S P E C I A L S T U D YS P E C I A L S T U D Y M O D U L EM O D U L E

COMPUTERSCIENCEXII

Manik Kumar PGT (Comp. Sc.)

K V Sonpur

How to use Study Material: I

twillbemuchbeneficialtothosestudentswhoneedspecialcareandattention.Iamsure,thoroughstudy and practicingsimilar patterns of questions ofthismaterialwillhelpsuchstudentssecure60%andabove.

However it isnot 100%substitutefor Textbook . Minimu

masetoffivequestionswithanswersfromeachquestionsoftheBoardPatternQuestionPaperareincluded,keepinginmindtheneedsandinterestoftargetgroup.

Conceptsineveryunithavebeenexplainedusingnotes/solutionstoquestions/guidelinesinasimplelanguage.

Practiceandpeergroupdiscussiononthisstudymaterialwilldefinitelyimprovetheconfidencelevelofthestudentswhentheysolvethequestions.

Nowyouarewelcometothe...content...

Weightagetodifferenttopics/contentunitsS.No

UNIT VSA1 Mark

SAI2Marks

SA II3Marks

LA4Marks

Total

1 Review ofC++ coveredinClassXI 1(1) 8(4) 3(1) 12(6)2 ObjectOrientedProgramminginC++

2(1) 4(1) 6(2)a)IntroductiontoOOP using C++b)Constructor&Destructor 2(1) 2(1)c)Inheritance 4(1) 4(1)

3 Data Structure &Pointers

2(1)3(1)

4(1)

3(1)a)Address Calculationb)StaticAllocation ofObjects 3(1) 5(2)c)DynamicAllocationofObjects 4(1)d)Infix&Postfix Expressions 2(1) 2(1)

4 DataFileHandlinginC++a) FundamentalsofFileHandlingb) TextFilec) BinaryFiles

1(1)2(1)

3(1)

1(1)2(1)3(1)

5 DatabasesandSQL2(1)

4(1)2(1)a)Database Concepts

b)StructuredQueryLanguage 2(1) 6(2)6 BooleanAlgebra

1(1)2(1)

3(1)

2(1)a)IntroductiontoBooleanAlgebra &Lawsb)SOP&POS 1(1)c)KarnaughMap 3(1)d)BasicLogicGates 2(1) 2(1)

7 Communication&OpenSource Concepts2(2)

4(1)2(2)a)IntroductiontoNetworking

b)Media,Devices,Topologies&Protocols 4(1)c)Security 2(2) 2(2)d)Webservers 1(1) 1(1)e)OpenSourceTerminologies 1(1) 1(1)

TOTAL 9(9) 26(13) 15(5) 20(5) 70(32)

Review of C++covered in Class XI

Questions based on HeaderFilesVery Short AnswerQuestions (1 mark)

Q1.Writethe namesofthe headerfilestowhichthefollowingbelong:

math.hQ2. Nametheheaderfile(s)thatshallbeneededforsuccessfulcompilationofthefollowingC++

code:voidmain(){

charsubject[30];strcpy(subject,”ComputerScience”);puts(subject);

}Ans: string.h

stdio.h

Q3. Nametheheaderfile(s)thatshallbeneededforsuccessfulcompilationofthefollowingC++code:voidmain(){

charname[20];gets(name);cout<<setw(20)<<name;

}Ans: iomanip.h

stdio.h

Q4.Nametheheaderfile(s)thatshallbeneededforsuccessfulcompilationofthefollowingC++code:voidmain(){

}Ans: ctype.h

stdio.h

chara,b;a=getchar();b= toupper(a)cout<<”\nThe uppercasecharacter of“<<a<<”is“<<b;

Note:Marksarenottobedeductedifanyadditionalheaderfileismentioned.Butthese headerfilesmustbewritten.

(i)isdigit()(vi)setw()

(ii)strcmp()(vii)exit()

(iii)fabs()(viii)tolower()

(iv)gets()(ix)ceil()

(v)eof()(x)feof()

(xi)strupr()(xvi)strstr()

(xii)atoi()(xvii)put()

(xiii)setprecision()(xviii)puts()

(xiv)floor()(xix)exp()

(xv)remove()(xx)free()

(xxi)fwrite() (xxii)write() (xxiii) setiosflags() (xxiv)sin() (xxv)abs()

Ans: (i)ctype.h(vi)iomanip.h

(ii)string.h(vii)process.h

(iii)math.h(viii)ctype.h

(iv)stdio.h(ix)math.h

(v)iostream.h(x)stdio.h

(xi)string.h(xvi)string.h

(xii)stdlib.h(xvii)iostream.h

(xiii)iomanip.h(xviii)stdio.h

(xiv)math.h(xix)math.h

(xv)stdio.h(xx)stdlib.h

(xxi)stdio.h (xxii)iostream.h (xxiii)iomanip.h (xxiv)math.h (xxv)stdlib.h/

Concept Questionsbased on C++Review (2 marks)

Q1. Whatisthedifference betweena keywordandanidentifierinC++? Giveexamplesof both.Ans:

Keywordisaspecialwordthathasaspecialmeaningandpurpose.Keywordsarereservedandarefew.Forexample:goto,for,while,if,elseetc.Identifiersaretheuser-definednamegiventoapartofaprogram.Identifiersarenotreserved.Itshouldbethename ofanykeyword.Forexample:name,stud, _myfile,opetc.

Q2.Whatisa referencevariable?Whatisitsusage?Ans:

Areferencevariableisanaliasnameforapreviouslydefinedvariable.Theusageofitisthatthesamedataobjectcanbereferredtobytwonamesandthesenamescanbeusedinterchangeably.

Q3. Writetwoadvantagesofusinginclude compilerdirectives.Ans:(i)The#includecompilerdirectiveletsusincludedesiredheaderfilesinourprogramwhichenables us

workwithalldeclaration/definitions/macrosinsidetheincludedheaderfile(s).(ii) Itsupportsmodularity.

Q4. Differentiate betweenaLogical ErrorandSyntaxError.Alsogivesuitableexamplesofeachin C++.

Ans: LogicalError:Error occurredduetoincorrectlogicappliedbythe programmer.SyntaxError:Erroroccurreddueto notfollowingthe proper grammar/syntaxofthelanguageOR the erroroccurredduetoviolating rules ofthe programminglanguageExample://Programtofindarea andperimeterofrectanglevoidmain(){

intA=10,B=20,AR, P;AR=2*(A*B); //LogicalError–WrongFormulaP=2*(A+B);cout<<A<<P>>endl; //SyntaxError–Useof>>withcout

}

Q5.Whatisthedifference betweenGlobalVariableandLocalVariable?Ans:

#include<iostream.h>floatNUM=900; //NUMisa globalvariablevoidLOCAL(intT){

intTotal=0;

//Totalisalocalvariablefor(intI=0;I<T;I++)Total+=I

;cout<<NUM+Total;}voidmain(){

Global Variable LocalVariable I

tisavariable,whichisdeclaredoutsideallthefunctions

Itisaccessiblethroughoutthepro

Itisavariable,whichisdeclaredwithinafunctionorwithinacompoundstatement

Itisaccessible onl

LOCAL(45); }

Q6.Whatisthedifference betweenObjectOrientedProgramming andProceduralProgramming?Ans:ObjectOrientedProgramming ProceduralProgramming EmphasisonData FollowsBottom-

Upapproachinprogramdesign Data hidingfeature

preventsaccidentalchangeindata Featureslike data

encapsulation

Emphasisondoingthings (functions) FollowsTop-

downapproachinprogramdesign

Presence ofGlobalvariablesincreasechancesofaccidentalchangeindataQ7. DifferentiatebetweenaCallbyValueandCallbyReference,givingsuitableexamplesofeach?

Ans:CallbyValue CallbyReference Th

ecalledfunctioncreatesitsowncopiesofthe original valuessent toit.

Anychangesthataremadeinthefunctionrun,changesinthe

Thecalledfunctionaccessesandworkswiththe originalvalues usingtheirreferences.

Anychangesthatoccur inthefunction run,changesinthe originalvaluesarereflected.

void change(intb){

b=10;}voidmain(){

inta=5;cout<<”\na=“<<a;change(a);cout<<”\na=“<<a;

}Outputwillbe:

voidchange(int&b){

b=10;}voidmain(){

inta=5;cout<<”\na=“<<a;change(a);cout<<”\na=“<<a;

}Outputwillbe:

Q8.Whatisaparameter?Differentiatebetweenanactualandaformalparameterwithanexample?Ans:Parameteristhevariable/

valuepassedtoafunctionorthevariablethatisusedastheincomingvaluesinafunction.Thevariables/valuespassedtoafunctionarecalledactualparameters.Thevariablesthatareusedastheincomingvaluesinafunctionarecalledformalparameters.For Example:

void change(intb) //bistheformalparameter{

b=10;}voidmain(){

inta=5;change(a);

//aistheactualparametercout<<”\na=“<<a;

}

Q9. Enlistanyfourjump statementswith theiruses.Ans:(i)goto: A gotostatementcantransferthe programcontrolanywhereintheprogram.

(ii) break:Abreakstatementenablesaprogramtoterminateoftheloop/block,skipping anycode inbetween.

(iii) continue:Abreakstatementenablesaprogramtoforcethenextiterationtotakeplace,skipping anycode inbetween.

(iv) return:Areturnstatementisusedtoreturnfromafunction.

Q10.How arethefollowing relatedtoone another?(i) arrayandstructure (ii)structure andclass

Ans:(i)Arrayisagroupofitemsofthesamedatatypeswhereasstructurebringstogetheragroupofrelated dataitems ofanydatatypes.

(ii) Structureisactuallyaclass(inC++)declaredwithkeywordstruct.Bydefault,allmembersarepublicinastructure;ontheotherhandallmembersareprivatebydefaultinaclass.

Questions based on ProgramErrors (2marks)Note

:Errorsareexplainedinthesolutions.Whilesolvingsuchtypeofquestion,astudentshouldcarefullygothroughcorrectnessofeachstatementandthelogicofthewholeprogram.Students areadvised topracticevariousquestionstodevelop theskilloffinding theerrors.

Q1. Rewrite thefollowing programafterremoving thesyntactical errors(ifany).Underlineeachcorrection.#include<iostream.h>intfunc(inty=10,int&x){

if(x%y=0)return++x;elsereturny--;}voidmain(){

intp=20, q=23;r= func(p,q);cout>>p>>q>>r;

}Ans: #include<iostream.h>

intfunc(int y , int&x) // violating theruleofDefaultargument{

if(x%y= = 0)return++x;elsereturny--; //==relationaloperator}voidmain(){

intp=20, q=23;intr= func(p,q); //rshouldbedeclaredcout << p << q << r ; //<<operator forcout

}Q2..Rewritethefollowingprogramafterremovingthesyntacticalerrors(ifany).Underlineeachcorrection.

#include<iostream.h>voidmain(){

intX[]={60,50,30,40},Y; count=4;cin>>Y;for(i=count-1;i>=0;i--)switch(i){

case1;

case2:cout<<Y *X;break;case3:cout<<Y+Y;

}}

Ans:#include<iostream.h>voidmain(){

intX[]={60,50,30,40},Y,count=4;/

/multipledeclarationseparatedbycommacin>>Y;for(int i =count-1; i>=0;i--) //ishouldbe declaredswitch(i){

case 1:; /

/caseshouldfollowby:case2:cout<<Y*X[i];break; //LvaluerequiredforXcase3:cout<<Y+ Y;

}}

Q4. Rewritethefollowingprogramafterremovingthesyntacticalerrors(ifany).Underlineeachcorrection.#include<iostream.h>str

uctPixels{intColor,Style;}voidShowPoint(PixelsP){cout<<P.Color,P.Style<<endl;}voidmain(){

PixelsPoint1=(5,3);ShowPoint(Point1);PixelsPoint2=Point1;Color.Point1+=2;ShowPoint(Point2);

}Ans:

#include<iostream.h>structPixels{ intColor,Style;}; //Definition ofstructurePixelsmustbeendedwith;voidShowPoint(PixelsP){ cout<<P.Color<<P.Style<<endl;} // In cascadingofcout,<<tobeusedvoidmain(){

PixelsPoint1={5,3}; //{}tobeusedtoinitialiseofmembersofthe objectShowPoint(Point1);PixelsPoint2=Point1;Point1.Color+=2;

//membertofollowedbytheobjectusing.operatorShowPoint(Point2);

}Question

s based on Finding Outputsusing random() ShortAnswerQuestions(2ma

rks)Note:random(n)generatesthe numbersrandomlyfrom0 ton–1.Forexample:random(20)generates

randomlyfrom0to19.Explanationsaregivenattheendofeachsolution.

