c and ds manuaL

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JYOTHISHMATHI COLLEGE OF ENGINEERING&TECHNOLOGY.

C and Data Structures Lab manua

Lab Manual for the Academic Year 2010-2011

I B.Tech

In!c"ar#e HOD $r%nc%a

$reared b'( Ar)*ed &Re*%e+ed b'(

Issued b'( +.e., Date(

D.RAMBABU 08-oct-2010

1


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No. Name of the Program Page No. !emar"

1 #ee" l.a$ Write a C program to find the sum of individual digits of a

positive integer.b$ A Fibonacci sequence is defined as follows: the first andsecond terms in the sequence are 0 and 1.Subsequent terms are found b adding the preceding two termsin the sequence. Write a C program to generate the first n termsof the sequence.c$ Write a C program to generate all the prime numbers between1 and n! where n is a value supplied b the user

1

" #ee" 2.a$ Write a C program to calculate the following Sum:Sum#1$%"&"' (%)&)'$%*&*'(%+&+'$%10&10'

b$ Write a C program to find the roots of a quadratic equation

11

, #ee" %a$ Write C programs that use both recursive and non$recursivefunctionsi- o find the factorial of a given integer.ii- o find the /C greatest common divisor- of two givenintegers.iii- o solve owers of 2anoi problem.

1+

) #ee" &

a$ he total distance travelled b vehicle in 3t4 seconds is givenb distance # ut(1&"at"where 3u4 and 3a4 are the initial velocitm&sec.- and acceleration m&sec"-. Write C program to find thedistance travelled at regular intervals of time given the values of3u4 and 3a4. he program should provide the fle%ibilit to theuser to select his own time intervals and repeat the calculationsfor different values of 3u4 and 3a4.b$ Write a C program! which ta5es two integer operands andone operator from the user! performs the operation and thenprints the result. Consider the operators (!$!6! &! 7 and useSwitch Statement-

,8

9 #ee" 'a$ Write a C program to find both the largest and smallest number in a listof integers.b$ Write a C program that uses functions to perform the following:i- Addition of wo atricesii- ultiplication of wo atrices

)*

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* #ee" (a$ Write a C program that uses functions to perform the followingoperations:i- o insert a sub$string in to a given main string from a given

position.ii- o delete n Characters from a given position in a given string.b$ Write a C program to determine if the given string is apalindrome or not.

9;

; #ee" )a$ Write a C program that displas the position or inde% in thestring S where the string begins! or < 1 if S doesn4t contain .b$ Write a C program to count the lines! words and characters ina given te%t.

;1

No. Name of the Program Page No. !emar"+ #ee" *

a$ Write a C program to generate =ascal4s triangle.b$ Write a C program to construct a pramid of numbers.

;8

8 #ee" +Write a C program to read in two numbers! % and n! and thencompute the sum of this geometricprogression:1(%(%"(%,(>>>>.(%n

For e%ample: if n is , and % is 9! then the program computes

1(9("9(1"9.=rint %! n! the sum=erform error chec5ing. For e%ample! the formula does not ma5esense for negative e%ponents < if n is less than 0. 2ave ourprogram print an error message if n?0! then go bac5 and read inthe ne%t pair of numbers of without computing the sum.Are an values of % also illegal @ f so! test for them too.

+*

10 #ee" 10a$ "4s complement of a number is obtained b scanning it fromright to left and complementing all the bits after the first

appearance of a 1. hus "4s complement of 11100 is 00100.Write a C program to find the "4s complement of a binarnumber.b$ Write a C program to convert a Boman numeral to itsdecimal equivalent.

80

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11 #ee" 11Write a C program that uses functions to perform the followingoperations:i- Beading a comple% number

ii- Writing a comple% numberiii- Addition of two comple% numbersiv- ultiplication of two comple% numbersote: represent comple% number using a structure.-

101

1" #ee" 12a$ Write a C program which copies one file to another.b$ Write a C program to reverse the first n characters in a file.ote: he file name and n are specified on the command line.-

10*

1, #ee" 1%

a$ Write a C program to displa the contents of a file.b$ Write a C program to merge two files into a third file i.e.! the contents of the first file followed b those of thesecond are put in the third file-

110

1) #ee" 1&Write a C program that uses functions to perform the followingoperations on singl lin5ed list.:i- Creation ii- nsertion iii- eletion iv- raversal

11)

19 #ee" 1'Write C programs that implement stac5 its operations- usingi- Arras ii- =ointers

1)*

1* #ee" 1(Write C programs that implement Dueue its operations- usingi- Arras ii- =ointers

1*0

No. Name of the Program Page No. !emar"

1; #ee" 1)Write a C program that uses Stac5 operations to perform thefollowing:i- Converting infi% e%pression into postfi% e%pressionii- Evaluating the postfi% e%pression

1;)

1+ #ee" 1*Write C programs that use both recursive and non recursive

functions to perform the following searching operations for ae value in a given list of integers :i- Ginear search ii- Hinar search

1+"

18 #ee" 1+Write a C program that implements the following sortingmethods to sort a given list of integers in ascending orderi- Hubble sortii- Selection sort

"0,

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"0 #ee" 20Write C program that implements the following sorting methodto sort a given list of integers in ascending order:i- Duic5 sort

"0;

"1 #ee" 21Write C program that implement the following sorting methodto sort a given list of integers in ascending order:i- erge sort

"19

"" #ee" 22Write C programs to implement the Gagrange interpolation andewton$ /regor forward interpolation.

""0

", #ee" 2%Write C programs to implement the linear regression andpolnomial regression algorithms.

"";

") #ee" 2&Write C programs to implement rapeIoidal and Simpsonmethods.

",*

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,bectie / 1

o find the sum of individual digits of a given number

ecrition/Sum of the individual digits means adding all the digits of a numberE%: 1", sum of digits is 1("(,#*

Algorithm/

Step 1: startStep ": read nStep ,: initialiIe the s#0Step ): if n?0 goto Step ;Step 9: if n'#0 goto Step * else goto step ;

Step *: store n710 value in pAdd p value to sAssign n&10 value to n/oto Step 9

Step ;: print the outputStep +:stop

*


7/259

lo3chart/

BJE FAGSE

BJE

;

SAB

S#0

BEA

F ?0

KJ=J

W2GE

' #0

SK= =#710

S#S(=

#&10


8/259

Program/

Linclude?stdio.hMmain-

N int n!s!pO clrscr-O printfPenter the vaue for n:QnP-O scanfP7dP!Rn-O s#0O ifn?0- printfPhe given number is not validP-O else N whilen'#0- &6 chec5 the given value #0 or not 6&

N p#n710O n#n&10O s#s(pO

printfPsum of individual digits is 7dP!s-O getch-O,utut/

1.Enter the value for n: ,,, Sum of individual digits is 8".Enter the value for n: );,, Sum of individual digits is 1;,. Enter the value for n: $111 he given number is not valid4oncluion /he program is error free

5I5A 6789ATI,N9/

1- What is the mean of sum of the individual digits@ Ans: Sum of the individual digits means adding each digit in a number

"- What is positive integer@Ans: if the integer value is grater than Iero then it is called positive integer,- efine preprocessor @Ans: Hefore compiling a process called preprocessing is done on the source code b aprogram called the preprocessor.

