# and put mar#.

. Derive a class result from test and sports class and define function void displa*.. et the value for get number' get mar#s and get score function through main function.

7. Call the displa* function in class result.

1. Stop the program.

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PROGRAM:

,include-iostream.hclass student

/protected:

int rollPnumber

public

void getPnumber"int a)/

rollPnumber0a

5void putPnumber"void)

/cout--(AQQ >A:--rollPnumber--4n5

5

class test :public student

/protected:

float part1'part2

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public:

void getPmar#s"float 8'float *)

/part108

part20*

5void putPmar#s"void)

/

cout--mar#s obtined:--4n--part10--part1--4n

--part20--part2--4n

5

5class sports

/

protected:

float scorepublic:

void getPscore"float s)/

score0s

5void putPscore"void)

/

cout--sports wt:--score--4n4n

55

class result:public test'public sports

/float total

public:

void displa*"void)5

void result::displa*"void)

/

total0part1+part2+scoreputPnumber")

putPmar#s")

putPscore")cout--total score:--total--4n

5

int main")/

result studentP1

studentP1.getPnumber"123!)

studentP1.getPmar#s"2.$'33.)

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studentP1.getPscore"%.)

studentP1.displa*")

return 5

OUTPUT:

(oll no: 123!

Oar#s obtained:art102.$

art2033

Total score: %%.$

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RESULT:

Thus to write a c++ program for h*brid inheritance was successfull* completed.

EX.NO:5 ARRAY IMPLEMENTATION OF LIST ADT

DATE:

AIM:To rite a C++ rogram for arra* implementation of list EDT.

ALGORITHM:

Step1: Create nodes first'last'ne8t'prev and cur then set the value as >RQQ.Step 2: (ead the list operation t*pe.

step 3: &f operation t*pe is create then process the following steps.

1. Ellocate memor* for node cur.2. (ead data in curs data area.

3. Essign cur node as >RQQ.

!. Essign first0last0cur.Step !: &f operation t*pe is &nsert then process the following steps.

1. Ellocate memor* for node cur.

2. (ead data in curs data area.

3. (ead the position the Data to be insert.!. Evailabilit* of the position is true then assing curs node as first and first0cur.

$. &f availabilit* of position is false then do following steps.

Step$: &f operation t*pe is delete then do the following steps.

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1. (ead the position .

2. Chec# list is 9mpt* .&f it is true displa* the message Qist empt*.

3. &f position is first.1. Essign cur as first.

2. Essign irst as first of node.

3. (eallocate the cur from memor*.!. &f position is last.

Step %: &f operation is traverse.

1. Essign current as first.2. (epeat the following steps until cursor becomes >RQQ.

Step : Stop the program.

PROGRAM:


,include-process.h

void create")void insert")

void deletion")

void search")void displa*")

int a'bI2J'n'd'e'f'i

void main")

/int c

char g0*

clrscr")

do/

cout--4n Oain Oenucout--4n 1.Create 4n 2.Delete 4n 3.Search 4n !.insert 4n $.Displa* 4n %.98it

cout--4n enter *our choice4n

cinc

switch"c)/

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case 1: create") brea#

case 2: deletion") brea#

case 3: search") brea#case !: insert") brea#

case $: displa*") brea#

case %: e8it") brea#default:

cout--The given number is not between 1;$4n

5cout--4nDo u want to continue 4n

cing

clrscr")

5while"g00*UU g00@)

getch")

5

void create")/

cout--4n 9nter the number4ncinn

for"i0i-ni++)

/cinbIiJ

5

5

void deletion")/

cout--9nter the limit u want to delete 4n

cindfor"i0i-ni++)

/

if"bIiJ00d)/

bIiJ0

5

55

void search")

/cout--9nter the limit 4n

cine

for"i0i-ni++)/

if"bIiJ00e)

/

cout--Malue found the position4n--bIiJ

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5

5

5void insert")

/

cout--enter how man* number u want to insert 4ncinf

for"i0i-fi++)

/cinbIn++J

5

5

void displa*")/cout--4n4n4n

for"i0i-ni++)

/

cout--4n4n4n--bIiJ5

5

OUTPUT:

Oain Oenu

1.Create

2.Delete3.Search

!.&nsert

$.Displa*%.98it

9nter *our choice

19nter the number

2

3

!Do u want to continue

n

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RESULT:

Thus' the arra* implementation of list EDT program has been written and e8ecuted

successfull*.

EX.NO:5 ARRAY IMPLEMENTATION OF LIST ADT

DATE :

AIM:

To rite a C++ rogram for arra* implementation of list EDT.