Q1. Inthefollowingprogram,ifthevalueofNgivenbytheuseris50,whatmaximumandminimum values the programcouldpossiblydisplay?#include<iostream.h>#include<stdlib.h>voidmain(){

intN,Guessme;randomize();cin>>N;Guessme=random(N)+5;cout<<Guessme<<endl;

}

Ans: Minimum: 5 Maximum:54Explanation:Sincerandom(50)givesanumberrandomlyfrom0to49.Ifitreturns0i.e.minimumforrandom(50),theminimumvalueforGuessmewillbe0+5=5.Ifitreturns49i.e.maximumforrandom(50),themaximumvalueforGuessmewillbe49+5=54.

Q2.Studythefollowingprogramandselect the possibleoutputfrom it:#include<iostream.h>#include<stdlib.h>constintMax=3;voidmain(){

randomize();intNumber;Number=50+random(Max);for(intP=Number; P>=50;P--)

cout<<P<<”#”;cout<<endl;

}(i) 53#52#51#50#(ii) 50#51#52#(iii) 50#51#(iv)51#50#

Ans:51#50#

Q3.Inthefollowingprogram,ifthevalueofNgivenbytheuseris20,whatmaximumandminimum values the programcouldpossiblydisplay?#include<iostream.h>#include<stdlib.h>voidmain(){

intN,Guessnum;randomize();cin>>N;

Explanation:Sincerandom(3)givesanumberrandomlyfrom0to2.So,thevaluesofNumberrangesfrom50(50+0)to52(50+2)andtheNumbershouldbedisplayedindescendingorderaccordingtotheprogramme.53isnotpossible,sothe answeris(iv).

Guessnum=random(N – 10)+10;cout<<Guessnum<<endl;

}Ans: MaximumValue:19 Minimum Value:10

Q4.

Inthefollowingprogram,ifthevalueofGuessenteredbytheuseris65,whatwillbetheexpectedoutput(s) fromthefollowing options(i),(ii),(iii)and(iv)?#include<iostream.h>#include<stdlib.h>voidmain(){

intGuess;randomize();cin>>Guess;for(intI=1;I<=4;I++){

New=Guess+random(I);cout<<(char)New;

}}

(i) ABBC(ii) ACBA(iii) BCDA(iv) CAB

DAns:(i)ABBC

Questions based on Finding Outputs ShortAnswerQuestions(2marks)

Note:Whilesolvingsuchtypeofquestion,astudentshouldcarefullygothroughthelogicofthewholeprogramandconceptsusedinit.Variouspatternsofquestionsaregivenbelow.Students areadvised topracticevariousquestionstodevelop theskilloffinding theoutput.

Explanation:InfirstiterationI=1andrandom(I)gives 0.So,New =65+0,henceoutputisA.InfirstiterationI=1,So,New=65+0,henceoutputisA.InseconditerationI=2andrandom(I)gives0or1.So,New=65or66,henceoutputisAorB.InthirditerationI=3andrandom(I)gives0to2.So,New=65or66or67,henceoutputisAorBorC.InfourthiterationI=4andrandom(I)gives0to3.So,New =65or66or67or68,henceoutputis AorBorCorD.Option(i)satisfiesallthe above only,soitisthe answer.

Q1. Findthe output ofthe

followingprogram:#include<iostream.h>structGame{

charMagic[20];intScore;};voidmain(){

Game M={“Tiger”,500};char*Choice;Choice=M.Magic;Choice[4]=’P’;Choice[2]=’L’;M.Score+=50;cout<<M.Magic<<M.Score<<endl;GameN=M;N.Magic[0]=’A’;N.Magic[3]=’J’;N.Score-=120;cout<<N.Magic<<N.Score<<endl;

}Ans: TiLeP550Q2. Find the output ofthe

followingprogram:#include<iostream.h>voidmain(){

intNumbers[]={2,4,8,10};int*ptr =Numbers;for(intC =0;C<3;C++){

cout<<*ptr<<“@”;ptr++;

}cout<<endl;for(C =0;C<4;C++){

(*ptr)*=2;--ptr;

}for(C =0;C<4;C++)

cout<<Numbers[C]<< “#”;cout<<endl;

}Ans : 2@4@8@

4#8#16#20#

Q3. Find the output ofthe followingprogram:

#include<iostream.h>voidfun(intp, int*q){

p= *(q)+= 2;}voidmain(){

intx,b=5, a[2]={10,20};for(x =1;x <3; x++)

{fun(a[x –1],&b);

cout<<“\n”<<a[x -1]<<”:“<<b;}

}

Ans: 10:720:9

Q4.Find the outputofthe followingprogram:#include <iostream.h>voidmain(){

longNUM=98534210;intf=0,s=0;do{

intrem= NUM %10;if(rem%2== 0)

f+=rem;else

s += rem;NUM /=10;

}while(NUM >0);cout<<“\n”<<f<<”–“<<s<<“= ”<<f– s;

}

Ans : 14 – 18= –4

Q5. Find the output ofthe followingprogram:

#include<iostream.h>inta=10;voidmain(){

voiddemo(int&,int, int*);inta=20,b=5;demo(::a,a,&b);cout<<”\n”<<::a<<”*”<<a<<”*”<<b;

}void demo(int&x,inty,int*z){

a=a+x; y=y*a;*z=a+y;cout<<”\n”<<x<<”*”<<y<<”*”<<*z;

}Ans: 20*400*420

20*20*420Q6.Find the outputofthe followingprogram:#include <iostream.h>#include<ctype.h>voidmain(){

char*name= "ThEbEStmeNwIN";for(inti=0;name[i]!='\0' ;i++){

if(islower(name[i]) )name[i]= toupper(name[i]);

elseif

(isupper(name[i]))if(i%2== 0)

name[i]––;else

name[i]= tolower(name[i–1]);}cout<<name;

Questions based on Finding Outputs Short Answer Questions (3 marks)

Q1. Findthe outputofthe followingprogram:#include<iostream.h>#include<string.h>structKEY{

charword[10];intcount;

};

voidchangekeyword(KEYsomekey);voidmain(){

KEYaKEY;strcpy(aKEY.word,“#define”);aKEY.count=10;cout<<aKEY.word<<“\t”<<aKEY.count<< “\n”;changekeyword(aKEY);cout<<aKEY.word<<“\t”<<aKEY.count<< “\n”;

}voidchangekeyword(KEYsomekey){

strcpy(somekey.word,“const”);somekey.count+= 1;cout<<somekey.word<<“\t”<<somekey.count<<“\n”;

}

Ans : #define 10#const11#define 10

Q2. Findthe outputofthefollowingprogram:#include<iostream.h>intmodify(inttemp=2){

if(temp %3== 0)temp =temp +1;

elsetemp =temp +3;

returntemp;}void doupdation(intm,int&n){

staticinti;i++;m=n+i;if(n>10)

n=modify();else

n=modify(n);cout<<m<<"; "<<n<<endl;

}voidmain(){

intx =8,y=20;doupdation(x,y);cout<<x<<";"<<y<<endl;doupdation(y,x);cout<<x<<";"<<y<<endl;doupdation(y,x);cout<<x<<";"<<y<<endl;

}Ans: 21;5

8;510;11

Q3. Find the output of thefollowingprogram:

#include <iostream.h>voidchang (int&x,inty){

inttemp;temp=x;x=y;y=temp;cout<<”\n”<<x<<” :“<<y;

}voidmain(){

inta=10,b;b=a++ +++a +++a +a+++2;cout<<”\n”<<a<<” :“<<b;chang(a,b);cout<<”\n”<<a++ <<” :”<<++b;

}

Ans : 14:5050:14

11;514;55;5

Q4.Givethe outputofthe followingprogram:#include<iostream.h>voidmain(){

inta,*b,**c,***d;a=12,b=&a,c=&b,d=&c;***d*=5;cout<<”\n”<<a<<”,”<<a+*b;(**c)+= 10;cout<<”\n”<<**c +***d;(***d)+=10;cout<<”\n”<<a+*b;

}Ans: 60, 120

140160

Q5. Givethe outputofthe followingprogram:#include<iostream.h>#include<string.h>#include<ctype.h>voidfunnystr(char*s,int n=2){

inti=n;while(i< strlen(s)){

}i=0;

s[i]='-';i=i+n;

while(s[i]!='\0'){

if(s[i]> 'A'&&s[i]<'P')s[i]=tolower(s[i]);

elseif(s[i]>'a'&&s[i]<'p'){

if(i%3 == 0)s[i]=tolower(s[i-1]);

} i++;}

}

elses[i]=tolower(s[i]);

voidmain(){

charstr[]="MiCroSoFT";funnystr(str,3);cout<<str;

}

Ans : mic–oS–fT

Introduction to OOP using C++ Objec

t Oriented Programming ConceptsShortAnswerQuestions

Q1. Defineobject.Ans.Objectisanidentifiableentitywithsome characteristicsandbehaviour.

Q2. Defineclass.Ans.A classisgroupofobjectsthatsharecommonproperties andrelationships.

Q3.Definedataabstraction.Ans.Abstractionreferstothe actofrepresentingessentialfeatureswithoutincludingthebackground details

orexplanations.

Q4.Whatis inheritance?Ans.Inheritancei

sthecapabilitiesofoneclassofthingstoinheritcapabilitiesorpropertiesfromanotherclass.

Q5.Definemodularity.Ans

.Modularityisthepropertyofasystemthathasbeendecomposedintoasetofcohesiveandlooselycoupledmodules.

Q6.Defineencapsulation.Ans.Thewrappingupofdataandoperations/

functions(thatoperateonthedata)intoasingleunit(calledclass )isknownasencapsulation .

Q7.Whatispolymorphism?Ans.Polymorphism isthe abilityforamessageordatatobeprocessedin morethanoneform.

Q8.HowpolymorphismisimplementedinC++?Ans.PolymorphismisimplementedinC++throughoverloadedfunctions,overloadedoperatorsandvirtual

functions.

Q9.HowinheritanceisimplementedinC++?Ans.InheritanceisimplementedinC++byspecifyingthebaseclassfromwhichthederivedclassofthe base

classisdefined.

Q10.Whatisdata hiding?HowitisimplementedinC++?Ans.Dat

ahidingisapropertywherebytheinternaldatastructureofanobjectishiddenfromtherestoftheprogram.DatahidingisimplementedinC++byprivateandprotectedmembersofclass.

Q11.How data abstraction canbeimplemented inC++?Ans.Dataabstractioncan beimplemented inC++byusing publicmembersofclass.

Q12.Whatisprogrammingparadigm?Ans.A programmingparadigmdefinesthe methodology ofdesigningandimplementingprogramsusing

the keyfeaturesandbuildingblocksofa programminglanguage.

Q13.Whatisproceduralprogrammingparadigm?Ans.Inproceduralprogrammingparadigm,dataissharedamongallthefunctions.Itsemphasisisondoing

thingsratherthanusing the data.

SOMEMPORTANTQUESTIONSTOREFRESHTHECONCEPT

Q14.Whatarethe advantagesofOOP?Ans:TheadvantagesofOOPareReusabilityofcode,Easeof

comprehension,easeoffabricationandmaintenance,easeofredesignandextension.Q15.Whatdoyouunderstandbyfunctionaloverloading?GiveanexampleillustratingitsuseinaC++

program.Ans:Afunctionnamehavingseveraldefinitionsthataredifferentiableby

thenumberortypesoftheirargumentsisknownasfunctionaloverloading.Forexample:floatvolume(floata){returna*a*a;}floatvolume(floata,floatb){returna*a*b;}floatvolume(floata,floatb,floatc){returna*b*c;}

CLASSES AND OBJECTS

Q1. Write fourattributesassociatedwithdeclarationofclasses.An

s.TheattributesassociatedwithdeclarationofclassesareDataMembers,MemberFunctions,ProgramaccesslevelsandClassTagname.