+


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,bectie/2

o print the Fibonacci series for 1 to n valueecrition

A fibonacci series is defined as followshe first term in the sequence is 0he second term in the sequence is 1he sub sequent terms 1 found b adding the preceding two terms inthe sequenceFormula: let t1!t"!>>>>tn be terms in fibinacci sequence

t1#0! t"#1tn#tn$"(tn$1>>where nM"

algorithm/

Step 1: startStep ": initialiIe the a#0! b#1Step ,: read n

Step ): if n## 1 print a go to step ;. else goto step 9Step 9: if n## " print a! b go to step ; else print a!bStep *: initialiIe i#,

i- if i?# n do as follows. f not goto step ;c#a(bprint ca#bb#cincrement valuegoto step *i-

Step ;: stop

8


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lo3chart/

SAB

A#0!b#1

Bead n

f n##1

Kutput af n##"

Kutput a!b Kutput a!b

#" i((

?n

C # a ( b

Kutput c

A # b H # c

Stop

rue False

False

rue

False

10


11/259

Program/

Linclude?stdio.hM

void main-N int a!b!c!n!iO clrscr-O printfPenter n valueP-O scanfP7dP!Rn-O a#0O b#1O ifn##1- printfP7dP!a-O else

ifn##"- printfP7d7dP!a!b-O else N printfP7d7dP!a!b-O

&&GKK= WGG BJ FKB " E GESS SEBES AS 2ESE WAS=BE ATACE

fori#"Oi?nOi((- N

c#a(bOprintfP7dP!c-Oa#bOb#cO

getch-O

,utut/

1. Enter n value : 90 1 1 " ,

". Enter n value : ;0 1 1 " , 9 +

,. Enter n value : $*0 1

4oncluion /he program is error free

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5I5A 6789ATI,N9/

1$ #hat i ibonacci erie :

An/ A fibonacci series is defined as followshe first term in the sequence is 0he second term in the sequence is 1he sub sequent terms 1 found b adding the preceding two terms inthe sequenceFormula: let t1!t"!>>>>tn be terms in fibinacci sequence

t1#0! t"#1tn#tn$"(tn$1>>where nM"

"- What are the various tpes of unconditional statements@ Ans: goto!Hrea5 and continue

,-What are the various tpes of conditional statements@

Ans: if ! if else !switch statements)- E%pand ?SK.2 M@Ans: standard input output header file

1"


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,bectie /%

o print a prime numbers up to 1 to n

ecrition/=rime number is a number which is e%actl divisible b one and itself onlE%: "! ,!9!;!>>>O

Algorithm/

Step 1: startStep ": read nStep ,: initialiIe i#1!c#0Step ):if i?#n goto step 9

f not goto step 10Step 9: initialiIe U#1Step *: if U?#1 do as the follow. f no goto step ;

i-if i7U##0 increment cii- increment Uiii- goto Step *

Step ;: if c## " print iStep +: increment iStep 8: goto step )Step 10: stop

1,


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lo3 chart/

Start

Bead n

# 1

?#n

((

V # 1 V((

V?#i

ffact##"

f 7 U ##0

Fact ((

true

false

true

false

Kutputfact

false

stop

false

1)


15/259

Program/

Linclude?stdio.hMLinclude?conio.hM

void main-N int n!i!fact!UO clrscr-O printfPenter the number:P-O scanfP7dP!Rn-O fori#1Oi?#nOi((- N fact#0O

&&2S GKK= WGG C2EC A K K HE =BE K. KB K. forU#1OU?#iOU((-

N ifi7U##0-fact((O

iffact##"- printfPQn 7dP!i-O getch -O

,utut/

Enter the number : 9" , 9

Enter the number : 10" , 9 ;Enter the number : 1"" , 9 ; 11

4oncluion /he program is error free

5I5A 6789ATI,N9/

1$ #hat i rime number :

An/ =rime number is a number which is e%actl divisible b one and itself onl

"-What is an algorithm@Ans : A step b step procedure is called algorithm

,-What is flow chart@Ans: A pictorial representation an algorithm is called a flow chart)-What is program@

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Ans : A collection of statements is called

,bectie/&

o find the roots of the quadratic equationecrition/

ature of roots of quadratic equation can be 5nown from the quadrant

# b"

$)acf b"$)ac M0 then roots are real and unequalf b"$)ac #0 then roots are real and equalf b"$)ac ?0 then roots are imaginar

Algorithm/

Step 1: startStep ": read the a!b!c valueStep ,: if b6b$)acM0 then

Boot 1# $b( powb6b$)6a6c-!0.9--&"6aBoot "# $b$powb6b$)6a6c-!0.9--&"6a

Step ): if b6b$)ac#0 then

Boot1 # Boot" # $b&"6a-Step 9: Ktherwise =rint maginar roots. /oto step ;.Step *: print rootsStep ;: stop

1*


17/259

lo3chart/

Start

Bead a!b!c

# powb6b$)6a6c-!0.9

f d M 0

B1 # $b(- & "6a--

B" # $b$- &"6a--

Kutput B1! B"

Stop

f d## 0

B1#$b & " 6 a -

B"# $b & " 6 a-

maginarroots

true

false

1;


18/259

Program/

Linclude?stdio.hMLinclude?math.hMvoid main-

Nfloat a!b!c!r1!r"!dOclrscr-OprintfPEnter the values for equation:P-OscanfP7f7f7fP!Ra!Rb!Rc-O&6 chec5 the condition 6&ifa##0-printfPEnter value should not be Iero P-OelseN d#b6b$)6a6cO

&6 chec5 the condition 6& ifdM0- N r1#$b(sqrtd-&"6a--O r"#$b$sqrtd-&"6a--O printfProots are realand unequalQnP-O printfP7fQn7fQnP!r1!r"-O

else ifd##0-

N r1#$b&"6a-O r"#$b&"6a-O printfProots are real and equalQnP-O printfProot#7fQnP!r1-O printfProot#7fQnP!r"-O

else printfProots are imaginarP-O getch-O

,utut/

1. Enter the values for equation: 1! *! 8 Boots are real and equal Boot# $,.0000 Boot# $,.0000

1+


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". Enter the values for equation: "! ;! * Boots are real and unequal Boot# $*.;9 Boot# $;."9,. Enter the values for equation: 1! "! ,

Boots are imaginar

4oncluion/he program is error free

5I5A 6789ATI,N9/

1$ What are various tpes of loop statements@Ans : While! do$ while! for loop statements

9- What is the difference between while and do$while statements@Ans: n while the condition will be chec5ed first and then enter into a loop. Hut in do$ while the statements will be e%ecuted first and then finall chec5 the

Condition.,- 2ow to find the roots of qudratric equtations @Ans: ature of roots of quadratic equation can be 5nown from the quadrant # b"$)ac

f b"$)ac M0 then roots are real and unequalf b"$)ac #0 then roots are real and equal

f b"$)ac ?0 then roots are imaginar

)- Gist out the C features @Ans: =ortabilit!fle%ibilit! wide acceptabilit etc..!

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,bectie/'

o calculate the sum. Sum#1$%"&"'( %)&)'$ %*&*'( %+&+'$ %10&10'

Algorithm/main program:Step 1: start

Step ": declare %!i!n!s#0!cStep ,: read % valueStep ): for i#0 ! n#0O i?#10O i#i("! n(( goto step 9Step 9: s#s(pow$1!n-6pow%!i-&facti--Step *: print s valueStep ;: stop

9ub rogram/Step 1: while %'#0 goto Step "Step ": #(%O %Step ,: return

Step ): return to main program

"0


21/259

lo3chart/

Start

Bead %!!n!c

S # 0

#0!n#0 i?#10

# ( "! n((

S#s(pow$1!n-6pow%!i- & facti--

=rint s

Stop

Sub =rogram

While %'#0

X#6%

%$$

Beturn

Fact -

Beturn to mainprogram

"1


22/259

Program/

Linclude?stdio.hMLinclude?math.hMlong factint-Ovoid main-

N int %!i!nO float s#0!cO clrscr-O printfPQn enter the value of %QtP-O scanfP7dP!R%-O

&6perform the looping operation6& fori#0!n#0Oi?#10Oi#i("!n((- s#s(pow$1!n-6pow%!i-&facti--O printfPQn the result is 7fP!s-O getch-O

&6 calling sub program6&long factint %-N long int #1O while%'#0- N #6%O %$$O return O,utut/

1.Enter the valueof % : 1 he result is 0.9)0,0"" Enter the value of %: " he result is $0.)1*1994oncluion/he program is error free5I5A 6789ATI,N9/

1- What is function @Ans: A function is a sub program it returns a value.