ALGORITHM:

Step1: Create nodes first'last'ne8t'prev and cur then set the value as >RQQ.

Step 2: (ead the list operation t*pe.Step 3: &f operation t*pe is create then process the following steps.


2. (ead data in curs data area.3. Essign cur node as >RQQ.

!. Essign first0last0cur.

Step !: &f operation t*pe is &nsert then process the following steps.1. Ellocate memor* for node cur.

2. (ead data in curs data area.

3. (ead the position the Data to be insert.!. Evailabilit* of the position is true then assing curs node as first and first0cur.

$. &f availabilit* of position is false then do following steps.

1. Essign ne8t as cur and count as Fero.

2. (epeat the following steps until count less than postion.1 .Essign prev as ne8t

2. >e8t as prev of node.

3. Edd count b* one.

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!. &f prev as >RQQ then displa* the message &>MEQ&D AS&T&A>.

$. &f prev not Vual to >RQQ then do the following steps.

1. Essign curs node as prevs node.2. Essign prevs node as cur.


1. (ead the position .2. Chec# list is 9mpt* .&f it is true displa* the message Qist empt*.

3. &f position is first.

1. Essign cur as first.2. Essign irst as first of node.

3. (eallocate the cur from memor*.

!. &f position is last.

1. Oove the current node to prev.2. curs node as >ull.

3. (eallocate the Qast from memor*.

!. Essign last as cur.

$. &f position is enter Oediate.1. Oove the cur to reVuired postion.

2. Oove the revious to curs previous position3. Oove the >e8t to curs >e8t position.

!. >ow Essign previous of node as ne8t.

$. (eallocate the cur from memor*.


1. Essign current as first.2. (epeat the following steps untill cur becomes >RQQ.

Step :Stop the program.

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PROGRAM:

,include-iostream.h

,include-conio.h,include-process.h

void create")

void insert")

void deletion")void search")

void displa*")

int a'bI2J'n'd'e'f'ivoid main")

/

int cchar g0*

clrscr")

do

/cout--4n Oain Oenu

cout--4n 1.Create 4n 2.Delete 4n 3.Search 4n !.insert 4n $.Displa* 4n %.98it

cout--4n enter *our choice4n

cincswitch"c)

/case 1: create") brea#

case 2: deletion") brea#

case 3: search") brea#

case !: insert") brea#case $: displa*") brea#

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case %: e8it") brea#

default:

cout--The given number is not between 1;$4n5

cout--4nDo u want to continue 4n

cingclrscr")

5

while"g00*UU g00@)getch")

5

void create")

/cout--4n 9nter the number4n

cinn

for"i0i-ni++)

/cinbIiJ

55void deletion")

/

cout--9nter the limit u want to delete 4ncind

for"i0i-ni++)

/

if"bIiJ00d)/

bIiJ0

555void search")

/

cout--9nter the limit 4ncine

for"i0i-ni++)

/

if"bIiJ00e)/

cout--Malue found the position4n--bIiJ

555void insert")

/

cout--enter how man* number u want to insert 4ncinf

for"i0i-fi++)

/

cinbIn++J

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5

5

void displa*")/

cout--4n4n4n

for"i0i-ni++)/

cout--4n4n4n--bIiJ

55

OUTPUT:

Oain Oenu1.Create

2.Delete

3.Search!.&nsert

$.Displa*

%.98it

9nter *our choice1

9nter the number

23

!

Do u want to continuen

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RESULLT:

Thus' the arra* implementation of list EDT program has been written and e8ecutedsuccessfull*.

EX.NO: LIN!ED LIST IMPLEMENTATION OF LIST ADT

DATE :

AIM:

To rite a C++ rogram for lin#ed list implementation of list EDT.

ALGORITHM:


Step 2: (ead the list operation t*pe.

step 3: &f operation t*pe is create then process the following steps.1. Ellocate memor* for node cur.

2. (ead data in curs data area.3. Essign cur lin# as >RQQ.

!. Essign first0last0cur.Step !: &f operation t*pe is &nsert then process the following steps.


2. (ead data in curs data area.3. (ead the position the Data to be inserting.

!. Evailabilit* of the position is true then assign curs lin# as first and first0cur.

$. &f availabilit* of position is false then do following steps.1. Essign ne8t as cur and count as Fero.

2. (epeat the following steps until count less than position.

1 .Essign prev as ne8t2. >e8t as prev of lin#.


!. &f prev as >RQQ then displa* the message &>MEQ&D AS&T&A>.