Q2. Definedatamembers andmemberfunctions.Ans.Datamembers areexactlylikethevariablesina structure.TheyarenormallymadeMemberfunction

isfunction definedwithina classthatactson thedatamembersinthe class.Theyarenormallymade public.

Q3. Writethe scoperules of members ofa class.Ans.Scope rules of membersofa classare:-

Private Thesemembers canbe accessed onlybythememberfunctions&friendfunctionsoftheclass.ItcannotbeinProtected Thesemembers canbe accessed onlybythememberfunctions&friendfunctionsoftheclass.ItcanbeinheriPublic Thesemembers canbe directlyaccessedbyanyfunction.

Q4. How manytypesoffunctions are usedinaclass?Ans.1.Accessorfunction2.Mutatorfunctions3.Managerfunctions

Q5. Whenwillyoumakeafunctioninline?Ans.Afunctionismadeinlinewhenitisverysmallanddoesnotreturnanyvalue.Alsoitdoesnot

containanyloopand staticvariables.Q6. Whatisscope resolutionoperator?Ans

.Thisoperator(::)isusedinsituationswhereaglobalvariableexistswiththesamenameasalocalvariable.Thisisalsousedwhenthememberfunctionsaredeclaredoutsidetheclassdefinition;thefunctionnameisprecededbytheclassnameandthebinaryscoperesolutionoperator(::).

Q7. Definestatic datamembers.Ans

.Staticdatamembersareinitializedtozerowhenthefirstobjectofitsclassiscreated.Nootherinitializationispermitted.Onlyonecopyofthatmemberiscreatedfortheentireclassandissharedbyallthe objectsforthatclass.

Q8. Howismemoryallocated toaclass anditsobjects?Ans.Whena classisdefined,memoryisallocatedforitsmemberfunctionsandtheyare stored

inthememory.When objectiscreated,separate memoryspaceisallocatedforitsdatamembers.Allobjects workwiththesamecopyof

memberfunctionsharedbyall.

BOARDPATTERNQUESTIONS:LongAnswer Questions (4marks)

Q1.Definea classemployeewiththefollowingspecifications;Privat

emembersofclassemployee:empno integerename 20charactersbasic,hra,da floatnetpay floatcalculate() Afunctiontocalculate basic+hra+dawithfloatreturn type

Publicmemberfunctionsofclassemployee:havedata() functiontoacceptvaluesforempno,sname,basic,hra,daandinvokecalculate() tocalculatenetpaydispdata() functiontodisplayallthedata members on thescreen .

Ans.classemployee{

intempno;charename[20];floatbasic,da,hra;floatnetpay;floatcalculate(){

return (basic+da+hra);}

public:voidhavedata(){

cout<<”Enteremployeeno”;cin>> empno;cout<<”\nEnter name“;cin>>ename;cout<<”\nEnter basicsalary”;cin>> basic;cout<<”\nEnter DA“;cin>>da;cout<<“\nEnterHRA“;cin>>hra;netpay= calculate();

}voiddispdata(){

cout<<“\nEmployeenois“<<empno<< endl;cout<<“\nNameis“<< ename << endl;cout<<“\nBasic Salaryis”<< basic << endl;cout<<“\nDAis“<<endl ;cout<<“\nHRA is“<<hra<<endl;cout<<“\nNetpayis“<<netpay<<endl;

}};

Q2. Definea classStudentforthefollowingspecifications:

PrivatemembersoftheStudentare:roll_nointegername arrayofcharactersofsize 20

class _starrayofcharactersofsize8marks arrayof integers ofsize 5percentagefloatcalculatethatcalculatesthe percentagemarks

Publicmembersofthe Studentare:readmarksreadsmarkandinvoke the calculate functiondisplaymarks prints thedata.

Ans.classStudent{

introll_no;charname[20];charclass_st[8];intmarks[5];floatpercentage;voidcalculate();public:

voidreadmarks();voiddisplaymarks();

};voidStudent::calculate(){

floattotal=0;for(intI=0; I<5; I+

+)total+=marks[I];percentage =total/5;

}voidStudent::readmarks(){

cout<<“\nEnter5marks:“;for(intI=0; I<5; I++)

cin>>marks[I];calculate();

}voidStudent::displaymarks(){

cout<<“\nThe5marksare:“;for(intI=0;I<5; I++)

cout<<marks[I]<< endl;cout<<“\nPercentage=“<<percentage << endl;

}Q3. Definea classDONORwiththefollowingspecifications :

Private:Donornumber integerName 20charactersBloodgroup 2characters

Public:Input() AfunctiontoacceptalltheinformationOutput() A functiontodisplayalltheinformationCheckgroup() Afunctionwith

char*returntoreturnBloodGroupDefineboththenumberfunctionswiththeirgiven description.

Ans.class DONOR{ intd_no;

charname[20];charb_grp[2];

public:voidinput(){

cout<<“\nEnterthe detailsasdonorno,nameandbloodgroup:“;cin>>d_no; gets (name );gets(b_grp);

}voidoutput( ){ cout<<“\nThedetailsare“;

cout<<“\nName :“<<name<<“\nDonorNo:<<d_no;cout<<“\nBloodGroup:“<< b_grp;

}char*checkgroup (){ return( b_grp);}

};Q4. Defineaclass TESTinC++ with

followingdescription:Private Members TestCodeoftypeinteger Descriptionoftypestring NoCandidateoftypeinteger CenterReqd(number ofcentersrequired)oftypeinteger AmemberfunctionCALCNTR()tocalculate

andreturnthenumberofcentersas(NoCandidates/100+1)Public Members

• A functionSCHEDULE()toallow usertoentervaluesforTestCode,Description,NoCandidate&callfunctionCALCNTR()tocalculatethenumberofCentres

• A functionDISPTEST()toallow usertoview thecontentofallthe data membersAns:class TEST{

public:

};

intTestCode;charDescription[20];intNoCandidate,CenterReqd;voidCALCNTR();

voidSCHEDULE();voidDISPTEST();

void TEST::CALCNTR(){

CenterReqd=NoCandidate/100+1;}voidTEST::SCHEDULE(){

cout<<”TestCode :”;cin>>TestCode;cout<<”Description :”;gets(Description);cout<<”Number :”;cin>>NoCandidate;CALCNTR();

}void TEST::DISPTEST(){

cout<<”TestCode

:”<<TestCode<<endl;cout<<”Description :”<<Description<<endl;cout<<”Number :”<<NoCandidate<<endl;;cout<<”Centres :”<<CenterReqd<<endl;;

}

SOMEMPORTANTQUESTIONSTOREFRESHTHECONCEPTCONSTRUCTORANDDESTRUCTOR

Q1.Whatisconstructor?Ans.Amemberfunctionwiththesamenameasitsclassiscalledconstructoranditisusedtoinitialize the

objectsofthatclass typewithalegalinitialvalue.Q2.Whatisdestructor?Ans

.Adestructorisamemberfunctionhavingsamenameasitsclassbutprecededby~signanditdeinitialisesanobjectbefore itgoesoutofscope.

Q3.Whataredifferenttypes ofconstructors?Ans

.ThedifferenttypesofconstructorsareDefaultconstructor,ParameterizedconstructorandCopyconstructor.

Q4. Whatis defaultconstructor?Ans.Aconstructorthatacceptsnoparameteriscalledthedefaultconstructor.Q5. Whatisparameterizedconstructor?Ans.Aconstructorthatacceptsparametersforitsinvocationisknownasparameterizedconstructor.Q6. Whatiscopyconstructor?Ans.Acopyconstructorisaconstructorthatdefinesandinitializesanobjectwithanotherobject.Ittakes

theformclassname(classname).Foracopyconstructortheremustbeadefaultconstructororaparameterizedconstructor.

BOARDPATTERNQUESTIONS-Short AnswerQuestions (2 marks)

Note: Twooptionswith thefollowing patternaregenerallyaskedinthe BoardExam.Variousoptionalquestionsaregivenbelow.

Q7.Answerthefollowing questionsaftergoing throughthefollowingclass:class Seminar{

public:

};

intTime;

Seminar(); //Function 1voidLecture() //Function 2{cout<<”Lecturesinthe seminaron”<<end1;}Seminar(int); //Function 3Seminar(Seminar&abc); //Function 4~Seminar() //Function 5{ cout<<”Voteofthanks”<<end1;

(i) InObjectOrientedProgramming,whatisFunction5referredasandwhendoesitgetinvoked/called?

Ans:Function5isreferredasdestructoranditisinvokedassoonasthescopeoftheobjectgetsover.(ii) InObjectOriente

dProgramming,whichconceptisillustratedbyFunction1,Function3andFunction4 alltogether?Ans: ConstructorOverloading (Polymorphism)

(iii) Whichcategoryofconstructor-Function1belongsto?WriteanexampleillustratingthecallsforFunction1.

Ans:DefaultConstructor.Exampletoinvokefunction1 SeminarS;(iv) Whichcategoryofconstructor-Function3belongsto?

WriteanexampleillustratingthecallsforFunction3.Ans: ParameterisedConstructor.Example toinvokefunction3 SeminarA(8);(v) Whichcategoryofconstructor-Function4belongsto?

WriteanexampleillustratingthecallsforFunction4.Ans:CopyConstructor.Exampletoinvoke function4 SeminarS2(S);Or

SeminarS2=S;(vi) Write an exampleillustrating thecallsforFunction3explicitly.Ans: SeminarA= Seminar(8);(vii) Write an exampleillustrating thecallsforFunction4explicitly.Ans: SeminarS2= Seminar(S);(viii) Write thecomplete definitionforFunction1 toinitializeTime as30.Ans: Seminar::Seminar()

{Time=30;}(ix) Writethe complete definitionforFunction3 toinitializeTimewith Mytime

asparametertotheFunction3.Ans: Seminar::Seminar(intMytime)

{Time= Mytime;}(x) Writethe complete definitionforFunction4.Ans: Seminar::Seminar(Seminar&abc)

{Time=abc.Time;}Q8.Answerthefollowing questionsaftergoing throughthefollowingclass:class

Complex{

public:

};

int x;inty;

Complex(); //Function 1voiddisp() //Function 2{cout<<”The Complexnumberis:“<<x<<” +“<<y<<”i”<<end1;}Complex(int,int); //Function 3Complex(Complex&abc); //Function 4

(i) Whichcategoryofconstructor-Function1belongsto?WriteanexampleillustratingthecallsforFunction1.

Ans:DefaultConstructor.Exampletoinvokefunction1 ComplexC;(ii) Whichcategoryofconstructor-Function3belongsto?

WriteanexampleillustratingthecallsforFunction3.Ans:Parameterised Constructor.Exampletoinvoke function3 Complex C(6,8);(iii) Whichcategoryofconstructor-Function4belongsto?

WriteanexampleillustratingthecallsforFunction4.Ans:CopyConstructor.Exampletoinvoke function4 Complex C2(C);Or

Complex C2=C;(iv) Write anexampleillustrating thecallsforFunction3explicitly.Ans: ComplexC =Complex(6,8);(v) Write anexampleillustrating thecallsforFunction4explicitly.Ans: ComplexC2=Complex(C);(vi) Write the complete definitionforFunction1 toinitializex as 10andyas 20.Ans: Complex::Complex()

SOMEMPORTANTQUESTIONSTOREFRESHTHECONCEPT

{x =10;y=20;}(vii) Write the complete definitionforFunction3 toinitializethedatamembers

withpandqasparametersto the Function3.Ans: Complex::Complex(intp,intq)

{x =p;y=q;}(viii) Write the complete definitionforFunction4.Ans: Complex::Complex(Complex&abc)

{x =abc.x; y=abc.y;

}

Inheritance

Q1. Write the reasonsbehind theintroductionof inheritanceinOOP.Ans.ThemajorreasonsbehindtheintroductionofinheritanceinOOPare:

(i)Itensurestheclosenesswiththerealworldmodels,(ii)ideaofreusability,(iii)transitivenatureofinheritance.