"- What is procedure @Ans: A procedure is a sub program it does not returns a value

,- What are the basic data tpes in C @Ans: int! char! float! double)- 2ow to define preprocessor @Ans: H using the L smbal E%: Linclude?stdio.hM

""


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,bectie/ (

=rograms that use recursive function to find the factorial of a given integer.

ecrition/

Factorial of a number is nothing but the multiplication of numbers from a given

number to 1

Algorithm/main programStep 1: startStep ": read nStep ,: call sub program as f#factn-Step ): print f valueStep 9: stop

9ub rogram/Step 1: initialiIe the f

Step ": if n# # 0 or n ## 1 return 1 to main program if not goto step ,Step ,: return n6factn$1- to main program

",


24/259

lo3chart/

Start

Bead n

Call subprogram F # factn-

output

Stop

9ub rogram

Fact -

f n#0 YY n#1

Beturn n6factn$1-


true

false

")


25/259

Program/

Linclude?stdio.hMLinclude?conio.hMint factint n-

N int fO ifn##0-YYn##1-- && chec5 the condition for the n value returnn-O else f#n6factn$1-O &&calculate the factorial of n returnf-Ovoid main-N int nO

clrscr-O printfPenter the number :P-O scanfP7dP!Rn-O printfPfactoria of number7dP!factn--O getch-O,utut/

1. Enter the number : 9 Factorial of number: 1"0". Enter the number : , Factorial of number: *,. Enter the number : 8 Factorial of number: $,0,,*

4oncluion/the program is error free

5I5A 6789ATI,N9/

1- What is the meaning of factorial number@An / Factorial of a number is nothing but the multiplication of numbers from a given

number to 1"- What is the meaning of recusive function @

Ans: A function call it self is called recursive function,- define librar functions @

Ans: he functions have alread been written! compiled and placed in libraries and arecalled librar functions.

)- efine formal parameters @Ans: Formal parameters are the parameters given in the function declaration as

function definition.

"9


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,bectie / )

=rogram that use non recursive function to find the factorial of a given integer.ecrition/

Factorial of a number is nothing but the multiplication of numbers from a givennumber to 1

E%: 9' #96)6,6"61# 1"0Algorithm/main programStep 1: startStep ": read nStep ,: call the sub program factn-Step ): print the f valueStep 9: stop9ub rogram/

Step 1: initialiIe the f#1Step ": if n##0 or n#1 return 1 to main program.f not goto step ,Step ,: perform the looping operation as follows

For i#1 i?#nO i((Step ): f#f6iStep 9: return f value to the main program

"*


27/259

lo3chart/

actorial nonrecurie

start

Bead i

Call subprogram Factn-

=rint outputTalue of fact

Stop

9ub rogram

Fact -

F # 1! i

f n ## 0 YYn ## 1

Beturn to main

program

# 1 i((

?#n

F # f 6 i

";


28/259

Program/

Linclude?stdio.hMLinclude?conio.hMint factint n- &&starting of the sub program

N int f#1!iO ifn##0-YYn##1-- && chec5 the condition for n value return1-O else fori#1Oi?#nOi((- && perform the looping operation for calculating the factorial f#f6iO

returnf-Ovoid main-N

int nO clrscr-O printfPenter the number :P-O

scanfP7dP!Rn-O printfPfactoria of number7dP!factn--O getch-O

,utut/

1.Enter the number: ; Factorial of number: 90)0". Enter the number: * Factorial of number: ;"0,. Enter the number: + Factorial of number: $"9"1*

4oncluion/

he program is error free

5I5A 6789ATI,N9/

1- What is meant b call b value @Ans: passing values to the function as arguments"- What is meant b call b reference @

Ans: passing address to the function as arguments,-define actual parameters @

Ans: he actual parameters often 5nown as arguments are specified in the function call.

"+


29/259

,bectie/ *

o find the /C of two given integers b using the recursive function

ecrition/

/C means /reatest Common ivisor. i.e the highest number which divides the

given numberE%: /C1"!")- is 1"Formula: /C# product of numbers& GC of numbers

Algorithm/main programStep 1: startStep ": read a!bStep ,: call the sub program /Ca!b- for print the valueStep ): stop

9ub rogram/

Step 1: if nMm return /Cn!m-Step ": if n##0 return m else goto step ,Step ,: return /C n!m7n-Step ): return to main program

"8


30/259

lo3chart/

Start

Bead a!b

Call sub program/#gcda!b-

=rint gcdvalue

Stop

/cd -

fnMm

Beturn/cd n!m-

f

n##0

Beturn m Call the same functionBeturn gcd n!m7n-


true false

trueF

,0


31/259

Program/

Linclude?stdio.hMLinclude?conio.hMint gcdrecursiveint m!int n- && starting of the sub program

N ifnMm- return gcdrecursiven!m-O

ifn##0- return mO

else return gcdrecursiven!m7n-O && return to the mainprogramvoid main-N int a!b!igcdO

clrscr-O printfPenter the two numbers whose gcd is to be found:P-O scanfP7d7dP!Ra!Rb-O printfP/C of a!b is 7dP!gcdrecursivea!b--O && return to the sub program getch-O,utut/

1. enter the two numbers whose gcd is to be found:9!"9 /C of a!b is : 9". enter the two numbers whose gcd is to be found:,*!9) /C of a!b is : 1+,. enter the two numbers whose gcd is to be found:11!1, /C of a!b is : 14oncluion/


5I5A 6789ATI,N9/

1- What is meaning of /C @Ans: /C means /reatest Common ivisor. i.e the highest number which divides

the given number"- efine scope of a variable @Ans: he scope of a variable can be define as the region over which the variable is

accessible,- Show an scope resolution operator @Ans: double colon//-

)- efine e%tent of a variable @ Ans: he period of time during which memor is associated with a variable is called

e%tent of the variable.

,1


32/259

,bectie/ +

o find the /C of two given integers b using the non recursive function

ecrition/

/C means /reatest Common ivisor. i.e the highest number which divides the

given numberE%: /C1"!")- is 1"Formula: /C# product of numbers& GC of numbers

Algorithm/

Step 1: startStep ": read a!bStep ,: call sub program g#/Ca!b-Step ): print the g valueStep 9: stopSub program:

Step 1: initialiIe the p#1! q! remainderStep ": remainder#p$p&q6q-Step ,: remainder#0 return q else goto step )Step ): /Cq!remainder- return to main program

,"


33/259

lo3chart/

start

Bead a! b

Call subprogram g#gcda!b-

output

stop

/cd -

Bemainder#p$p&q6q-

fremainder##0

/cdq!remainder- Beturn q

Beturn to main program

false rue

,,


34/259

Program/

Linclude?stdio.hMLinclude?conio.hMLinclude?math.hMint gcdnonrecursiveint m!int n-

N int remainderO remainder#m$m&n6n-O ifremainder##0- return nO else gcdnonrecursiven!remainder-O

void main-N

int a!b!igcdO clrscr-O printfPenter the two numbers whosegcd is to be found:P-O scanfP7d7dP!Ra!Rb-O printfP/C of 7dP!gcdnonrecursivea!b--O getch-O

,utut/

1. enter the two numbers whose gcd is to be found:9!"9 /C of a!b is : 9". enter the two numbers whose gcd is to be found:,*!9) /C of a!b is : 1+,. enter the two numbers whose gcd is to be found:11!1, /C of a!b is : 1

4oncluion/


5I5A 6789ATI,N9/

1$What is meaning of /C @Ans:

/C means /reatest Common ivisor. i.e the highest number which divides thegiven number

,)


35/259

,bectie /10

o solve the towers of 2anoi problem b using the recursive functionecrition/

owers of 2anoi problem means we have three towers

2ere source !intermediate and destination are the three towers. We have totransfer all the dis5s from source to destination towers. 2ere the restriction is not

to place a big dis5 on smaller one . for this we use intermediate tower. Finall thearrangements in the destination tower must be as same as the dis5s in the sourcetower at first.