$. &f prev not Vual to >RQQ then do the following steps.1. Essign curs lin# as prevs lin#.

2. Essign prevs lin# as cur.


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2. Essign irst as first of lin#.


1. Oove the current node to prev.

2. curs lin# as >ull.3. (eallocate the Qast from memor*.


$. &f position is enter Oediate.

1. Oove the cur to reVuired position.2. Oove the revious to curs previous position

3. Oove the >e8t to curs >e8t position.

!. >ow assign previous of lin# as ne8t.

$. (eallocate the cur from memor*.Step %: &f operation is traverse.

1. Essign current as first.2. (epeat the following steps until cur becomes >RQQ.

Step : Stop the program.

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PROGRAM:

,include-iostream.h

,include-conio.h,include-stdlib.h

class list

/struct node

/

int data

node RQQ)

/

p0new node

p;data08p;lin#0>RQQ

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5

else

/V0p

while"V;lin#X0>RQQ)

V0V;lin#t0new node

t;data08

t;lin#0>RQQV;lin#0t

5

cout--4n4n&nserted successfull* at the end..

disp")5

void list:: insbeg"int 8)

/

node


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cout--4n4n9lement *ou entered --8-- is not found..

5

void list:: delbeg")/

cout--4n4nThe list before deletion:

disp")node RQQ)

/

cout--4n4nThere is no data in the list..return

5

if"V;lin#00>RQQ)/

p0V;lin#

delete Vreturn

5

while"V;lin#;lin#X0>RQQ)

V0V;lin#V;lin#0>RQQ

return

5list::Wlist")

/

node RQQ)

/

V0p;lin#

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delete p

p0V

55

void list::disp")

/node RQQ)

/

cout--V;data--4nV0V;lin#

55

void list :: insne8t"int value'int position)

/node


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int list::see#"int value)

/

node S9(T&A>

cout--4n4n1.&nsertion at begining4n2.&nsertion at the end

cout--4n3.&nsertion between two >odescout--4n4n9nter ur choice:

cinps

cout--9nter the value to insert:cinv

switch"ps)

/case 1:

l.insbeg"v)

brea#

case 2:

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l.inslast"v)

brea#

case 3:cout--4n9nter the position to insert the value:

cinp

l.insne8t"v'p)brea#

default:

cout--4nThe choice is invalidreturn

5

brea#

case 2:clrscr")

cout--4n1.Delete the first element4n2.Delete the last element

cout--4n3.9nter the element to delete from the list

cout--4n4n9nter ur choice:cinps

switch"ps)/

case 1:

l.delbeg")cout--4nThe list after deletion:

l.disp")

brea#

case 2:l.dellast")

cout--4nThe list after deletion:

l.disp")brea#

case 3:

l.disp")cout--4n9nter the element to delete :

cinv

l.delelement"v)

cout--4nThe list after deletion:l.disp")

brea#

default:cout--4nThe option is invalid...

brea#

5brea#

case 3:

clrscr")

l.disp")

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OUTPUT:

Singl* Qin#ed Qist1.Create

2.&nsert

3.Delete

!.98it9nter @our Choice : 1

9nter The Data: 1

11.Create

2.&nsert

3.Delete!.98it

9nter @our Choice : 2

9nter The Data: 39nter The osition: 1

3

1

1.Create2.&nsert

3.Delete

!.98it9nter @our Choice : 3

9nter The osition : 2

Qist &s 9mpt*

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RESULT:

Thus' the lin#ed list implementation of list EDT for singl* lin#ed list program has beenwritten and e8ecuted successfull*.

EX.NO:7 CURSOR IMPLEMENTATION OF LIST ADT

DATE :

AIM:

To rite a C++ rogram for Cursor implementation of list EDT.

ALGORITHM:


Step 2: (ead the list operation t*pe.

Step 3: &f operation t*pe is create then process the following steps.


3. Essign cur lin# as >RQQ.!. Essign first0last0cur.

Step !: &f operation t*pe is &nsert then process the following steps.


3. (ead the position the Data to be inserting.

!. Evailabilit* of the position is true then assign curs lin# as first and first0cur.$. &f availabilit* of position is false then do following steps.

1. Essign ne8t as cur and count as Fero.

2. (epeat the following steps until count less than position.1 .Essign prev as ne8t2. >e8t as prev of lin#.


!. &f prev as >RQQ then displa* the message &>MEQ&D AS&T&A>.$. &f prev not Vual to >RQQ then do the following steps.

1. Essign curs lin# as prevs lin#.

2. Essign prevs lin# as cur.Step$: &f operation t*pe is delete then do the following steps.




2. Essign irst as first of lin#.


1. Oove the current node to prev.

2. curs lin# as >ull.3. (eallocate the Qast from memor*.