Q2.Whatarethedifferentforms of inheritance?Ans.Thedifferentformsofinheritanceare(i)SingleInheritance,(ii)MultipleInheritance,(iii)Hierarchical

Inheritance,(iv)Multilevel Inheritance,(v)HybridInheritance.Q3

.Howdoestheaccessofinheritedmembersdependupontheiraccessspecifiersandthevisibilitymodesofthe baseclass?

Ans.Accessspecifier

Visibilitymodepublicinheritance protectedinh

eritanceprivate inheritance

publicmemberin base class publicinderivedclass protectedinderivedclass

privateinderivedclass

protectedmemberinbaseclass

protectedinderivedclass

protectedinderivedclass

privateinderivedclass

privatememberinbaseclass

hiddeninderived class hiddeninderivedclass

hiddeninderivedclass

Q4. Write thedifferentways ofaccessibilityof base classmembers.Ans.

AccessSpecifier Accessiblefromown class

Accessiblefromderivedclass (Inheritable)

Accessiblefromobjectsoutsideclasspublic Yes Yes Yes

protected Yes Yes Noprivate Yes No No

Q5.Howisthe size ofaderivedclassobjectcalculated?Ans.Thesizeofaderivedclassobjectisequaltothesumofsizesofdatamembersinbaseclassandthe

derivedclass.Q6

.Inwhatorderareclassconstructorsandclassdestructorscalledwhenaderivedclassobjectiscreatedordestroyed?

Ans.Whentheobjectofaderivedclassiscreated,firstlytheconstructorofthebaseclassisinvokedandthen,the constructorofthe derivedclassisinvoked.Onthe otherhand,whenthe objectofa derivedclassis destroyed,firstlythedestructorofthederivedclassis invokedandthen,the destructorofthebaseclassisinvoked.

Note: Observethefollowing questionsandtheir solutions.Theabove concepts are implemented.Practice ofsuchquestionswilldefinitelyclearyourdoubtsandimproveyourconfidencelevel.Discussionwithyourfriends anddistinguishthe varietiesofthese questions.

BOARDPATTERNQUESTIONSLongAnswerQuestions(4Marks)

Q1.Answerthe questions(i)to(iv) basedonthefollowing:

classbus:privateheavyvehicleclassvehicle{

protected:in

twheels;intpa

ssenger;

classheavyvehicle: protectedvehicle{

intdiesel_petrol;protected:

intload;

{

public:cha

rmake[20];voidfetchdata(

);voiddisplaydata();

public:voidinputdata();voidoutputdata();

};

public: };voidreaddata(int,int);voidwritedata();

};

i) Write themember(s)thatcanbeaccessedfromtheobjectofbus.ii) Write the data member(s)thatcanbeaccessedfromthe functiondisplaydata().iii) How manybytes are requiredbyanobjectofbusandheavyvehicle classesrespectively?iv) Is thememberfunctionoutputdata( )accessible tothe objectsofthe classheavyvehicle?

Ans : (i)fetchdata(),displaydata() (ii)make,load,passanger(iii)forthe objectof bus –28bytes,forthe objectofheavyvehicle–8bytes (iv)No

Q2.Answerthe questions(i)to(iv) basedonthefollowing:classlivingbeing{

charspecification[20];intaverageage;

public:voidread();voidshow();

};

classape:privatelivingbeing{

intno_of_organs;intno_of_bones;

protected:intiq_level;

public:voidreadape();voidshowape();

classhuman:publicape{

charrace[20];charhabitation[30];

public:voidreadhuman();voidshowhuman();

};

};

(i) Write thememberswhichcanbeaccessedfromthememberfunctionsofclasshuman.(ii) Write themembers,whichcanbeaccessedbyanobjectofclass human.(iii) Whatisthe size ofan object(inbytes)ofclasshuman?(iv) Write theclass(es)whichobjectscanaccessread()declaredinlivingbeingclass.

Ans : (i)race,habitation,iq_level,readhuman(),showhuman(),readape(),showape()(ii) readhuman(),showhuman(),readape(), showape()(iii) 78bytes (iv)livingbeing

Q3.Answerthe questions(i)to(iv) basedonthefollowing:classparent{

charname[20];protected:

intson;public:

voidinputdata();voidoutputdata();

};

classfather: protectedparent{

intdaughter;protected:

intbaby;public:

voidreaddata();voidwritedata();

};

classmother:publicfather{

intchild;public:

voidfetchdata();voiddispdata();

};

(i)In caseoftheclassfather,whatis thebaseclassoffatherandwhatis the derivedclassof father?(ii) Write thedatamember(s)thatcanbeaccessedfromfunctiondispdata().(iii) Write thememberfunction(s),whichcanbeaccessedbyanobjectofmotherclass.(iv) Isthememberfunction outputdata()accessible to the objectsoffatherclass?

Ans : (i)base classof father–parent,derivedclass offather–mother

(ii) child,baby,son(iii) fetchdata(),dispdata(), readdata(),writedata() (iv)No

Data Structure & Pointers Address Calculation

ShortAnswerQuestions(3marks)FormulaeofRowMajor&ColumnMajorareusedinthe givenquestions.Kindlygothroughit.Q1.Anarrayx[30]

[10]isstoredinthememorywitheachelementrequiring4bytesofstorage.Ifthebaseaddressofxis4500,findoutmemorylocationsofx[12][8]andx[2][4],ifthecontentisstoredalongtherow.

Ans:Given, W=4, N =30, M =10, Base(x)RowMajorFormula:

=4500

Loc(x[12][8])=Base(x)+W*(M*I+J) Loc(x[2][4]) =Base(x)+W*(M*I+J)=4500+4*(10 *12+8)=4500+4*128=4500+512=5012

=4500+4*(10*2+4)=4500+4*24=4500+96=4596

Q2.AnarrayP[20][30]isstoredinthememoryalongthecolumnwitheachoftheelementoccupying4bytes,findouttheBaseAddressofthearray,ifanelementP[2][20]isstoredatthe memorylocation5000.

Ans:Given,W=4,N=20,M=30, Loc(P[2][20])=5000ColumnMajorFormula:

Loc(P[I][J])=Base(P)+W*(N*J+I)Loc(P[2][20])=Base(P)+4*(20*20+2)Base(P)=5000– 4*(400+2)

=5000–1608=3392

Q3.AnarrayARR[5][25]isstoredinthememorywitheachelementoccupying4bytesofspace.AssumingthebaseaddressofARRtobe1000,computetheaddressofARR[5][7],whenthearrayisstoredas :(i)Row wise (ii) Columnwise.

Ans:(i) Rowwise:Given, W=4, N=5,M

=25,Base(ARR)=1000RowMajorFormula:

Loc(ARR[5][7])=Base(ARR)+W*(M*I+J)=1000+4*(25*5+7)=1000+4*132=1000+528=1528

(ii) Columnwise :Given, W=4, N=5,M

=25,Base(ARR)=1000ColumnMajorFormula:

Loc(ARR[5][7])=Base(ARR)+W*(N*J+I)=1000+4*(5*7+5)=1000+4*40=1000+160=1160

Static Allocation ofObjects ShortAnswerQuestions(2marks)Note

:Practicethewaytowritethefunctiondefinitionwherearrayanditssizearepassedasarguments.Practicewiththeconceptsofaccessingtheelementsoftheone/twodimensionalarray.Applythesuitablelogictosolvethe givenproblemandwrite the codingof it.

Q1. WriteafunctioninC++ tofindsumofrowsfromatwodimensionalarray.Ans:voidMatAdd(intA[][],intN,intM){

for(intR=0;R<N;R++){

intSumR=0;for(intC=0;C<M;C++)

SumR+=A[C][R];cout<<SumR<<endl;

}}

Q2. WriteafunctioninC++tofindthesumofbothleftandrightdiagonalelementsfromatwodimensionalarray(matrix).

Ans: voidDiagSum(intA[][ ], intN){

intSumD1=0,SumD2=0;for(intI=0;I<N;I++){

SumD1+=A[I][I];SumD2+=A[N-I-1][I];

}cout<<”SumofDiagonal1:”<<SumD1<<endl;cout<<”SumofDiagonal2:”<<SumD2<<endl;

}Q3.WriteafunctioninC+

+whichacceptsanintegerarrayanditssizeasargumentsandreplaceselementshavingevenvalueswithitshalfandelementshavingoddvalueswithtwiceitsvalue.eg:

ifthearraycontains: 3,4,5,16,9thenthefunctionshouldberearrangedas6,2,10,8, 18Ans:

voidcalc(intx[], intm){

for(inti=0;i<m;++i){

if(x[i]%2==0)x[i]=x[i]/2;

else

}}

x[i]=x[i]*2;

Q4.WriteauserdefinedfunctioninC++whichintakes onedimensionalarrayandsizeof arrayasargumentanddisplaytheelementswhichareprime.If1Darrayis 10, 2,3, 4,5, 16,17,23Thenprime numbersinabove arrayare:2,3,

5,17,23Ans:voidprimedisp(intx[],intm){

intnf;for(inti=0;i<m;++i){

nf=0;for(intj=1;j<=x[i];j++)

if(x[i]%j== 0)nf++;

if(nf== 2)cout<<”\n”<<x[i];

}}

Q5.WriteafunctioninC++whichacceptsanintegerarrayanditssizeasarguments/parametersand exchanges the valuesatalternate locations.example:ifthearrayis8,10,1,3,17,90,13,60thenrearrange thearrayas 10,8,3,1,90,17,60,13Ans:

voidexchange(intx[], intm){

inty;for(inti=0;i<m;i=i+2){

if(m-i== 1) //ifm(no.ofelement)isoddlastelementi=m;//shouldnotalterits position.

else{

}}

}

`

y=x[i];x[i]=x[i+1];x[i+1]=y;

Short Answer Questions (3 marks) Q1.Write a user defined function DispTen(int A[][4],int N,int M) In c++ to find and display all the numbers, which are divisible by 10. For example if the content of array is:12 20 132 10 30The output should be20 10 30

Ans: void DispTen(int A[ ][4], int N, int M){int i, j ;for(i=0 ; i<N ; i++){for(j=0 ; i<M ; j++){if(A[i][j]%10= =0)cout<<A[i][j]<<” “ ;} } }

Q2Write a function ALTERNATE (int A[ ] [3], int N, int M) in C++ to display all alternate elements from two-dimensional array A (staring from A [0] [0]). For example:If the array is containing:

23 54 7637 19 2862 13 19The output will be23 76 19 62 19Ans:void ALTERNATE (int A [] [3], int N, int M)

{int T=0;for (int I=0 ; I<N; I++)for (int J=0 ; J<M ; J++){if (T%2= =0)cout<<A[I] [J]<<" ";T++ ;}}

Q3. Write a DSUMO function in C++ to find sum of Diagonal Elements from aNxN Matrix. (Assuming that the N is a odd number)Ans.

void DSUM (int A[] [100], int N){int SUMR =0, SUML=0;for (int i=0; i<N; i++){SUMR = SUMR + A[i] [i] ;SUML = SUML + A[i] [N-1-i] ;}cout<< "Sum of Right Diagonal Elements = "<<SUMR<<end1;cout<< "Sum of Left Diagonal Elements = "<<SUML<<end1;}

Q5.Define a function SWAPCOL( ) in C++ to swap (interchange) the first column elements with the last column elements, for a two dimensional integer array passed as the argument of the function.

Example: If the two dimensional array contains2 1 4 91 3 7 75 8 6 37 2 1 2After swapping of the content of 1st column and last column, it should be:9 1 4 27 3 7 13 8 6 52 2 1 7

Ans:void SWAPCOL(int A[ ][100], int M, int N){int Temp, I;for (I=0;I<M;I++){Temp = A[I][0];A[I][0] = A[I][N-1];A[I][N-1] = Temp;}}

Q6.WriteafunctioninC++whichacceptsanintegerarrayanditssizeasargumentsandassignthe elementsinto atwodimensionalarrayofintegersinthefollowingformatIfthearrayis1,2,3,4,5,6 ifthearrayis1,2,3Theresultant2Darrayis Theresultant2D arrayis123456 123012345 012001234 001000123000012000001

Ans://Logic:Conditionforputtingthevalueisthe position (i<=j) of2DarrayotherwiseputzerovoidChange2Darray(intx[],intsize){

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

intk=0;for(intj=0;j<size;j++){

if(i<=j){

}else

y[i][j]=x[k];k++;

y[i][j]=0;}

}for(i=0;i<size;i++){

cout<<”\n”;for(intj=0;j<size;j++)

cout<<y[i][j]<<"";}

}Q7.WriteafunctioninC+

+whichacceptsanintegerarrayofdoubledimensionalwithitssizeasargumentsanddisplaysthetotalnumbersofodd,evenandprimenumbersinthearray.Example : ifthefollowingintegerarraywillbepassed tothefunction,i.e.