Algorithm/ main rogram

Step 1: startStep ": initialiIe the source#a! intermediate#c! destination # dStep ,: read nStep ): call the sub program 2anoi recursion n value!a !b! c-Step 9: stopSub program:

Step 1: if n## 1 call the sub program 2anoi recursion num$1! a! c! b-

Step ": print the output from a to bStep ,: call the sub program 2anoi recursionnum$1! b! c! a-Step ): return to main program

Source intermediate destination

,9


36/259

lo3chart/

SAB

SKJBCE # AEBEAE # CESAK # H

BEA n

Call subprogram2anoinum!source!intermediate!destination- A!C! H

Stop

=rint A!C


hanoirecursive -

f num##1

=rint A!C

Call ubprogram2anoinum$1!H!C!A--

true false

Call sbgroram

um$1!source ZA[

,*


37/259

Program/

Linclude?stdio.hMLinclude?conio.hMvoid 2anoirecursionint num!char ndl1!char ndl"!char ndl,-N

ifnum##1- N printfPove top dis5 from needle 7c to needle 7cP!ndl1!ndl"-O returnO 2anoirecursionnum$1!ndl1!ndl,!ndl"-O printfPove top dis from needle 7c to needlle 7cP!ndl1!ndl"-O 2anoirecursionnum$1!ndl,!ndl"!ndl1-O

void main-

N int noO clrscr-O printfPEnter the no. of dis5 to be transferred:P-O scanfP7dP!Rno-O ifno?1- printfPQn here\s nothing to moveP-O else printfPQn recursiveP-O 2anoirecursionno!\A\!\H\!\C\-O getch-O,utut/

1. Enter the no. of dis5 to be transferred :, ove top dis5 from needle a to needle b

ove top dis5 from needle a to needle c ove top dis5 from needle b to needle c ove top dis5 from needle a to needle b ove top dis5 from needle c to needle a ove top dis5 from needle c to needle b ove top dis5 from needle a to needle b4oncluion/

he program is error free5I5A 6789ATI,N9/

1$ #hat i uroe of to3er of ;anoi :

An/ We have to transfer all the dis5s from source to destination towers. 2ere therestriction is not to place a big dis5 on smaller one . for this we use intermediatetower. Finall the arrangements in the destination tower must be as same as thedis5s in the source tower at first.

,;


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,bectie /11

o solve the towers of 2anoi problem b using the non recursive functionecrition/

owers of 2anoi problem means we have three towers

2ere source !intermediate and destination are the three towers. We have totransfer all the dis5s from source to destination towers. 2ere the restriction is not

to place a big dis5 on smaller one . for this we use intermediate tower. Finall thearrangements in the destination tower must be as same as the dis5s in the sourcetower at first.

Algorithm/

Step 1: startStep ": declare the noStep ,: read the no value

Step ): if no?1- =rint nothing to moveElse =rint nonrecursion

Step 9:2anoi non recursionno!.A4!4H4!4C4-Step *:stopSub program:

Step 1: eclare num!sndl!indl!dndl! st5n] ^!st5sndl] ^!st5dndl ] ^!st5add] ^! temp!top!addStep ": declare the top#JGGStep ,: one:

fnum##1-then=rint the out put value/oto four

Step ): two: op#top(1St5n]top^#numSt5sndl]top^#sndlSt5indl]top^#indlSt5dndl]top^#dndlSt5add]top^#,um#num$1

Source intermediate destination

,+


39/259

Sndl#sndlemp#indlndl#dndlndl#temp/oto one. /oto step ,

Step 9: hree:=rint the outputop#top(1St5n]top^#numSt5sndl]top^#sndlSt5indl]top^#indlSt5dndl]top^#dndlSt5add]top^#9um#num$1temp#sndl

sndl#indlndl#tempndl#dndl/oto one. /oto step ,

Step *:Four:

ftop##JGG-Beturn to main programum# st5n]top^Sndl# st5sndl]top^ndl# st5indl]top^ndl#st5dndl]top^Add#st5add]top^op#top$1fadd##,-/oto three. /oto step 9Elsefadd##9-/oto four. /oto step *.

Step ;: return to main program

,8


40/259

lo3 chart/

)0

Start

Bead no

f no? 1 =rint nothing tomove

true false

=rintnonrecursion

2anoinonrecursion no!.A4!4H4!4C4-

Stop


41/259

2anoi nonrecursion -

eclare num!sndl!indl!dndl! st5n] ^!st5sndl] ^!st5dndl ] ^!st5add] ! temp!top!add

op # JGG

f num##1=rint the value

true

ftop#JGG

um# st5n]top^Sndl# st5sndl]top^ndl# st5indl]top^

ndl#st5dndl]top^Add#st5add]top^op#top$1

fadd##,

false

fadd##9

op#top(1St5n]top^#numSt5sndl]top^#sndlSt5indl]top^#indlSt5dndl]top^#dndlSt5add]top^#,um#num$1Sndl#sndlemp#indl

ndl#dndlndl#temp

one

four

two

=rint value

op#top(1St5n]top^#numSt5sndl]top^#sndlSt5indl]top^#indlSt5dndl]top^#dndlSt5add]top^#9

um#num$1temp#sndlsndl#indlndl#tempndl#dndlop#top(1

)1


42/259

Program/

Linclude?stdio.hMLinclude?conio.hMvoid 2anoinonrecursionint num!char sndl!char indl!char dndl-N

char st5n]100^!st5sndl]100^!st5indl]100^!st5dndl]100^!st5add]100^!tempOint top!addOtop#JGGOone: ifnum##1- N printfPQn ove top dis5 from needle 7c to needle 7cP!sndl!dndl-O goto fourO two: top#top(1O

st5n]top^#numO st5sndl]top^#sndlO st5indl]top^#indlO st5dndl]top^#dndlO st5add]top^#,O num#num$1O sndl#sndlO temp#indlO indl#dndlO dndl#tempO goto oneO

three: printfPQn ove top dis5 from needle 7c to needle 7cP!sndl!dndl-O top#top(1O st5n]top^#numO st5sndl]top^#sndlO st5indl]top^#indlO st5dndl]top^#dndlO st5add]top^#9O num#num$1O temp#sndlO sndl#indlO indl#tempO dndl#dndlO goto oneO

four: iftop##JGG- returnO

)"


43/259

num#st5n]topÔ sndl#st5sndl]topÔ indl#st5indl]topÔ dndl#st5dndl]topÔ add#st5add]topÔ

top#top$1O ifadd##,- goto threeO else ifadd##9- goto fourO void main-N int noO clrscr-O printfPEnter the no. of diss to be transferred:P-O

scanfP7dP!Rno-O ifno?1- printfPQn here\s nothing to moveP-O else printfPQn nonrecursiveP-O 2anoinonrecursionno!\A\!\H\!\C\-O getch-O

,utut/

1.Enter the no. of diss to be transferred:, nonrecursiveove top dis5 from needle A to needle Cove top dis5 from needle A to needle Hove top dis5 from needle C to needle Hove top dis5 from needle A to needle Cove top dis5 from needle H to needle Aove top dis5 from needle H to needle Cove top dis5 from needle A to needle C4oncluion/he program is error freed

5I5A 6789ATI,N9/

1$ #hat i uroe of to3er of ;anoi :

An/ We have to transfer all the dis5s from source to destination towers. 2ere therestriction is not to place a big dis5 on smaller one . for this we use intermediatetower. Finall the arrangements in the destination tower must be as same as thedis5s in the source tower at first.