51


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$. &f position is enter Oediate.

1. Oove the cur to reVuired position.2. Oove the revious to curs previous position

3. Oove the >e8t to curs >e8t position.

!. >ow assign previous of lin# as ne8t.$. (eallocate the cur from memor*.


1. Essign current as first.2. (epeat the following steps until cur becomes >RQQ.

52


53/96

PROGRAM:


,include-process.h

,define ma8 $struct node

/

int data

int ne8t5

t*pedef struct node >AD9

int avail'list 0 ;1>AD9 cursIma8J

void initial")

/int i

avail 0

for"i0i-ma8;1i++)cursIiJ.ne8t0i+1

cursIiJ.ne8t0;1

5

void create")/

int n'i'item'temp

cout--4n9nter the no of elements: cinn

==cout--n

if"n0ma8)cout--4n SiFe e8ists

else

/

initial")if"avail00;1)

/

cout--4nThere is no space to insert

e8it")5

list 0 availif"n00ma8;1) avail 0 ;1

else avail0n+1

cout--4n9nter the elements one b* one:

for"i0i-ni++)/

53


54/96

cout--4n 9nter the --i+1--th element

cinitem

cout--4t==cout--item

cursIi+1J.data0item

5cursInJ.ne8t0;1

5

5void displa*")

/

int i

cout--4nCursor space: cout--4n4nEvail 0 --avail-- 4t

cout--Qist 0 --list--4n

cout--PPPPPPPPPPPPPPP

cout--4n DETE >9?T 4ncout--PPPPPPPPPPPPPPP

i0while"i-ma8)

/

cout--4n--cursIiJ.data-- 4t --cursIiJ.ne8tcout--4nPPPPPPPPPPPPPPP

i++

5

5void insbeg")

/

int item'icout--4n9nter the item to be inserted:

cinitem

==cout--itemcout--4n

i0avail

avail0cursIavailJ.ne8t

cursIiJ.data0itemcursIiJ.ne8t0cursIlistJ.ne8t

cursIlistJ.ne8t0i

5void insend")

/

int item'icout--4n9nter the item to be inserted:

cinitem

==cout--item

cout--endl

54


55/96

i0list

while"cursIiJ.ne8tX0;1)

i0cursIiJ.ne8tcursIiJ.ne8t0avail

avail0cursIavailJ.ne8t

i0cursIiJ.ne8tcursIiJ.data0item

cursIiJ.ne8t0;1

5void insint")

/

int item'i'pos'count'temp

cout--4n9nter the item to be inserted: cinitem

==cout--item

cout--endl

cout--4n9nter the position of the item: cinpos

==cout--poscout--endl

i0list

count01while"count-pos)

/

i0cursIiJ.ne8t

count0count+15

temp0avail

avail0cursIavailJ.ne8tcursItempJ.data0item

cursItempJ.ne8t0cursIiJ.ne8t

cursIiJ.ne8t0temp5

void delbeg")

/

int ii0cursIlistJ.ne8t

cursIlistJ.ne8t0cursIiJ.ne8t

cursIiJ.ne8t0availavail0i

cursIavailJ.data0

if"cursIlistJ.ne8t00;1)/

cursIlistJ.ne8t0avail

avail0list

list0;1

55


56/96

5

5

void delend")/

int i'prev

i0listwhile"cursIiJ.ne8tX0;1)

/

prev0ii0cursIiJ.ne8t

5

cursIprevJ.ne8t0;1


cursIavailJ.data0

if"cursIlistJ.ne8t00;1)

/cursIlistJ.ne8t0avail

avail0listlist0;1

5

5void delint")

/

int pos'i'count'prev

cout--4n9nter the position: cinpos

==cout--pos

cout--endli0list

count01

while"count-0pos)/

prev0i

i0cursIiJ.ne8t

count0count+15

cursIprevJ.ne8t0cursIiJ.ne8t


cursIavailJ.data0

if"cursIlistJ.ne8t00;1)/

cursIlistJ.ne8t0avail

avail0list

list0;1

56


57/96

5

5

void main")/

int ch

clrscr")do

/

cout--4n


58/96

else delend")

brea#

case :if"list00;1) cout--4n>o element to delete

else delint")

brea#case :

displa*")

brea#case 7:

cout--4n9nd of the operation

brea#

default:cout--4n9nter onl* 1 to 7:

5

5while"chX07)

5

OUTPUT:

Q&ST1. Create

58


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60/96

PPPPPPPPPPPPPPP

$ ;1

PPPPPPPPPPPPPPPQ&ST

1. Create

2. &nsert at begin3. &nsert at end

!. &nsert at intermediate

$. Delete at begin%. Delete at end

. Delete at intermediate

. Displa*

7. 98it9nter *our choice:%

%

Q&ST

1. Create2. &nsert at begin

3. &nsert at end!. &nsert at intermediate

$. Delete at begin

%. Delete at end. Delete at intermediate

. Displa*

7. 98it

9nter *our choice:

Cursor space:

Evail 0 ! Qist 0 PPPPPPPPPPPPPP

DETE >9?T

PPPPPPPPPPPPPP 1

PPPPPPPPPPPPPPP

2 2

PPPPPPPPPPPPPPP3 3

PPPPPPPPPPPPPPP

! ;1PPPPPPPPPPPPPPP

;1

PPPPPPPPPPPPPPPQ&ST

1. Create

2. &nsert at begin

3. &nsert at end

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$. Delete at begin


. Displa*

7. 98it9nter *our choice:2

2

9nter the item to be inserted: %%

Q&ST

1. Create



$. Delete at begin


. Displa*7. 98it

9nter *our choice:

Cursor space:

Evail 0 ;1 Qist 0

PPPPPPPPPPPPPP

DETE >9?TPPPPPPPPPPPPPP

!

PPPPPPPPPPPPPPP2 2

PPPPPPPPPPPPPPP

3 3PPPPPPPPPPPPPPP

! ;1

PPPPPPPPPPPPPPP

% 1PPPPPPPPPPPPPPP

Q&ST


3. &nsert at end

!. &nsert at intermediate$. Delete at begin

%. Delete at end


. Displa*

61


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7. 98it

9nter *our choice:

9nter the position: 2

2

Q&ST1. Create

2. &nsert at begin


$. Delete at begin

%. Delete at end

. Delete at intermediate. Displa*

7. 98it

9nter *our choice:

Cursor space:

Evail 0 1 Qist 0 PPPPPPPPPPPPPP

DETE >9?T

PPPPPPPPPPPPPP !

PPPPPPPPPPPPPPP

;1

PPPPPPPPPPPPPPP3 3

PPPPPPPPPPPPPPP

! ;1PPPPPPPPPPPPPPP

% 2

PPPPPPPPPPPPPPPQ&ST

1. Create

2. &nsert at begin


$. Delete at begin


. Displa*

7. 98it9nter *our choice:3

3

9nter the item to be inserted: 7

7

62


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Q&ST

1. Create



$. Delete at begin%. Delete at end


. Displa*7. 98it

9nter *our choice:

Cursor space:Evail 0 ;1 Qist 0

PPPPPPPPPPPPPP

DETE >9?T

PPPPPPPPPPPPPP !

PPPPPPPPPPPPPPP7 ;1

PPPPPPPPPPPPPPP

3 3PPPPPPPPPPPPPPP

! 1

PPPPPPPPPPPPPPP

% 2PPPPPPPPPPPPPPP

Q&ST


3. &nsert at end

!. &nsert at intermediate$. Delete at begin

%. Delete at end


. Displa*7. 98it

9nter *our choice:$

Q&ST1. Create

2. &nsert at begin


$. Delete at begin

%. Delete at end


63


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. Displa*

7. 98it

9nter *our choice:

Cursor space:

Evail 0 ! Qist 0 PPPPPPPPPPPPPP

DETE >9?T

PPPPPPPPPPPPPP 2

PPPPPPPPPPPPPPP

7 ;1

PPPPPPPPPPPPPPP3 3

PPPPPPPPPPPPPPP

! 1

PPPPPPPPPPPPPPP ;1

PPPPPPPPPPPPPPPQ&ST

1. Create



$. Delete at begin


. Displa*

7. 98it9nter *our choice:!

!

9nter the item to be inserted: 2121

9nter the position of the item: 2

2

Q&ST1. Create

2. &nsert at begin


$. Delete at begin


. Displa*

7. 98it

9nter *our choice:

64


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Cursor space:

Evail 0 ;1 Qist 0 PPPPPPPPPPPPPP

DETE >9?T

PPPPPPPPPPPPPP 2

PPPPPPPPPPPPPPP

7 ;1PPPPPPPPPPPPPPP

3 !