Thenthe outputshouldbe: Thetotaloddnumbers are:13Thetotaloddnumbers are:12Thetotaloddnumbers are:10

Ans: voidnumcheck(intarr[][ ],intm,intn){

inti,j, oddt=0, event=0,primet=0,nf,k;for(i=0;i<m;i++){

for(j=0;j<n;j++){

if(arr[i][j]%2==0)event++;

else

nf=0;oddt++;

for(k=1; k<=arr[i][j]; k++){

if(arr[i][j]%k==0)nf++;

}if(nf==2)

primet++;}

}cout<<”\nThetotaloddnumbers are:“<<oddt;cout<<”\nThetotalevennumbersare:“<<event;cout<<”\nThetotalprimenumbers are:“<<primet;

}

6 4 13 19 57 3 8 11 519 12 23 4 621 29 18 9 1028 5 12 2 6

Dynamic Allocation of Objects LongAnswerQuestions (4marks )Note

:InsertionatthebeginningofLinkedList(PushoperationinStack),InsertionattheendofLinkedList(InsertioninaQueue),DeletionfromthebeginningofLinkedList(PopoperationinStackaswellasDeletionofnodeinaQueue)areinthesyllabus.So,onlythelogicofthesethreefunctionsshouldbepracticed.Thelogicandthewaytosolvethesefunctionsaregivenbelow.Practicethem.

1. Write a function QDELETE( ) in c++ to perform delete operation on a Linked Queue, which contains passenger no and passenger name. Consider the following definition of node in the codestruct node{long int Pno ;char Pname[20];node *Link ;}

Ans. class Queue{NODE *Front, *Rear;public:Queue ( ) {Front = NULL; Rear = NULL; }void QueAdd ( );void QueDel ( );void QueDis ( );~Queue();} ;void Queue: :QueDel ( ){if (Front!=NULL){NODE *Temp=Front;cout<<Front->Pno<< " " ;cout<<Front->Pname<< "Deleted";Front=Front->Link;delete Temp;if (Front==NULL) Rear=NULL;}elsecout<<"Underflow ! Queue is empty. .";}Infix& PostfixExpressions Short Answer Questions (2 marks)

Q1. Evaluatethefollowingpostfixnotationofexpression:25 8 3 – / 6 * 10 +Ans:

Operator Scanned StackContent25 258 25,83 25,8,3- 25,5/ 56 5,6* 3010 30,10+ 40

So, theansweris40.

(Infix to Postfix) Q2.ConvertA+(B*C–(D/E))*Fintopostfixformshowingstackstatusaftereverystep.Ans:

StepNo. SymbolScanned Stack Expression1 A ( A2 + (+ A3 ( (+( A4 B (+( AB5 * (+(* AB6 C (+(* ABC7 – (+(- ABC *8 ( (+(-( ABC *9 D (+(-( ABC *D10 / (+(-(/ ABC *D11 E (+(-( ABC *DE12 ) (+( ABC *DE/–13 ) ( ABC *DE/– +14 * (* ABC *DE/– +15 F (* ABC *DE/– +F16 ) ABC *DE/– +F*

So, thepostfixform is: ABC *DE/–+F*Q3.ConvertNOTAOR NOTB ANDNOTCintopostfixform.Ans:

StepNo. SymbolScanned Stack Expression1 (2 NOT (NOT3 A (NOT A4 OR (OR ANOT5 NOT (ORNOT ANOT6 B (ORNOT ANOTB7 AND (ORAND ANOTBNOT8 NOT (ORANDNOT ANOTBNOT9 C (ORAND ANOTBNOTC NOT10 ) ANOTBNOTC NOTANDOR

So, thepostfixform is: ANOTBNOTC NOTAND OR

Q4. Evaluate the following POSTFIX notation. Show status of Stack after every step of evaluation (i.e. after each operator): True, False, NOT, AND, False, True, OR, AND

Ans. Element Scanned Stack StatusTrue TrueFalse True, FalseNOT True, TrueAND TrueFalse True, FalseTrue True, False, TrueOR True, TrueAND TrueFinal Answer: True

SOMEMPORTANTPOINTSTOREFRESHTHECONCEPT

DATA FILEHANDLING

1.TypeofLink StreamClass Example DeclarationFile-to-Memory ifstream ifstreamfin;Memory-to-File ofstream ofstreamfout;File-to-Memory/Memory-to-File fstream fstreamfinout;

2. Syntaxofopen()-<streamtype>.open(<data file>,<filemodes>);Example:-

fin.open(“MARK.DAT”,ios::in |ios::binary);fout.open(“MARK.DAT”,ios::out|ios::app|ios::binary);

3. Differentfilemodesare : ios::in,ios::out,ios::ate,ios::app,ios::trunc,ios::binary,ios::nocreate,ios::noreplace.

4. Syntaxofread()-<streamtype>.read((char*)&buffer,sizeof(buffer));Example : fin.read((char *)&stud, sizeof(stud)); //studisanobject/class

5. Syntaxofwrite()-<streamtype>.write((char*)&buffer,sizeof(buffer));Example : fout.write((char*)&stud,sizeof(stud));//studisanobject/class

6. DetectingEOF:Example :-fin.eof()7. Commonforms offilepointerfunctions:

(a) seekg()–<istreamobject>.seekg(long,seek_dir);Example :-fin.seekg(30,ios::beg);

(b) seekp()–<ostreamobject>.seekp(long,seek_dir);Example :-fout.seekp(–1*sizeof(stud),ios::cur);

(c) tellg()/tellp()returns thepositioninbytesofget_pointer/put_pointer.

(d) Differentdefinitionsofseek_dirareios::beg(referstobeginningofthe file),ios::cur(referstocurrentpositioninthefile),ios::end(refers toendofthefile).

8. Closingafile:stream_object.close();Example :fin.close();Very Short AnswerQuestions (1 mark) Q1. ObservetheprogramsegmentgivenbelowcarefullyandfilltheblanksmarkedasStatement1and

Statement2 usingseekg()/seekp()functionsforperforming therequired task.

#include<fstream.h>classItem{

intIno;charItem[20];public://Functionto search anddisplaythe contentofa particularrecordvoidSearch(int);//Functionto modifythecontentofaparticularrecordnumbervoidModify(int);

};voidItem::Search(intRecNo){

fstreamFile;File.open(“STOCK.DAT”,ios::binary|ios::in);

//Statement1File.read((char*)this,sizeof(Item));cout<<Ino<<”==>”<<Item<<endl;File.close();

}voidItem::Modify(intRecNo){

fstreamFile;File.open(“STOCK.DAT”,ios::binary|ios::in|ios::out);cout>>Ino;cin.getline(Item,20);

//Statement2File.write((char*)this,sizeof(Item));File.close();

}Ans : Statement1 – File.seekg((RecNo–1)*sizeof(Item),ios::beg);Statement2–

File.seekp((RecNo–1)* sizeof(Item),ios::beg);

Q2.ObservetheprogramsegmentgivenbelowcarefullyandfilltheblanksmarkedasStatement1andStatement2usingseekg()andtellg()functionsforperformingtherequired task.#include <fstream.h>class Employee{intEno;charEname[20];pub

lic://Functionto countthetotalnumber ofrecordsintCountrec();};intItem::Countrec(){

fstreamFile;File.open(“EMP.DAT”,ios::binary|ios::in);

//Statement1

intBytes = //Statement2

intCount=Bytes/sizeof(Item);File.close();returnCount;

}Ans: Statement1–

File.seekg(0,ios::end);Statement2–File.tellg();

Q3.Afilenamedas“STUDENT.DAT”containsthestudentrecords,i.e.objectsofclassstudent.AssumingthatthefileisjustopenedthroughtheobjectFILEoffstreamclass,intherequiredfilemode,writethecommandtopositiontheputpointertopointtothebeginningofthesecondrecordfromthelastrecord.

Ans :FILE.seekp(–2* sizeof(student),ios::end);

Q4. A filenamedas“EMPLOYEE.DAT”contains the studentrecords,i.e.objectsofclassemployee.Assumingthat thefileisjustopenedthroughtheobjectFILEof fstreamclass,intherequiredFilemode,writethecommand topositionthegetpointertopoint toeighthrecordfromthebeginning.

Ans :File.seekg(8,ios::beg);

Short AnswerQuestions ( 2 marks) Q1. WriteafunctioninC++ tocountthenumberofalphabetspresentinatextfile“DATA.TXT”.Ans:

void count( ){

ifstreamfile(“DATA.TXT”);charch;inttotal=0;while(file.get(ch)){

if(isalpha(ch))total++;

}cout<<”\nTheTotalnoofalphabetsare”<<total;file.close();

}

Q2. Write afunctioninC++ tocount thenumberof vowelspresentina textfile“STORY.TXT”.

Ans: void vowelcount(){

ifstreamfile(“STORY.TXT”);int n=0;while(file.get(ch)){

if(ch==’a’||ch==’A’||ch==’e’||ch==’E’||ch==’i’||ch==’I’||ch==’o’||ch==’O’||ch==’u’||ch==’U’)

n++ ;}cout<<”\nTotalno. ofvowelsare”<<n;file.close();}

Q3. Write afunctioninC++ tocount thenumberofwords ina textfileNOTES.TXT.Ans:void wordcount( ){

ifstreamf1(“NOTES.TXT”);charch;inttot=0;f1.seekg(0);while(f1.get(ch)){

if(ch==’ ‘||ch==’.’||ch==’?’)tot++;}cout<<”\ntotalno. ofwordsare“<<tot;f1.close();

}Q4. Write a functioninC++ tocount thenumberof lines presentina text file“STORY.TXT”.Ans: void CountLine()

{ifstreamFIL(“STORY.TXT”);intLINES=0;charSTR[80];while(FIL.getline(STR,80)

)LINES++;cout<<”No.ofLines:”<<LINES<<endl;FIL.close();

}

Q5. Write a function CountYouMe( ) in c++ which reads the contents of a text file story.txt and counts the words You and Me (not case senitive)For example, if the file containsYou are my best friend.You and me make a good team.The function should display the output asCount for you: 2Count for me: 1Ans: void COUNTYouMe ( )

{ifstream Fil ("story.txt") ;char STR [10] ;int countyou = 0 ;int countme = 0;while (!Fil.eof ( )){Fil>>STR;if (strcmpi (STR, "Me") = =0 )countme++ ;

if(strcmpi (STR, "You") = =0)countyou++;}cout<<"Count for You : "<<countyou<<end1;cout<<"Count for Me : "<<countme<<end1;Fil.close( ) ; //Ignore

}Q6. Write a function in C++ to read the content of a text file "DELHI.TXT' and display all those lines on screen, which are either starting with 'D' or starting with 'M'.Ans. void DispDorM()

{if stream File("DELHI.TXT");char Str[80];while(File.getline(Str,80)){if(Str[0]='D' || Str[0]='M')cout«Str«endl;}File.close(); //Ignore}

Short Answer Questions (3 marks)

Q1.WriteafunctioninC++toread&displaythedetailsofallthememberswhosemembershiptypeis‘L’or‘M’fromabinaryfile“CLUB.DAT”.AssumethebinaryfilecontainsobjectofclassCLUB.class CLUB{

intmno;charmname[20];chartype;

public:voidregister ();voiddis();charwhattype(){ return type;}

};

Ans: voidshow(){

ifstream file;CLUBc1;chars;file.open(“CLUB.DAT”,ios::in|ios::binary);while(file.read((char*)&c1,sizeof(c1))){

s = c1.whattype();if((s==’L’)|| (s==’M’))

c1.dis();}File.close();

}

Q2. GivenabinaryfilePHONE.DATcontainingrecordsofthefollowingstructuretype.class Phonlist{

public:

charadd;charName[20];chararea[5];

voidregister();voidshow();intcheck(charac[]){

returnstrcmp(area,ac)}

};Q3. Write afunctionTRANSFER()inC++,thatwould copyallrecordswhicharehavingarea as”DEL”fromPHONE.DATto PHONBACK.DAT.