2$ #hat i an arra< :

Ans: An arra is a sequence of memor location of same data tpe.

),


44/259

,bectie/ 12

he total distance travelled b vehicle in \t\ seconds is given b distance #ut(1&"at" where \u\ and \a\ are the initial velocit m&sec.- and accelerationm&sec"-. Write C program to find the distance travelled at regular intervals oftime given the values of \u\ and \a\. he program should provide the fle%ibilit to

the user to select his own time intervals and repeat the calculations for differentvalues of \u\ and \a\.ecrition/

he total distance travelled b vehicle in \t\ seconds is given b distance #ut(1&"at" where \u\ and \a\ are the initial velocit m&sec.- and accelerationm&sec"-.

Algorithm/

Step 1:Start

Step" : Bead t !dt

Step ,: Set i to 1

Step ):Set 5 to dt

Step 9: Bead u!a

Step *: set s to u65(0.96d6565

Step ;: Write s

Step +: f5?#t- and i#1 thenHeginStep +.1 go to step *

And Else

HeginStep +." :readStep +., :ifU#0- thenHegin

Step +.,.1:Set to 0End

ElseHegin

Step +.,.": Set to 1Step +.,.,: go to step )

EndStep 8: Stop

Step 10: End

))


45/259

lo3chart/

)9


46/259

Program/

Linclude?stdio.hMmain-

N int a!u!t!t1!t"!iOfloat sOclrscr-OprintfPEEB 2E TAGJES KF a!u!t!t1!t":P-OscanfP7d7d7d7d7dP!Ra!Ru!Rt!Rt1!Rt"-Ofori#t1Oi?#t"Oi#i(t- && performing the looping operation for time intervalsN

s#u6i-(0.96a6i6i-O && calculate the total distanceprintfPQnQnthe distance travelled in 7d seconds is 7f P!i!s-O

getch-O

Inut=,utut/

1.EEB 2E TAGJES KF a!u!t!t1!t":1 " , 1 9 the distance travelled in 1 seconds is ".900000

the distance travelled in ) seconds is 1*.000000".EEB 2E TAGJES KF a!u!t!t1!t":0 1 " , ) the distance travelled in , seconds is ,.000000concluion/he program is error free

5I5A 6789ATI,N9/

1- 2ow man tpes of arras are there @Ans: hree tpes. he are one dimensional !two dimensional and multi dimensionalarrs

)*


47/259

,bectie/1%

wo integer operands and one operator form user! performs the operationand then prints the result.

Consider the operators (!$!6! &! 7 and use Switch Statement-ecrition/o ta5e the two integer operands and one operator from user to performthe some arithmetic operations b using the following operators li5e(!$!6! &! 7E%: "(,#9

Algorithm/

Step 1: Start

Step ": Bead the values of a!b and operator

Step ,: if the operator is 3(4 thenB#a(b/o to step +Hrea5

Step ): Else if the operator is 3$3 thenB#a$b/o to step +

Step 9: Else if the operator is 363 thenB#a6b/o to step +

Step *: Else if the operator is 3&3 thenB#a&b/o to step +

Step ;: Else if the operator is 373 thenB#a7b/o to step +

Step +: write B

Step 8:End

);


48/259

lo3chart/

)+


49/259

Program/

Linclude?stdio.hMmain-N

char opOfloat a!b!cOclrscr-OprintfPenter two operands:P-OscanfP7d7dP!Ra!Rb-OprintfPenter an operator:P-OscanfP 7cP!Rop-Oswitchop- && used to select particular case from the userNcase \(\:printfPsum of two numbers 7"d 7"d is: 7dP!a!b!a(b-O

brea5O

case \$\:printfPsubtraction of two numbers 7"d 7"d is:7dP!a!b!a$b-Obrea5O

case \6\:printfPproduct of two numbers 7"d 7"d is:7dP!a!b!a6b-O

brea5O case \&\:printfPquotient of two numbers 7"d 7"d is:

7dP!a!b!a&b-O brea5O

case \7\:printfPreminder of two numbers 7"d 7"d is:7dP!a!b!c-O

brea5O default:printfPplease enter correct operatorP-O

brea5O getch-O

Inut=,utut/

1.enter two operands:" , enter an operator:( sum of two numbers " , is: 9

".enter two operands:, ) enter an operator: $ subtraction of two numbers , ) is: $1

,.enter two operands:, 9

)8


50/259

enter an operator:6 product of two numbers , 9 is: 19

).enter two operands:9 " enter an operator:&

quotient of two numbers 9 " is: "9. enter two operands:9 " enter an operator:7 reminder of two numbers 9 " is: 1

concluion/he program is error free

5I5A 6789ATI,N9/

1- What are the various tpes of arithemetic operators@

Ans: addition (-! multiplication6-! subtraction $-! division&- ! modulo7-."- What are the tpes of relational operators @ Ans: less than?-! grater thanM-! less than or equal to?#-!equal to##-! etc..!

,- ,- What are the tpes of logical operators @Ans: logical A RR-! logical KBYY-! logical K'-

90


51/259

,bectie / 1&

o find both the largest and smallest number in a list of integersecrition/

his program contains n number of elements! in these elements we can find thelargest and smallest numbers and displa these two numbers

Algorithm/Step 1: startStep ": read nStep ,: initialiIe i#0Step ): if i?n do as follows. f not goto step 9

Bead a]i^ncrement i/oto step )

Step 9: min#a]0^! ma%#a]0^Step *: initialiIe i#0Step ;: if i?n do as follows. f not goto step +

f a]i^?minAssign min#a]i^ncrement i goto Step ;

Step +: print min!ma%Step 8: stop

91


52/259

Bead n!a]i^!min!ma%!

in # a]0^ a% # a]0^

=rintmin!ma%

# 0 i((

? n

a%?min

A]i^Mma

%

in#a]i^

a%#a]i^

# 0

?n i((

stop

Bead a]i^

Start

true

false

true

false

false

true

lo3chart/

9"


53/259

Program/

Linclude?stdio.hMvoid main-N int a]10^!i!n!min!ma%O

clrscr-O printfPenter the arra siIe:P-O scanfP7dP!Rn-O printfPEnter the elements of arraP-O fori#0Oi?nOi((- && read the elements of an arra scanfP7dP!Ra]i^-O min#a]0Ô ma%#a]0Ô fori#0Oi?nOi((-&& read the elements of an arra N ifa]i^?min-&& chec5 the condition for minimum value

min#a]iÔ ifa]i^Mma%-&&chec5 the condition for ma%imum value ma%#a]iÔ printfPma%imum value is:7dQnP!ma%-O printfPminimum value is:7dQnP!min-O getch-O,utut/

1.enter the arra siIe:) Enter the elements of arra ,* 1, " )9 ma%imum value is:)9 minimum value is:"".enter the arra siIe:9 Enter the elements of arra * " 1 , + ma%imum value is:+ minimum value is:1,.enter the arra siIe:9 Enter the elements of arra$* 8 $8 " 9 ma%imum value is:8 minimum value is:$8concluion/the program is error free5I5A 6789ATI,N9/

1$ What is an arra:Ans: he collection of similar elements is called arra

"- 2ow man tpes of arras are there @Ans: hree tpes. he are one dimensional !two dimensional and multi dimensionalarrs

9,


54/259

,bectie/1'

o perform the addition of two matricesecrition/

program ta5es the two matri%es of same siIe and performs the addition an alsota5es the two matri%es of different siIes and chec5s for possibilit of

multiplication and perform multiplication if possible.algorithm/Step 1: startStep ": read the siIe of matrices A!H < m!nStep ,: read the elements of matri% AStep ): read the elements of matri% HStep 9: select the choice for ou want. f ou select case 1 then goto matric

addition. Else goto Step ;.Step *: print Sum of matri% A and HStep ;: if ou select case " then goto matri% multiplicationStep +: chec5 if n#p! if not print matrices can not be multiplied