PPPPPPPPPPPPPPP

! 1PPPPPPPPPPPPPPP

21 3

PPPPPPPPPPPPPPP

Q&ST1. Create

2. &nsert at begin


$. Delete at begin

%. Delete at end

. Delete at intermediate. Displa*

7. 98it

9nter *our choice:7

RESULT:

Thus' the cursor implementation of list EDT for singl* lin#ed list program has been

written and e8ecuted successfull*.EX.NO:" STAC! ADT# ARRAY IMPLEMENTATION

DATE:

AIM:

To write a C++ rogram to Stac# EDT implementation using arra*

ALGORITHM:

Step 1: Define a stac# siFe.

Step 2: (ead the stac# operation.

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Step 3: (ead the stac# element.

Step !: Chec# the stac# operation is ush or op.

Step $: &f operation is push then chec# the stac# status.i. &f stac# status is over flow we canLt push the element in to stac#.

ii. Atherwise we can add the data into stac# .

iii. Oove top to ne8t position.Step %: Stop the program.

PROGRAM:

,include-iostream.h

,include-conio.h,include-stdlib.h==using namespace std

class stac#

/

int st#I$Jint top

public:

66


67/96

stac#")

/

top0;15

void push"int 8)

/if"top !)

/

cout --stac# over flowreturn

5

st#I++topJ08

cout --inserted --85

void pop")

/

if"top -)/

cout --stac# under flowreturn

5

cout --deleted --st#Itop;;J5

void displa*")

/

if"top-)/

cout -- stac# empt*

return5

for"int i0topi0i;;)

cout --st#IiJ -- 5

5

main")

/Clrscr")

int ch

stac# stwhile"1)

/

cout --4n1.push 2.pop 3.displa* !.e8it4n9nter ur choicecin ch

switch"ch)

/

case 1: cout --enter the element

67


68/96

cin ch

st.push"ch)

brea#case 2: st.pop") brea#

case 3: st.displa*")brea#

case !: e8it")5

5

return ")5

OUTPUT:

1.push

2.pop

3.displa*

!.e8it9nter ur choice2

stac# under flow

1.push

2.pop3.displa*

68


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70/96

RESULT:

Thus' the stac# EDT; arra* implementation program has been written and e8ecuted

Successfull*.

EX.NO:$ STAC! ADT# LIN!ED LIST IMPLEMENTATION

DATE :

AIM:

To rite a C++ rogram to Stac# EDT implementation using lin#ed list

ALGORITHM:

Step 1: create a list.

i) Create a new empt* node top.ii) (ead the stac# element and store it in tops data area.

iii) Essign tops lin# part as >RQQ "i.e. top;lin#0>RQQ).

iv) Essign temp as top "i.e. temp0top).

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Step 2: (ead ne8t stac# operation.

i) &f it is Create then go to step1.

ii) &f it is ush then it process following stepsa) Chec# Oain memor* for node creation.

b) Create a new node top.

c) (ead the stac# element and store it in tops data area.d) Essign tops lin# part as temp "i.e. top;lin#0temp).

e) Essign temp as top "i.e. temp0top).

iii) &f it is pop then it process following stepsa) &f top is >RQQ then displa* stac# is empt*.

b) Atherwise assign top as temp "i.e. top0temp' bring the top to top position)

c) Essign temp as temps lin#. "i.e. temp0temp;lin#' bring the temp to tops previous

position).d) Delete top from memor*.

iv) &f it is traverse then process the following steps

a) Hring the top to stac#Ls top position"i.e. top0temp)

b) (epeat until top becomes >RQQi) Displa* the tops data.

ii) Essign top as tops lin# "top0top;lin#).

PROGRAM:

,include-iostream.h


==using namespace std

class node

/public:

class node


72/96

int tos

public:

stac#")/

tos0;1

5void push"int 8)

/

if "tos - )/

head 0new node

head;ne8t0>RQQ

head;data08tos ++

5

else

/node


73/96

5

5

void pop")/

node RQQ)/

temp0temp;ne8t

5

temp;ne8t0>RQQ5

5main")

/

Clrscr")stac# s1

int ch

while"1)

/cout --4n1.RSB4n2.A4n3.D&SQE@4n!.9?&T4n enter ur choice:

cin ch

switch"ch)/

case 1: cout --4n enter a element

cin chs1.push"ch)

brea#

case 2: s1.pop")brea#

case 3: s1.displa*")brea#

case !: e8it")

55

return ")

5

73


74/96

OUTPUT:

1.RSB2.A

3.D&SQE@

!.9?&T

enter ur choice:1

enter a element23

1.RSB

2.A

3.D&SQE@!.9?&T

enter ru choice:1

enter a element%

74


75/96

1.RSB

2.A3.D&SQE@

!.9?&T

enter ru choice:3

23 %

1.RSB

2.A

3.D&SQE@

!.9?&Tenter ru choice:2

1.RSB

2.A3.D&SQE@


23

1.RSB

2.A

3.D&SQE@!.9?&T

enter ru choice:2

1.RSB

2.A

3.D&SQE@!.9?&T

enter ru choice:2

stac# under flow

1.RSB

2.A3.D&SQE@

!.9?&T

enter ru choice:!