Ans: void transfer(){

ifstreamFin;OfstreamFout;Phonlistph;Fin.open(“PHONE.DAT”,ios::in|ios::binary);Fout.open(“PHONBACK.DAT”,ios::out |ios::binary);while(Fin.read((char*)&ph,sizeof(ph))){ if(ph.check(“DEL”)== 0)

Fout.write((char*)&ph,sizeof(ph));}Fin.close();Fout.close();

}

Q4.WriteafunctioninC++tomodifythenameofastudentwhoserollnumberisenteredduringtheexecutionoftheprogram.Searchtheacceptedrollnumberfromabinaryfile“STUDENT.DAT”andmodify thename,assumingthebinaryfileiscontainingtheobjectsofthefollowing class.class STUD{

public:

intRno;charName[20];

intsrno(){returnRno;}voidEnter(){gets(Name);}voidDisplay(){cout<<Rno<<Name<<endl;}};

Ans: void{

Modifyname()fstreamFile;File.open(“STUDENT.DAT”,ios::binary|ios::in|ios::out);STUDs;intmrno;cout<<”\nEntertheroll numbertomodify:“;

if(s.srno(){cout<<”nModifythes.Enter();File.seekp(sizeof(s),ios::cur);File.write((char&s,e.close() ; }

Databases andSQL ShortAnswerQuestions Q1.Whatisadatabasesystem?Whatarethe advantagesprovidedbyadatabase system?Ans:

Adatabaseisacollectionofinterrelateddataandadatabasesystemisbasicallyacomputerbasedrecordkeepingsystem.Theadvantagesprovidedbyadatabasesystemare(i)reduceddataredundancy(ii)Controlleddatainconsistency(iii)Shareddata(iv)Standardizeddata(i) Secureddata(vi)Integrateddata.

Q2.Whatarethe variouslevels ofdatabase abstractioninadatabasesystem?Ans:Th

evariouslevelsofdatabaseabstractioninadatabasesystemare(i)Internal(Physical)Level(ii)ConceptualLevel(iii)External(View)Level.

Q3.WhatisDataIndependence?WhatarethelevelsofDataIndependence?Ans:DataIndependencei

stheabilitytomodifyaschemedefinitioninonelevelwithoutaffectingaschemedefinitioninthenexthigherlevel.ThetwolevelsofDataIndependenceare(i)PhysicalDataIndependence(ii)LogicalDataIndependence.

Q4.Whatisa view? How isituseful?Ans:

Aviewisavirtualtablethatdoesnotreallyexistinitsownrightbutisinsteadderivedfromoneormoreunderlyingbasetable(s).Itprovidesanexcellentwaytogivepeopleaccesstosome butnotalloftheinformationina table.

Q5.Whatisrelation?Define therelationaldatamodel?Ans:

Arelationisatablehavingatomicvalues,uniquerowsandunorderedrowsandcolumns.Therelationaldatamodelrepresentsdataandrelationshipsamongdatabyacollectionoftablesknownasrelation,eachofwhichhas anumber ofcolumnswith uniquenames.

Q6.Define:(i)Tuple(ii)Attribute (iii)Degree(iv)Cardinality.Ans:

(i)Tuple–Therowsoftables(relations)arereferredastuples.(ii) Attribute – Thecolumns oftables(relations)arereferredasattributes.

(iii) Degree–The numberofattributesina relation(table)iscalleddegree ofa relation.(iv) Cardinality–Thenumber oftuplesina

relation(table)iscalledcardinalityofarelation.Q7.WhatdoyouunderstandbyPrimaryKey&CandidateKeys?Ans:Anattributeorsetattributeswhichareusedtoidentify

atupleuniquelyisknownasPrimaryKey.Ifatablehasmorethanonesuchattributeswhichidentifyatupleuniquely thanallsuchattributesareknownasCandidate Keys.

Q8.Definealternatekey.Whatdoyouunderstandbyforeign key?Ans:Acandidatekeythatisnottheprimarykeyiscalledanalternatekey.Anon-

keyattribute,whosevaluesarederivedfromtheprimarykeyofsomeothertable,isknownasforeignkeyin its currenttable.

Q9.WhatisData DefinitionLanguage? Giveexamplesofsome DDLcommands.Ans:Th

eSQLDataDefinitionLanguageprovidescommandsfordefiningrelationschemas,deletingrelations,creatingindexesandmodifyingrelationschemas.TheexamplesofsomeDDLcommandsareCREATETABLE,ALTERTABLE,DROPTABLE,DROPVIEWcommandsetc.

Q10.WhatisData ManipulationLanguage? Giveexamplesofsome DMLcommands.Ans

:TheSQLDataManipulationLanguageincludesaquerylanguagetoinsert,deleteandmodifytuplesinthedatabase.TheexamplesofsomeDMLcommandsareSELECT,INSERT,DELETE,UPDATE commands etc.

Q11.WhatisSQL?Write the variousprocessingcapabilitiesofSQL.Ans:SQ

Lisalanguagethatenablesyoutocreateandoperateonrelationaldatabases.ThevariousprocessingcapabilitiesofSQLaredatadefinitionlanguage,datamanipulationlanguage,view definition,authorization,integrityandtransaction control.

SYNTAXof some SQLCOMMANDS : 1. SELEC

TcolumnlistFROM<tablename>WHERE<condition>GROUPBY<columnname(s)>HAVING<searchcondition>ORDER BY column name;

2. INSERTINTO<table name>[<columnlist>]VALUES(<value>,<value>,...);3. DELETE FROM<table name>[WHERE<predicate>];4. UPDATE<tablename>SET<columnname>=<newvalue>[WHERE<predicate>];5. CREAT

ETABLE<tablename>(<columnname><datatype>[(size)]<columnconstraint>,<columnname><datatype>[(size)]<columnconstraint>,...<tableconstraint> (<columnname>[<columnname>...])...);

6. CREATEVIEW<viewname>[(<columnname>,<columnname>,...)]AS<SELECTcommand>;

7. ALTERTABLE<tablename>ADD/MODIFY<columnname><datatype><size>;8. DROPTABLE<table name>;9. DROPVIEW<viewname>;

LongAnswer Questions

Q1.ConsiderthefollowingtablesACTIVITYandCOACH.Write SQLcommandsforthestatements(i)to(iv)andgive outputs forSQLqueries(v)to(viii).

Table:ACTIVITYACode ActivityName ParticipantsN

umPrizeMoney ScheduleDate

1001 Relay100x4 16 10000 23-Jan-20041002 Highjump 10 12000 12-Dec-20031003 ShotPut 12 8000 14-Feb-20041005 LongJump 12 9000 01-Jan-20041008 DiscussThrow 10 15000 19-Mar-2004

Table:COACHPCode Name ACode1 AhmadHussain 10012 Ravinder 10083 Janila 10014 Naaz 1003

(i) TodisplaythenameofallactivitieswiththeirAcodesindescendingorder.(ii) Todispla

ysumofPrizeMoneyforeachoftheNumberofparticipantsgroupings(asshownincolumnParticipantsNum10,12,16).

(iii) Todisplaythecoach’snameandACodesinascendingorderofACodefromthetableCOACH.

(iv)Todisplaythe contentofthe GAMEStablewhose ScheduleDateearlierthan01/01/2004inascendingorderofParticipantNum.

(v) SELECTCOUNT(DISTINCT ParticipantsNum)FROM ACTIVITY;(vi) SELECTMAX(ScheduleDate),MIN(ScheduleDate)FROM ACTIVITY;(vii) SELECTSUM(PrizeMoney)FROMACTIVITY;(viii) SELECTDISTINCTParticipantNumFROMCOACH;

Ans:(i) SELECTActivityName,ACode FROMACTIVITYORDERBYAcode DESC;(ii) SELECT SUM(PrizeMoney),ParticipantsNum FROM ACTIVITY GROUP

BYParticipantsNum;(iii) SELECTName,ACodeFROM COACHORDER BY ACode;(iv) SELECT*FROMACTIVITYWHEREScheduleDate<’01-Jan-

2004’ORDERBYParticipantsNum;(v) 3(vi)19-Mar-2004 12-Dec-2003(vii) 54000(viii) 1

6 10 12

Q2. Considerthe followingtablesGAMESandPLAYER.WriteSQLcommandsforthestatements(i)to(iv)andgive outputs forSQLqueries(v)to(viii).

Table:GAMESGCode GameName Number PrizeMoney ScheduleDate101 CaromBoard 2 5000 23-Jan-2004102 Badminton 2 12000 12-Dec-2003103 Table Tennis 4 8000 14-Feb-2004105 Chess 2 9000 01-Jan-2004108 LawnTennis 4 25000 19-Mar-2004

Table:PLAYERPCode Name Gcode1 NabiAhmad 101

2 RaviSahai 1083 Jatin 1014 Nazneen 103

(i) TodisplaythenameofallGames withtheirGcodes.(ii) Todisplaydetailsofthosegames whicharehavingPrizeMoneymorethan7000.(iii)Todisplaythe contentofthe GAMEStableinascendingorderofScheduleDate.(iv) TodisplaysumofPrizeMoneyforeachoftheNumber ofparticipation

groupings(asshownincolumnNumber2or4).(v) SELECTCOUNT(DISTINCT Number)FROMGAMES;(vi) SELECTMAX(ScheduleDate),MIN(ScheduleDate)FROM GAMES;(vii) SELECTSUM(PrizeMoney)FROMGAMES;(viii) SELECTDISTINCTGcodeFROM PLAYER;

Ans:(i) SELECTGameName,Gcode FROM GAMES;(ii) SELECT* FROMGAMESWHEREPrizeMoney>7000;(iii) SELECT*FROMGAMESORDERBY ScheduleDate;(iv) SELECTSUM(PrizeMoney),NumberFROMGAMESGROUPBYNumber;(v) 2(vi)19-Mar-2004 12-Dec-2003(vii) 59000(viii) 1

01 103 108

Q3.ConsiderthefollowingtablesHOSPITAL.GiveoutputsforSQLqueries (i)to(iv)andwriteSQLcommandsfor the statements(v)to(viii).

No Name Age Department Dateofadmin Charge Sex1 Arpit 62 Surgery 21/01/06 300 M2 Zayana 18 ENT 12/12/05 250 F3 Kareem 68 Orthopedic 19/02/06 450 M4 Abhilash 26 Surgery 24/11/06 300 M5 Dhanya 24 ENT 20/10/06 350 F6 Siju 23 Cardiology 10/10/06 800 M7 Ankita 16 ENT 13/04/06 100 F8 Divya 20 Cardiology 10/11/06 500 F9 Nidhin 25 Orthopedic 12/05/06 700 M10 Hari 28 Surgery 19/03/06 450 M

(i) SelectSUM(Charge)fromHOSPITALwhere Sex=’F’;(ii) SelectCOUNT(DISTINCTDepartment) fromHOSPITAL;(iii) SelectSUM(Charge)fromHOSPITALgroup byDepartment;(iv) SelectNamefromHOSPITALwhereSex=’F’ANDAge>20;(v) To showallinformationaboutthepatients whosenamesarehavingfourcharactersonly.(vi) Toreduce Rs 200fromthe charge of femalepatientswhoareinCardiologydepartment.(vii) Toinsertanew rowinthe above tablewith

thefollowingdata:11,‘Rakesh’,45,‘ENT’,{08/08/08},1200,‘M’

(viii) Toremovetherows fromtheabovetable where ageofthepatient>60.