Step 8: Ktherwise perform the multiplication of matricesStep 10: =rint the resultant matri%Step 11: Stop

9)


55/259

start

Bead the siIeof A!H matri%m!n

#0

? r1 i((

eclare a]^]^!b]^

]^!c]^]^!ch!!U!5!m!n!p!q!r1!c1

Bead the choice ch

f ch?#"RchM0

=rintvalidchoice

Switchch

Bead the siIeof A matri%m!n

Bead A]i^]U^

V#0

U? c1 Ui((

#0

? r1 i((

U#0

U?c1 U((

Bead H]i]U^

#0

? r1 i((

U#0

U?c1 U((

Bead A]i^]U^

Bead siIe ofmatri% H: p.q

#0

? p i((

H

Case 1Case "

lo3 chart/

99


56/259

H

#0

? r1 i((

U#0

U? c1 U((

=rint A]i^]U^(H]i^]U^

Stop

H

U#0

U? q U((

Bead H]i^]U^

fn##p

atri% cannotbe mutiplied

i#0

i?m i((

U#0

U? q U((

C]i^]U^#0

50

U? n 5((

C]i^]U^#C]i^]U^(A]i^]5^6H]5^]U^

i#0

i?m i((U((

U#0

U?q U((

=rint C]i^]U^

A

9*


57/259

Program/

Linclude?stdio.hM

void main-Nint ch!i!U!m!n!p!q!5!r1!c1!a]10^]10^!b]10^]10^!c]10^]10Ôclrscr-OprintfP666666666666666666666666666666666666P-OprintfPQnQtQtEJP-OprintfPQn6666666666666666666666666666666666P-OprintfPQn]1ÂK KF WK ABCESP-OprintfPQn]"^JG=GCAK KF WK ABCESP-OprintfPQn]0Ê_P-OprintfPQn6666666666666666666666666666666666P-O

printfPQnQtEnter our choice:QnP-OscanfP7dP!Rch-O

ifch?#" R chM0-N

printfPTalid ChoiceQnP-O

switchch-N

case 1: printfPnput rows and columns of A R H atri%:P-O scanfP7d7dP!Rr1!Rc1-O

printfPEnter elements of matri% A:QnP-O fori#0Oi?r1Oi((- N forU#0OU?c1OU((- scanfP7dP!Ra]i^]U^-O

printfPEnter elements of matri% H:QnP-O fori#0Oi?r1Oi((- N forU#0OU?c1OU((- scanfP7dP!Rb]i^]U^-O printfPQn #####atri% Addition#####QnP-O fori#0Oi?r1Oi((- N ForU#0OU?c1OU((- printfP79dP!a]i^]U^(b]i^]U^-O

9;


58/259

printfPQnP-O brea5O

case ":

printfPnput rows and columns of A matri%:P-O scanfP7d7dP!Rm!Rn-O printfPnput rows and columns of H matri%:P-O scanfP7d7dP!Rp!Rq-O ifn##p- N printfPmatrices can be multipliedQnP-O printfPresultant matri% is 7d67dQnP!m!q-O printfPnput A matri%QnP-O read`matri%a!m!n-O printfPnput H matri%QnP-O

&6Function call to read the matri%6& read`matri%b!p!q-O &6Function for ultiplication of two matrices6& printfPQn #####atri% ultiplication#####QnP-O fori#0Oi?mO((i-

forU#0OU?qO((U- N c]i^]U^#0O for5#0O5?nO((5-

c]i^]U^#c]i^]U^(a]i^]5^6b]5^]U^O

printfPBesultant of two matrices:QnP-O write`matri%c!m!q-O &6end if6&

else N

printfPatrices cannot be multiplied.P-O

&6end else6& brea5O

case 0: printfPQn Choice erminatedP-O e%it-O brea5O

default: printfPQn nvalid ChoiceP-O

9+


59/259

getch-O

&6Function read matri%6&

int read`matri%int a]10^]10^!int m!int n- N int i!UO fori#0Oi?mOi((-

forU#0OU?nOU((-scanfP7dP!Ra]i^]U^-O return 0O

&6Function to write the matri%6&int write`matri%int a]10^]10^!int m!int n-

N int i!UO fori#0Oi?mOi((- N

forU#0OU?nOU((-printfP79dP!a]i^]U^-OprintfPQnP-O

return 0O

,utut/

1.666666666666666666666666666666666666 EJ6666666666666666666666666666666666]1^AK KF WK ABCES]"^JG=GCAK KF WK ABCES]0^E_6666666666666666666666666666666666 Enter our choice:1Talid Choicenput rows and columns of A R H atri%:""Enter elements of matri% A:"""

98


60/259

"Enter elements of matri% H:"""

"

#####atri% Addition##### ) ) ) )666666666666666666666666666666666666 EJ6666666666666666666666666666666666]1ÂK KF WK ABCES]"^JG=GCAK KF WK ABCES]0Ê_

6666666666666666666666666666666666 Enter our choice:"Talid Choicenput rows and columns of A matri%:",nput rows and columns of H matri%:""atrices cannot be multiplied.

666666666666666666666666666666666666 EJ6666666666666666666666666666666666]1ÂK KF WK ABCES]"^JG=GCAK KF WK ABCES]0Ê_6666666666666666666666666666666666 Enter our choice:"Talid Choicenput rows and columns of A matri%:""nput rows and columns of H matri%:""matrices can be multipliedresultant matri% is "6"nput A matri%"""

*0


61/259

"nput H matri%"""

"

#####atri% ultiplication#####Besultant of two matrices: + + + +4oncluion /he program is error free

5I5A 6789ATI,N9/

1- What is condition for performing an matric addition @

Ans: program ta5es the two matri%es of same siIe and performs the addition"- What is condition for performing an matric addition @Ans: he two matri%es of different siIes and chec5s for possibilit of multiplicationand perform multiplication if possible

*1


62/259

,bectie /1(

Functions to insert a sub string into given main string from a given position

ecrition/in this program we need to insert a string into another string from a specified

position.Algorithm/Step 1: startStep ": read main string and sub stringStep ,: find the length of main stringr-Step ): find length of sub stringn-Step 9: cop main string into sub stringStep *: read the position to insert the sub string p-Step ;: cop sub string into main string from positionp$1Step +: cop temporar string into main string from position p(n$1Step 8: print the strings

Step 10: stop

*"


63/259

Start

Bead the strings A

R H

# 0

? r

C]i^ # A]i^

((

S # n ( rK # p ( n

#p ((

? s

_# C]i^

f t ? n

A

H

lo3 chart/

*,


64/259

A

A]i^ # H]t^

#t(1

A]o^#%

K # o ( 1

H

=rint output

Stop

*)


65/259

Program/

Linclude ?stdio.hMLinclude ?conio.hMLinclude ?string.hM

void main-N

char a]10Ôchar b]10Ôchar c]10Ôint p#0!r#0!i#0Oint t#0Oint %!g!s!n!oOclrscr-O

putsPEnter First String:P-Ogetsa-OputsPEnter Second String:P-Ogetsb-OprintfPEnter the position wherethe item has to be inserted: P-OscanfP7dP!Rp-Or # strlena-On # strlenb-Oi#0O

&& Coping the input string into another arrawhilei ?# r-N

c]i^#a]iÔi((O

s # n(rOo # p(nO

&& Adding the sub$stringfori#pOi?sOi((-N

% # c]iÔift?n-N

a]i^ # b]tÔ t#t(1O

a]o^#%Oo#o(1O

*9


66/259

printfP7sP! a-Ogetch-O

,utut/

1.enter first string: computer".enter second string: gec,.enter the position where the item has to be inserted:, comgecputerconcluion /the program is error free

5I5A 6789ATI,N9/

1- What is string @Ans: A string is an collection of characters"- Which command is used to combined the two strings @Ans: Strcat-,- Which command is used to cop the strings @Ans: H using the strcp- function copies one string to another

**


67/259

,bectie/1)

o delete n characters from a given position in a given stringecrition/

in this program we need to delete a string from the given string at a specifiedposition.