75


76/96

RESULT:

Thus' the stac# EDT; lin#ed list implementation program has been written and e8ecutedSuccessfull*.

EX.NO: 1%.1 &UEUE ADT# ARRAY IMPLEMENTATION

DATE:

AIM:

To write a C++ program to Yueue EDT implementation using arra*

ALGORITHM:

Step 1: &nitialiFe the Vueue variables front 0 and rear 0 ;1

Step 2: (ead the Vueue operation t*pe.

Step 3: Chec# the Vueue operations status.i). &f it is &nsertion then do the following steps

1. Chec# rear - VueuePsiFe is true increment the rear b* one and read the Vueue

element and also displa* Vueue. otherwise displa* the Vueue is full.

2. o to step2.ii). &f it is deletion then do the following steps

1. Chec# rear- front is true then displa* the Vueue is empt*.

2. Oove the elements to one step forward "i.e. move to previous inde8 ).

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3. Decreases the rear value b* one "rear0rear;1).

!. Displa* Vueue

$. o to step2.

PROGRAM:

,include-iostream.h

,include-conio.h

,include-stdlib.h==using namespace std

class Vueue

/

int Vueue1I$Jint rear'front

public:

Vueue")/

rear0;1

front0;15

void insert"int 8)

/

if"rear !)

77


78/96

/

cout --Vueue over flow

front0rear0;1return

5

Vueue1I++rearJ08cout --inserted --8

5

void delet")/

if"front00rear)

/

cout --Vueue under flowreturn

5

cout --deleted --Vueue1I++frontJ

5void displa*")

/if"rear00front)

/

cout -- Vueue empt*return

5

for"int i0front+1i-0reari++)

cout --Vueue1IiJ-- 5

5

main")/

Clrscr")

int chVueue Vu

while"1)

/

cout --4n1.insert 2.delet 3.displa* !.e8it4n9nter ur choicecin ch

switch"ch)

/case 1: cout --enter the element

cin ch

Vu.insert"ch)brea#

case 2: Vu.delet") brea#

case 3: Vu.displa*")brea#

case !: e8it")

78


79/96

5

5

return ")5

OUTPUT:

1.insert

2.delete3.displa*

!.e8it

9nter ur choice1

enter the element21

inserted21

1.insert

2.delete

3.displa*


enter the element22

inserted22

1.insert2.delet

79


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3.displa*

!.e8it

9nter ur choice1

enter the element1%

inserted1%

1.insert

2.delete3.displa*

!.e8it

9nter ur choice3

21 22 1%

1.insert

2.delet3.displa*


deleted21

1.insert

2.delet

3.displa*!.e8it

9nter ur choice3

22 1%

1.insert2.delet

3.displa*

!.e8it

9nter ur choice :!

80


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RESULT:

Thus' the Vueue EDT; arra* implementation program has been written and e8ecutedsuccessfull*.

EX.NO:1%.2 &UEUE ADT# LIN!ED LIST IMPLEMENTATION

DATE:

AIM:

To rite a C++ rogram to Yueue EDT implementation using lin#ed list

ALGORITHM:

Step 1: &nitialiFe the Vueue variables front 0 and rear 0 ;1

Step 2: (ead the Vueue operation t*pe.

Step 3: Chec# the Vueue operations status.i). &f it is &nsertion then do the following steps

3. Chec# rear not eVual to null is true increment the rear b* one and read the Vueue

element and also displa* Vueue. otherwise displa* the Vueue is full.!. o to step2.

ii). &f it is deletion then do the following steps

%. Chec# rear- front is true then displa* the Vueue is empt*.

. Oove the elements to one step forward "i.e. move to previous inde8 ).. Decreases the rear value b* one "rear0rear;1).

7. Displa* Vueue

1. o to step2.