Ans:(i) 1200(ii)4(iii)1050

70011501300

(iv) Dhanya(v) SELECT *FROMHOSPITALWHERENAMELIKE“____”;(vi) UPDATEHOSPITALSETCHARGE=CHARGE–

200WHERE(DEPARTMENT=‘CARDIOLOGY’ ANDSEX=‘F’);(vii) INSERTINTO HOSPITALVALUES(11,‘Rakesh’,45,‘ENT’,{08/08/08},1200,‘M’);(viii) DELETEFROMHOSPITALWHEREAGE>60;

Q4.Considerthefollowingtables BOOKS.WriteSQLcommandsforthe statements(i)to(iv)andgiveoutputs forSQLqueries(v)to(viii).

Table :BOOKSB_Id Book_Name Author_Name Publisher Price Type QuantityC01 FastCook Lata Kapoor EPB 355 Cookery 5F01 TheTears WilliamHopkins First 650 Fiction 20T01 MyC++ Brain&Brooke FPB 350 Text 10T02 C++ Brain A.W.Rossaine TDH 350 Text 15F02 Thuderbolts Anna Roberts First 750 Fiction 50

i). TolistthenamesfrombooksofTexttype.ii). Todisplaythenamesandpricefrombooks inascendingorderoftheirprice.iii).

ToincreasethepriceofallbooksofEPB publishers by50.iv). Todisplaythe Book_Name,QuantityandPriceforallC++ books.v).

Selectmax(price)frombooks;vi). Selectcount(DISTINCTPublishers)frombooks wherePrice >=400;vii). SelectBook_Name,Author_Namefrombooks where Publishers =‘First’;viii). Selectmin(Price)frombooks wheretype=‘Text’;

Ans: (i)SELECTBook_Name FROMBOOKSWHEREType=‘Text’;(ii) SELECTBook_Name,Price FROMBOOKSORDERBY Price;(iii) UPDATEBOOKSSETPrice=Price +50WHEREPublisher=‘EPB’;(iv) SELEC

TBook_Name,Quantity,PriceFROMBOOKSWHEREBook_NameLIKE‘%C++%’;(v)750(vi) 2(vii) TheTears

WilliamHopkinsThuderboltsAnna Roberts

(viii) 350

Q5.ConsiderthetablesITEMS&COMPANY.WriteSQLcommandsforthe statements(i)to(iv)and give theoutputsforSQLqueries(v)to(viii).

Table:ITEMSID PNAME PRICE MDATE QTYT001 Soap 12.00 11/03/2007 200T002 Paste 39.50 23/12/2006 55

T003 Deodorant 125.00 12/06/2007 46T004 HairOil 28.75 25/09/2007 325T005 ColdCream 66.00 09/10/2007 144T006 ToothBrush 25.00 17/02/2006 455

Table: COMPANYID COMP CityT001 HLL MumbaiT008 Colgate DelhiT003 HLL MumbaiT004 Paras HaryanaT009 Ponds NoidaT006 Wipro Ahmedabad

i). TodisplayPNAME,PRICE* QTYonlyforthecityMumbai.ii).

Todisplayproductname,companyname&priceforthoseitemswhichIDsareequaltotheIDsofcompany.

iii). Todeletetheitems producedbefore 2007.iv). Toincreasethequantityby20forsoapandpaste.v). SELECTCOUNT(*)FROMITEMS WHEREITEMS.ID=COMPANY.ID;

vi). SELECTPNAMEFROMITEMSWHEREPRICE=SELECTMIN(PRICE)FROMITEMS;vii). SELECTCOUNT(*)FROM COMPANYWHERECOMPLIKE“P____”;

viii). SELECTPNAMEFROMITEMSWHEREQTY<100;

Ans:(i)SELECTPNAME,QTY*PRICEFROMITEMSWHEREITEMS.ID=COMPANY.IDANDCOMPANY.City=’Mumbai’;

(ii) SELECTPNAME,COMP,PRICEFROMITEMS,COMPANYWHEREITEMS.ID= COMPANY.ID

(iii) DELETEFROMITEMSWHEREMDATE<{01/01/2007};(iv) UPDAT

EITEMSSETQTY=QTY+20WHEREPNAME=‘Soap’ORPNAME=‘Paste’;(v) 4(vi) Soap(vii) 2(viii) Paste

Deodorant

Boolean Algebra Introductio

n to BooleanAlgebra & Laws BooleanAlgebraRules1 0+X=X Propertiesof02 0. X =0

3 1+X=1 Propertiesof14 1. X =X5 X+X=X IdempotenceLaw6 X.X=X7 (X)=X InvolutionLaw8 X+X=1 ComplementarityLaw

9 X.X=010 X+Y=Y+X CommutativeLaw11 X.Y=Y. X12 X+(Y+Z)=(X+Y)+Z AssociativeLaw13 X(YZ)=(XY)Z14 X (Y+Z)=XY +XZ Distributive Law15 X+YZ=(X+Y)(X+Z)16 X+ XY=X Absorption Law17 X. (X+Y)=X18 X+XY=X+Y ThirdDistributive Law19 (X+Y)=X.Y DeMorgan’sTheorem20 (X.Y)=X+Y

Short Answer Questions (2 marks)

Q1.Stateand verifyDemorgan'sLaws algebraically.Ans: Demorgan'sLawsare :(i)(X+Y)'= X'.Y' (i)

(X.Y)'=X'+Y'Verification(X+Y)'.(X+Y)= X'.Y'.(X+Y)0=X'.Y'.X+X'.Y'.Y0=X'.X.Y'+ X'.00=0.Y'+00=0+00=0L.H.S= R.H.S

(X.Y)'+(X.Y)=(X'+Y')+ (X.Y)1=(X'+Y'+X).(X'+Y'+ Y)1=(X'+X+Y'). (X'+1)1= (1+Y'). 11=1.11=1L.H.S= R.H.S

Q2.StateandalgebraicallyverifyAbsorptionLaws.Ans: AbsorptionLawsare:(i)X+X.Y =X (ii)X.

(X+Y)=XVerification:L.H.S=X+X.Y

=X.1+X.Y= X.(1 + Y)=X.1=X=R.H.S

L.H.S=X.(X+Y)=X.X+X.Y=X+ X.Y= X.(1 + Y)=X.1=X=R.H.S

Q3.StateDistributive Laws.Writethedualof:X+XY=X+YAns:Distributivelaws are:(i)X.(Y+Z)=X.Y +X.Z (ii)X+Y.Z=(X+Y).

(X+Z)ThedualformofX+XY=X+YisX. (X+Y)=X.YQ4.StateanyoneformofAssociativelaw andverifyitusing truthtable.Ans: Associativelawstatesthat: (i)X+ (Y +Z)= (X+Y)+Z (ii)X.

(Y.Z)=(X.Y).ZVerificationofX + (Y+Z)=(X+Y)+Z using truthtableX Y Z Y+Z X+Y X+(Y+Z) (X +Y)+Z0 0 0 0 0 0 00 0 1 1 0 1 10 1 0 1 1 1 10 1 1 1 1 1 11 0 0 0 1 1 11 0 1 1 1 1 11 1 0 1 1 1 11 1 1 1 1 1 1

Q5.Statetheprincipleofdualityinbooleanalgebra andgivethedualofthefollowing:X.Y.Z+X.Y.Z+X.Y.Z

Ans: Principleofdualitystatesthatfromeveryboolean relation,anotherbooleanrelationcanbederivedby: (i)changingeachOR(+)signtoanAND(.) sign

(ii) changingeachAND (.)sign toanOR(+) sign(iii) replacing each1 by0andeach0 by1.

Thedual formofX.Y.Z+X.Y.Z+X.Y.Zis(X+Y+Z) . (X+Y+Z).(X+Y+Z)

Q6.Explainabouttautologyandfallacy.Ans:Ifresultofanylogical statementorexpressionis always TRUEor1,itiscalledtautology.

Ifresultofanylogical statementorexpressionis alwaysFALSEor0,itiscalledfallacy.Forexample,1+X=1isatautologyand 0.X=0 is afallacy.

SOP & POSVery Short AnswerQuestions (1 Mark)

Q1.Writethe POSformofaBooleanfunctionF,whichis represented inatruthtable asfollows:U V W F0 0 0 10 0 1 00 1 0 10 1 1 01 0 0 11 0 1 01 1 0 11 1 1 1Note:

InPOS,consideronly thecasesthosecontaining0inFcolumn.Forthesamerows,writesumofthelogicalvariables(Ex.Ui.e.0inUcolumn,U'i.e.1forUcolumn).Thenwritetheproductformofallsuchsumof logicalterms.Ans:F(U,V,W)=(U+V+ W').(U+V'+W').(U'+V+W')

Q2.Writethe SOPformofaBooleanfunctionG,whichis represented inatruthtable asfollows:X Y Z G0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 01 0 1 11 1 0 11 1 1 0Note:

InSOP,consideronly thecasesthosecontaining1inGcolumn.Forthesamerows,writeproductofthelogicalvariables(Ex.X'i.e.0inXcolumn,Xi.e.1forXcolumn).Thenwritethesumformofall suchproductoflogicalterms.Ans: G(X,Y,Z)=(X'.Y'. Z)+( X'.Y.Z')+(X.Y'. Z)+(X.Y. Z')

Q3.Convert thefollowingfunctionintocanonicalSOPform.F(A, B,C,D)= (1,4,6, 8,11, 13)

Ans:F(A,B,C,D)=A'B'C'D+A'B C'D'+A'BC D'+AB'C'D' +AB'C D+ABC'D(Note:Here1 =0001 A'B'C'D ,4=0100 A'BC'D',13 =1101 ABC'D ...)

Q4.Convert thefollowingfunctionintocanonicalSOPform.F(A, B,C,D)= (0, 3,7, 9, 10, 14)

Ans:F(A,B,C,D)=(A+B+C+D).(A+B+C'+D').(A+B'+C'+D').(A'+B+C+D').(A'+B+C'+D).(A'+B'+C'+D)

1

14 5

15

11412

11

CDAB

A'B'

C'D'1

C'D CD1

CD'1

032

A'B 1176

AB 1 113

AB' 1 1 18910

1 3 2

0

4

0 0 07

12 13

015 14

08

9

C+DA+BC+D C+D' C'+D' C'+D

A+B 00

A+B'5 6

A'+B'

01110

(Note :Here0=0000 A+B+C+D ,3 =0011 A+B+C'+D',10= 1010 A'+B+C'+D...)Q5.Convert thefollowingfunctiontoitsequivalentSOPshorthandnotation.

F(A, B,C,D)= (0, 3,7, 9, 10, 14)Ans: TheequivalentSOPshorthandnotationof F( A,B,C,D)

= (0,3,7,9,10,14)isF(A,B,C,D)= (1,2,4,5,6,8,11,12,13,15)Q6.Convert thefollowingfunctiontoitsequivalentPOSshorthandnotation.

F(A, B,C,D)= (1,4,6, 8,11, 13)Ans: The equivalentPOSshorthandnotationof F( A,B, C,D) = (1,4,6,8,11,13)

isF(A,B,C,D)= (0,2,3,5,7,9,10,12,14,15)Karnaug

h Map Short Answer Questions (3 marks)

Q1.ReducethefollowingBooleanExpressionusingK-Map:F( A, B,C,D) = (0,2,3,4,6,7,8,9,10,12, 13, 14)

Ans:

Thereare1octet, 2quadsaftereliminatingthe redundantgroups.Octet(m0,m2,m4,m6,m8,m10,m12,m14)reduces toD'Quad(m2,m3,m6,m7) reduces toA'CQuad (m8,m9,m12,m13) reduces

toAC 'Hence,F(A,B,C, D)=D'+A'C +AC'Q2.ReducethefollowingBooleanExpressionusingK-

Map:F(A, B, C,D)= ( 0,3,5,6,7,8,11,15)Ans:

A'+B

Thereare1quadand 3pairs aftereliminatingtheredundantgroups.Quad (M3,M7,M11,M15) reduces toC'+ D'Pair(M5, M7) reduces toA+B '+D'Pair(M6, M7) reduces toA+B '+C'Pair(M0, M8) reduces toB+C +D

Hence,F(A,B,C, D)=(C'+D').(A+B '+D').(A+B '+C').(B+C +D)

C’D’ C’D CD CD’A’B’ 1

01

1 3 2

A’B 1 1 17 64

AB12

114

AB’8 9 11

110

5

113 15

Q3.ReducethefollowingBooleanExpressionusingK-Map:F( A,B,C,D) = ( 2,3,4,5,10,11,13,14, 15)

Ans:CD

AB C'D' C'D CD CD'

A'B

'A'B

1 10 1 3 2

1 14 5 7 6

AB

AB'

112 13

1 115 14

1 18 9 11 10

Thereare2quadsand 2pairsaftereliminatingtheredundantgroups.Quad 1(m2,m3,m10,m11) reduces toB'CQuad 2 (m10,m11,m14,m15)reducestoACPair1(m4,m5)

reduces toA'BC 'Pair1(m13,m15) reduces toABD

Hence,F(A,B,C, D)=B'C+AC +A'BC'+ABD

Q4.ReducethefollowingBooleanExpressionusing K-Map:F(A,B,C,D)= (0,1,4,5,7,10,13,15)Ans:Thereare2quadsandone blockaftereliminatingtheredundantgroups.