Algorithm/Step 1: startStep ": read stringStep ,: find the length of the stringStep ): read the value of number of characters to be deleted and positionedStep 9: string cop part of string from position to end! and position(number of

characters to end-Step *: stop

*;


68/259

Start

Bead string

Bead position! no ofcharacters

elchar string! n! pos-

Stop

Subprogramelchar -

f a(b$1`?#

strlen%--

StrcpR%]b$1^!R%]a(b$1^-

=uts%-

Beturn tomainprogram

lo3 chart/

*+


69/259

Program/


void delcharchar 6%!int a! int b-O

void main-N char string]10^O int n!pos!pO clrscr-O

putsPEnter the stringP-O getsstring-O printfPEnter the position from where to deleteP-O

scanfP7dP!Rpos-O printfPEnter the number of characters to bedeletedP-O scanfP7dP!Rn-O delcharstring! n!pos-O getch-O

&& Function to delete n charactersvoid delcharchar 6%!int a! int b-N if a(b$1- ?# strlen%--

N strcpR%]b$1^!R%]a(b$1^-O puts%-O ,utut/

1.enter the string nagaraUu Enter the position from where to delete:) Enter the number of charcters to be deleted, nagUu". enter the string 5aliraUu Enter the position from where to delete:0 Enter the number of charcters to be deleted) BaUu4oncluion/the program is error free

*8


70/259

5I5A 6789ATI,N9/

1- Which command is used to delete the strings @Ans: delstr-O"- What are the various tpes of string functions @Ans: Strcat-! strcp-! delstr-! substr- !strlen-etc..!

;0


71/259

,bectie/ 1*

o determine if the given string is a palindrome or not

ecrition/

if the reverse of a string is equal to original string then it is called palindrome

Algorithm/

Step 1:startStep ": read the stringStep ,: store reverse of the given string in a temporar stringStep ): compare the two stringsStep 9: if both are equal then print palindromeStep *: otherwise print not palindromeStep ;: stop

;1


72/259

Start

Bead string

fspalindromestring-

spalindromestring-

=rintpalindrome

true

=rint notpalindrome

Stop

lo3 chart/

;"


73/259

spalindrome -

Enum Hoolean matched# true

f len## 0

Beturn 0

Geft # 0

Bight#len$1

Whileleft?rightRRmatched

fstring]left^'#string]right^-

atched#false

true

Geft ((Bight $$

Beturnmatched


;,


74/259

Program/

Linclude?stdio.hMLinclude?string.hM

enum HooleanNfalse!trueOenum Hoolean s=alindromechar string]^-N int left!right!len#strlenstring-O enum Hoolean matched#trueO

iflen##0- return 0O left#0O

right#len$1O

&6 Compare the first and last letter!second R second last R so on 6& whileleft?rightRRmatched- N

ifstring]left^'#string]right^- matched#falseO else N left((O right$$O return matchedO

int main-N char string])0^O clrscr-O printfP6666=rogram to test if the given string is a palindrome6666QnP-O printfPEnter a string:P-O scanfP7sP!string-O ifs=alindromestring-- printfPhe given string 7s is a palindromeQnP!string-O else printfPhe given string 7s is not a palindromeQnP!string-O getch-O return 0O

;)


75/259

,utut/

1. Enter the string:malaalam he given string malaalam is a palindrome". Enter the string:india he given string india is not a palindrome


5I5A 6789ATI,N9/

1- What is meant b palindrome @Ans: f the reverse of a string&number is equal to original string& number then it iscalled palindrome.

"- What is the use of gets- function @Ans: o read the string at a time

,- What is the use of puts- function @Ans: o write the string at a time

;9


76/259

,bectie/ 1+

=rogram that displas the position or inde% in the string S where the string

begins ! or $1 if S doesn4t contain Algorithm/Step 1: startStep ": read the string and then displaedStep ,: read the string to be searched and then displaedStep ): searching the string in string S and then perform the following steps

i. found#strstrS!-ii. if found print the second string is found in the first string at the

position. f not goto step 9Step 9: print the $1Step *: stop

;*


77/259

Start

nitialiIe s] ^!t] ^! found variables

Bead first string

ispla thestring

Bead string to besearched

ispla thestring

Found#strstrs!t-

f found

=rint $1=rint the stringFound$s

Stop

no es

lo3 chart/

;;


78/259

Program/

Linclude?stdio.hMLinclude?string.hMLinclude?conio.hM

void main-N char s],0^! t]"0^O

char 6foundO clrscr-O

&6 Entering the main string 6& putsPEnter the first string: P-O getss-O

&6 Entering the string whose positionor inde% to be displaed 6& putsPEnter the string to be searched: P-O getst-O

&6Searching string t in string s 6& found#strstrs!t-O iffound- printfPSecond String is found in the First String at 7d position.QnP!found$s-O else printfP$1P-O getch-O

,utut/

1.enter the first string: 5ali Enter the string to be seareched: li second string is found in the first string at"position".enter the first string: nagaraUu Enter the string to be seareched: raUu second string is found in the first string at)position,.enter the first string: nagarUuna Enter the string to be seareched:

ma $14oncluion/he program is error free

;+


79/259

5I5A 6789ATI,N9/

1- What is the difference between printf- and puts- @Ans: puts- is used to displa the string at a time and it doesn4t ta5e an integersvalues but printf- ta5es an values as defined b the user

"- define pointer variable @Ans: pointer variables are defined as variable that contain the memor addresses ofdata or e%ecutable code.,- What is use of the strcmp- function @Ans: his function compares two strings character b character and returns a value 0if both strings are equal and non Iero value if the strings are different.

;8


80/259

,bectie/ 20

o count the lines!words R charcters in a given te%tecrition/

n this program we have to count the no of lines! no of words and no ofcharacters in a given program or given te%t b using the string function

Algorithm/Step 1: StartStep ": Bead the te%t until an empt lineStep ,: Compare each character with newline char 3Qn4 to countno of linesStep ): Compare each character with tab char 3QtQ4 or space char 3 3 to count no

of wordsStep 9: Compare first character with JGG char 3Q04 to find the end of te%tStep *: o of characters # length of each line of te%tStep ;: =rint no of lines! no of words! no of charsStep +: Stop

+0


81/259

Start

nitialiIe end#0!chars#0!words#0!lines#0

WhileEnd##0

C # 0

true

f ctr#getchar--'#4Qn4

false

Gine]c^#4Q04

Gine]c((^#ctr

true

fline]0^#4Q04

true

Words (( # 0

(( line]i^'#4Q0Q

false

f line]i^##4 3YYGine]i^##4Qt4

Gines((Chars(#strlenline-

Words ((

=rint lines!Words!chars

stop

lo3 chart/

+1


82/259

Program/

Linclude ?stdio.hMmain-N

char line]+1^! ctrO

int i!c! end # 0!characters # 0!words # 0!lines # 0O

printfPEX 2E E_.QnP-OprintfP/TE KE S=ACE AFEB EAC2 WKB.QnP-OprintfPW2E CK=GEE! =BESS \BEJB\.QnQnP-Owhile end ## 0-N

&6 Beading a line of te%t 6&

c # 0Owhilectr#getchar-- '# \Qn\-line]c((^ # ctrO

line]c^ # \Q0\O&6 counting the words in a line 6&ifline]0^ ## \Q0\-

brea5 OelseN

words((Ofori#0O line]i^ '# \Q0\Oi((-

ifline]i^ ## \ \ YY line]i^ ## \Qt\-words((O

&6 counting lines and characters 6&lines # lines (1Ocharacters # characters ( strlenline-O

printf PQnP-OprintfPumber of lines # 7dQnP! lines-OprintfPumber of words # 7dQnP! words-OprintfPumber of characters # 7dQnP! characters-O