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PROGRAM:


,include-stdlib.h

==using namespace stdclass node

/

public:class node


83/96

if "rare - )

/

head 0new nodehead;ne8t0>RQQ

head;data08

rare ++5

else

/node


84/96

cout --Vueue under flow

return

5if"front 00 rare)

/

front 0 rare 0;1head0>RQQ

return

5front++

head0head;ne8t

5

5main")

/

Clrscr")

Vueue s1int ch

while"1)/

cout --4n1.RSB4n2.A4n3.D&SQE@4n!.9?&T4n enter ru choice:

cin chswitch"ch)

/

case 1:

cout --4n enter a elementcin ch

s1.push"ch) brea#

case 2: s1.pop")brea#case 3: s1.displa*")brea#

case !: e8it")

55

return ")

5

84


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OUTPUT:

1.RSB

2.A

3.D&SQE@!.9?&T

enter ru choice:1

enter a element23

1.RSB

2.A3.D&SQE@

!.9?&T

enter ru choice:1

enter a element$!

1.RSB

2.A

3.D&SQE@!.9?&T

enter ru choice:3

23 $!

85


86/96

1.RSB

2.A3.D&SQE@

!.9?&T

enter ru choice:2

1.RSB

2.A3.D&SQE@

!.9?&T

enter ru choice:2

1.RSB

2.A

3.D&SQE@


Vueue under flow

1.RSB

2.A

3.D&SQE@

!.9?&Tenter ur choice:!

86


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88/96

PROGRAM:

,include-iostream.h


==using namespace std

void insert"int'int )void delte"int)

void displa*"int)

int search"int)int search1"int'int)

int treeI!J't01's'8'i

main")

/clrscr")

int ch'*

for"i01i-!i++)treeIiJ0;1

while"1)

/

cout --1.&>S9(T4n2.D9Q9T94n3.D&SQE@4n!.S9E(CB4n$.9?&T4n9nter *ourchoice:

cin chswitch"ch)

/

case 1:

cout --enter the element to insertcin ch

88


89/96

insert"1'ch)

brea#

case 2:cout --enter the element to delete

cin 8

*0search"1)if"*X0;1) delte"*)

else cout--no such element in tree

brea#case 3:

displa*"1)

cout--4n

for"int i0i-032i++)cout --i

cout --4n

brea#

case !:cout --enter the element to search:

cin 8*0search"1)

if"* 00 ;1) cout --no such element in tree

else cout --8 -- is in --* --positionbrea#

case $:

e8it")

55

5

void insert"int s'int ch )/

int 8

if"t001)/

treeIt++J0ch

return

580search1"s'ch)

if"treeI8Jch)

treeI2


90/96

treeI8J0;1

else if"treeI2


91/96

/

cout --no element in tree

return ;15

if"treeIsJ00;1)

return s=2if"treeIsJ ch)

search1"2S9(T2.D9Q9T9

3.D&SQE@

!.S9E(CB$.9?&T

9nter *our choice:3

no element in tree:123!$%71112131!1$1%111722122232!2$2%222733132

1. &>S9(T2.D9Q9T9

3.D&SQE@

!.S9E(CB$.9?&T

9nter *our choice:1

9nter the element to insert 1

1.&>S9(T

2.D9Q9T9

3.D&SQE@!.S9E(CB

$.9?&T9nter *our choice: !

9nter the element to search: 1

1 is in 1 position

91


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93/96

PROGRAM:

,include-iostream.h

,include-conio.hclass YuiSort

/

int i'6'pivotpublic:

int n'aI2J

void Vuic#"int aIJ'int left'int right)

void swap"int aIJ'int i'int 6)5

void YuiSort :: Vuic#"int aIJ'int first'int last)

/if"first-last)

/

pivot0aIfirstJi0first

60last

while"i-6)

/while"aIiJ-0pivotGGi-last)

i++

while"aI6J0pivotGG6first)

6;;if"i-6)

swap"a'i'6)5

swap"a'first'6)

Vuic#"a'first'6;1)

Vuic#"a'6+1'last)5

93


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5

void YuiSort :: swap"int aIJ'int i'int 6)

/int temp

temp0aIiJ

aIiJ0aI6JaI6J0temp

5

void main")/

YuiSort ob6

clrscr")

cout--4n4nYR&CZ SA(Tcout--4n4n9nter the limit :

cinob6.n

clrscr")

cout--4n4n9nter the element4n4nfor"int i0i-ob6.ni++)

cinob6.aIiJob6.Vuic#"ob6.a''ob6.n;1)

cout--4n4nThe sorted list is 4n4n

for"i0i-ob6.ni++)cout--ob6.aIiJ--

getch")

5

94


95/96

OUTPUT:

9nter the limit: $

9nter the elements$

!3

2

1

The sorted list is1 2 3 ! $

95


96/96

RESULT:

Thus' the Vuic# sort program has been written and e8ecuted successfull*.

ds & oops lab manual.doc

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