Quad 1(m0,m1,m4,m5) reduces toA'C 'Quad 2(m0,m2, m4,m6) reduces toBDm10=AB'C D'Hence,F(A,B,C, D)=A'C '+B D+AB 'C D'

Page 54/70

Basic LogicGates Note: Agateissimplyanelectronic circuitwhichoperateson oneor moresignalstoproduce

anoutputsignal. The three basiclogicgates areNOT gate,ORgate, ANDgate.NORgate,NAND gate,XORgate,XNORgatecanbederivedfromthesethree basicgates.NORandNAND gates arecalleduniversalgates because anycombinationalcircuitispossible usingNORorNAND gates.Practicedifferentquestionsbasedonlogicgates.

Short Answer Questions (2 marks) Q1.WritetheequivalentBooleanExpressionforthefollowingLogic Circuit.

P

Q

Ans: F(P, Q)=(P’+Q). (P+Q’)Q2.WritetheequivalentBooleanExpressionforthefollowingLogic Circuit.

U

V

Ans: F(U,V)=U’.V+U.V’

Q3.DrawtheLogic CircuitDiagramforthefollowing BooleanExpression:F=(A.C)+( A.B)+(B.C)

Ans:A

B

FC

Communication &OpenSource Concepts Very Short AnswerQuestions

Q1.Whatisa network?Ans : A

networkisinterconnectedcollectionofautonomouscomputersthatcanshareandexchangeinformation.

Q2.Mention the advantages of networking.Ans : The advantagesofnetworkingareResourcesharing,Reliability,Reducedcost,Fastcommunication.

Q3.Defineserver.Mention thetypesofservers.Ans:

Acomputerthatfacilitatesthesharingofdata,software,andhardwareresourcesonthenetworkistermedasserver.The twotypesofserversare non-dedicated&dedicatedservers.

Q4.WhatisNIU? What does theMAC address refer to?Ans:ANetworkInterfaceUnitis an interpreter that helps establish communication between the server

and workstations.TheMAC address refers to the physical address assigned by NIC manufacturer.

Q5.Whatarethedifferenttypesofswitchingtechniques?Ans:Th

edifferenttypesofswitchingtechniquesarecircuitswitching,Messageswitchingandpacketswitching.Q6.Whatisacommunication channel?Give examplesofguidedmedia&unguidedmedia.Ans

:Communicationchannelmeanstheconnectingcablesthatlinkvariousworkstations.TheexamplesofguidedmediaareTwisted-paircables,Coaxialcables,Fibre-opticscables.Theexamplesofunguidedmediaaremicrowave,radiowave,satellite,infraredwave etc.

Q7.Define the term Bandwidth.Giv any one unit of Bandwidth.Ans:Bandwidth is referred to the volume of information per unit of time that a transmission

medium(like an Internet connection)can handle.OR

Theamountofdatathatcanbetransmittedinafixedamountoftimeisknownasbandwidth.Fordigitaldevices,thebandwidthisusuallyexpressedinbitspersecond(bps)orbytespersecond.Foranalogdevices,the bandwidthisexpressedincyclespersecond,orHertz(Hz).

Q8.Expand:HTTP,FTP, TCP/IP,SLIP, PPP, NCP,GSM,TDMA,CDMA,WLL,PSTN.Ans: HTTP:HypertextTransferProtocol, FTP:FileTransferProtocolTCP/IP:TransmissionControlProtocol/InternetProtocol

SLIP:SerialLineInternetProtocol, PPP: Point toPointProtocolNCP:NetworkControlProtocol,

GSM:GlobalSystemforMobileTDMA:Time DivisionMultipleAccess, CDMA: CodeDivisionMultiple AccessWLL:WirelessLocalLoop PSTN: PublicSwitchedTelephone Network

Q9.Whatarethedifferenttypesof networks?Ans: Thedifferenttypesofnetworks are :

(a) LocalAreaNetwork (LAN): This computernetworkisconfinedtoalocalisedarea suchasanoffice oracampus.

(b) MetropolitanAreaNetwork(MAN):Thisisthenetworkthatlinkscomputerfacilitieswithinacity.

(c) WideAreaNetwork(WAN):Thisnetworkspreadsoverlargedistanceslikeacrosscountries orevencontinents.

Q10.Whatdoyoumeanbynetworktopology?Whatarethemostpopulartopologies?Ans

:Topologyreferstothewayinwhichtheworkstationsattachedtothenetworkareinterconnected.ThemostpopulartopologiesareBusTopology,RingTopology,StarTopology,TreeTopology.

Q11.Whatarethe advantagesofE-mail?Ans:TheadvantagesofE-

mailareLowcost,Speed,Wastereduction,Easeofuse,Recordmaintenance,Patience.Q12.What is WWW? What are the WWW attributes?Ans:The WorldWideWeb(WWW) is a set of protocols that allow you to access any document on

the NET through an aming system based on URLs. The various WWW attributes are:User-friendly,Multimedia documents,Hypertext and Hyperlinks,Interactive,Frames.

Q13.Whatiswebbrowser? Giveanytwoexamples ofwebbrowsers.Ans:AwebbrowserisaWWWclientthatnavigatesthroughtheWWWanddisplayswebpages.

ExamplesofwebbrowsersareInternetExplorer,Netscape NavigatorandMosaic.Q14.Definetheterms:Website,Webpage,Homepage,Webportal.Ans: Website :Alocationon anetserveriscalleda website.

Webpage:AdocumentthatusesHypertextTransferProtocoliscalleda Webpage.Homepage :Thetop-levelwebpage ofawebsiteiscalledHome page.Webportal:The websitewhichhostsotherwebsitesiscalledWebportal.

Q15.WhendoyoupreferXMLoverHTMLandwhy?Ans:Th

efirstbenefitofXMListhatbecauseyouarewritingyourownmarkuplanguage,youarenotrestrictedtoalimitedsetoftagsdefinedbyproprietaryvendors.Ratherthanwaitingforstandardsbodiestoadopttagsetenhancements(aprocesswhichcantakequitesometime),orforbrowsercompaniestoadopteachother'sstandards,withXML,youcancreateyourownsetoftags atyourownpace.

Q16.Compareauthorisationwithauthentication.Ans

:Authorisationdetermineswhethertheserviceproviderhasgrantedaccesstothewebservicetotherequestor.Authorisationisperformedbyaskingtheuseralegalloginid.Authenticationensuresthateachentityinvolvedinusingawebserviceiswhatactuallyclaimstobe.Itistermedaspasswordprotectionastheauthoriseduserisaskedtoprovideavalid password.

Q17.WhatisFirewall?Howfirewallprotectsournetwork?Ans:Th

esystemdesignedtopreventunauthorisedaccesstoorfromaprivatenetworkiscalledfirewall.Itisadeviceorsetofdevicesconfiguredtopermit,deny,encrypt,decrypt,orproxyall(inandout)computertrafficbetweendifferentsecurity domainsbaseduponasetofrulesandothercriteria.

Q18.Whatarecookies?Ans:Cookies ar messages that a Webserver transmits to a Webbrowser so that the Webserver can

keep track of the user’s activity on a specificWebsite.Q19.How is a cracker different from a hacker? If someone hashacked yourWebsite,to whom

you lodge the Complain?Ans:Acracker is a malicious programmer who breaks into securesystem where as a hacker is more

interested in gaining know ledge about computersystems.If someonehashackedourWebsite thecomplainthas to be lodged with thePolice unde rITAct.

Q20.WhatisCyberlaw?Ans:Cyberlaw is a generic term,which refers to all the legal and regulatory aspects of Internet and

the WorldWideWeb.Q21.Expand:(i)FLOSS (ii)FSF (iii)OSI

(iv)W3CAns:(i)FLOSS= Free Libre andOpenSource Software

(ii) FSF= Free SoftwareFoundation (iii)OSI=OpenSource Initiative(iv)W3C =WorldWideWebConsortium

Q22.Whatisa computervirus? Write thebasic types ofviruses.Ans

:Computervirusisamaliciousprogramthatrequiresahostandisdesignedtomakeasystemsick.BroadlythreetypesofvirusesareFileinfectors,BootSectorVirusesandMacroviruses.

Q23.Definethe terms :Trojan Horses,Worm,Spam.Ans:TrojanHorses:Itiscodehiddeninaprogramsuchasagameorspreadsheetthatlookssafetorunbuthas

hiddensideeffects.

Worm:Awormisa programdesignedtoreplicate.Spam:Itrefers to electronic junk mail or junk news group posting.

Q24. What is cloud computing? Write any two features of it.Ans. The practice of using a network of remote servers hosted on the Internet to store, manage, and process data, rather than a local server or a personal computerFeatures : 1. On demand self service 2. Broad network access

3. Resource pooling 4. Rapid elasticity 5. Measured service 6. Non dependence on local hardware

BlockDBlock A

BlockB

BlockC Block

DBlock

B

BlockA

BlockA

BlockB

Q1.KnowledgeSupplementOrganisationhassetupitsnewcentreatMangaloreforitsofficeandwebbasedactivities.Ithas4blocksofbuildingsasshown inthe diagram below:

BlockC

Centre to centredistancesbetweenvarious blocks NumberofComputersBlackAtoBlockB 50mBlockBtoBlock C 150mBlock C toBlock D 25mBlockAtoBlock D 170mBlockBtoBlock D 125mBlockAtoBlock C 9

BlackA 25BlockB 50Block C 125BlockD 10

(a) Suggestacablelayout ofconnectionsbetweentheblocks.(b) Suggestthemostsuitableplace(i.e.block)tohousetheserverofthisorganisationwithasuitablereason.(c) Suggesttheplacementofthefollowing devices with justification

(i) Repeater (ii) Hub/Switch(d) Th

eorganizationisplanningtolinkitsfrontofficesituatedinthecityinahillyregionwherecableconnectionisnotfeasible,suggestaneconomicwaytoconnectitwithreasonablyhighspeed?

Ans :(a)

OR

BlockC

BlockD

Layout2Sincethe distancebetween BlockAand BlockBisquiteshort

(b) Themostsuitableplace/blocktohousetheserverofthisorganisationwouldbeBlockC,asthisblockcontains themaximumnumberof computers, thus decreasing thecablingcostformostofthecomputersaswellasincreasingtheefficiencyofthemaximumcomputersinthe network.

(c) (i)ForLayout1,sincethecablingdistancebetweenBlocksAandC,andthatbetweenBandCarequitelarge,soarepeatereach,wouldideallybeneededalongtheirpathtoavoidlossofsignalsduring thecourseofdataflow intheseroutes.

Forlayout2,sincethedistancebetweenBlocksAandCislargesoarepeaterwouldideallybeplacedinbetweenthis path(ii)Ahub/switcheachwouldbeneededinalltheblocks,tointerconnectthegroupofcablesfromthedifferentcomputersineachblock.

(d) Themosteconomicwaytoconnectitwithareasonablehighspeedwouldbetouseradiowavetransmission,a

Layout1

stheyareeasytoinstall,cantravellongdistances,andpenetratebuildingseasily,sotheyarewidely usedforcommunication,bothindoorsandoutdoors.Radiowavesalsohavetheadvantageofbeingomnidirectional,whichistheycantravelinallthedirectionsfromthesource,sothatthetransmitterandreceiverdonot havetobecarefullyalignedphysically