,utut

1.EX 2E E_. /TE KE S=ACE AFEB EAC2 WKB. W2E CK=GEE! =BESS \BEJB\. Admiration is a ver short$lived passion. Admiration involves a glorious obliquit of vision. Alwas we li5e those who admire us but we do not

+"


83/259

li5e those whom we admire. Fools admire! but men of sense approve. umber of lines # 9 umber of words # ,* umber of characters # "09


5I5A 6789ATI,N9/

1- What is use of strlen- @Ans: to read a string length"- what is the use of getc- function @Ans: o read the character one b one.,- What is the use of strstr - @

Ans: he function strstr- searches one string for the occurrence of another.t accepts

two strings as parameters and searches the first string for an occurrence of thesecond

+,


84/259

,bectie 21/

o generate =ascal4s triangleecrition/

=ascal4s triangle which is used for a coefficient in the equation in

polnominals.Alogrithm/ Step 1: Start

Step ": nitialiIe m#0Step ,: Bead nStep ): f m?n goto step 9.if not goto step 1"Step 9: initialiIe i#)0$mStep *: f iM0 is true do as follows. f not goto step ;

i. print white spaceii. decrement i

iii. goto Step *

Step ;: nitialiIe U#0Step +: f U#m do as follows. f not goto Step 10i- ifU##0YYm##0-ii- nitialiIe b#1 if not b#b6m$U(1-&Uiii- =rint white space! b .iv- /oto Step 8

Step 8: increment U! goto Step +Step 10: print new line controlStep 11: increment m! goto step )Step 1": Stop

+)


85/259

lo3 chart/

Start

Bead p

#0 ? p

((

trueB#)0$ r$$

rM0

=rint whitespace

true

_#0 %?#i

_((

false

true

f %##0YY

##0

falsetrue

H#1 H#b6i$%(1-&%

=tint b

F

=rintnewlin

e

Stop

false

+9


86/259

Program/

Linclude?stdio.hMLinclude?conio.hMvoid main-

N int i!p!r!%!binom#1O clrscr-O

printfPenter the how man lines to printP-O scanfP7dP!Rp-O

i#0O whilei?p- && chec5 the condition

N forr#)0$iOrM0Or$$- && perform the looping operation until 0 printfP P-O for%#0O%?#iO%((-

N if%##0-YYi##0-- && chec5 the conditionbinom#1Oelsebinom#binom6i$%(1-&%OprintfP7dP!binom-OprintfP P-O

printfPQnP-O i((O

getch-O

,utut/

1.enter the how man lines to print9 1 1 1 1 " 1 1 , , 1 1 ) * ) 1

".enter the how man lines to print, 1 1 1 1 " 14oncluion/the program is error free

+*


87/259

5I5A 6789ATI,N9/

1$ What is meant b =ascal4s triangle @

An/=ascal4s triangle which is used for a coefficient in the equation in polnominals"-define structure @

Ans: A structure in c is a heterogenous user efined data tpe. A structure ma containdifferent data tpes.t groups variables into a single entit.

+;


88/259

,bectie 22: o construct a pramid of numbersecrition/

n this program the we have to construct output in the pramid shape mannerAlgorithm/

Step 1: Start Step": initialiIe the num!!! %#,9 Step,: read the num Step):perform the loop operation

For#0O?#numO((- Step9:/oto%%!(1- Step*: perform the loop operation for displaing digits towards the left and right Fori#0$Oi?#Oi((- Step;: print absi-O Step+: %#%$"O Step8: Stop

++


89/259

lo3 chart/

Start

nitialiIenum!!!%#,9

Bead num

X # 0 ?#num

((

true

/oto%%!(1-

# 0 i((

?#

=rintabsi-

_#%$"

Stop

+8


90/259

Program/


void main-N int num!i!!%#,9O

clrscr-OprintfPQnEnter the number to generate the pramid:QnP-OscanfP7dP!Rnum-O

for#0O?#numO((-N

&6%$coordinate!$coordinate-6& goto%%!(1-O

&6for displaing digits towards the left and right of Iero6& fori#0$Oi?#Oi((-

printfP7,dP!absi--O %#%$,O

getch-O

,utut/

1.enter the number: 0) 1 0 1 " 1 0 1 " , " 1 0 1 " , ) , " 1 0 1 " , )

".enter the number: 0 , 1 0 1 " 1 0 1 " , " 1 0 1 " ,4oncluion/he program is error free5I5A 6789ATI,N9/

1- What is the use of dot operator in structures @Ans: he use of dot.- operator to access the members of a structure independentl.he dot operator connects a member with the structure variable."- efine unions @Ans: A union is a data tpe in c which allows the overla of more than one variable inthe same memor area.

80


91/259

,bectie 2%/

o read in two numbers % and n and then compute the sum of this geometricprogression 1(%(%"(%,(>>>.(%n

ecrition/n this program we have to read the two numbers and the calculate the sum of this

geometric progression in above mention .Algorithm/Step 1: StartStep ": read values of % and n! sum$1! i#1Step ,: chec5 for n R _

i- if n?#0 YY %?#0ii- print values are not validiii- read values of % and n

Step ): perform the loop operationi- fori#1Oi?#nOi((- then followsii- sum#sum(pow%!i-

Step 9: print sumStep *: Stop

81


92/259

Start

nitialiIe

Sum!!%!n

Bead %!n

fn?#0YY%?#0

=rint notvalid

#1 i?#n

((

Sum#sum(pow%!i-

=rint sum

Stop

false

truefalse

Sum # 1

true

lo3 chart/

8"


93/259

Program/


Linclude?math.hM

void main-N

int s`sum!i!%!nO

clrscr-OprintfPEnter the values for % and n:P-OscanfP7d 7dP!R%!Rn-O

ifn?#0 YY %?#0-

N printfPTalue is not validQnP-OelseN

printfPTalue is validQnP-O s`sum#1O fori#1Oi?#nOi((-

N s`sum#s`sum(pow%!i-O

printfPSum of series#7dQnP!s`sum-O

getch-O

,utut/

1.Enter the values for % and n:" , Talue is valid Sum of series#19".Enter the values for % and n:) 8 Talue is valid Sum of series#"1+)9,.Enter the values for % and n:0 1 Talue is not valid4oncluion/the program is error free

8,


94/259

5I5A 6789ATI,N9/

1- what are the difference between structures and unions @Ans: 2ere the maUor difference is with in the structure all elements must be allocatedmemor. Hut in union highest memor allocation must be allocated the all these

elements.

8)


95/259

,bectie 2&/

o convert the given binar number to "4s complementecrition/

n this program the given binar number is first covert the numbers 0 to1 and

1 to 0. And finall add the 1 to the converted number. hen we will get the"4s complement number.Algorithm/ main program

Step 1: StartStep ": declare the subprogram Zcomplementchar 6a-[Step ,: initialiIe the variable iStep ): read the binar numberStep 9: perform the loop operation. if it is true then follows. if not goto step ;

i- fori#0Oa]i^'#4Q04Oi((-ii- ifa]i '#404RRa]i^'#414- then displaed the number is not valid.

enter the correct number.

iii- E%it the loopStep *: call sub program 3complemta-4Step ;: stop9ub rogram/

Step 1: initialiIe the variable !c#0!b]1*0Step ": 1#strlena-Step ,: perform the loop operation. if it is true then follows. if not goto

i-fori#l$1OiM#0Oi$$-ii-ifa]i^##404- then b]i^#414 elseiii-b]i^#404

Step ): fori#l$1OiM#0Oi$$- is truei- ifi##l$1- thenii- ifb]i^##404- then b]i^#414 elseiii- b]i^#404!c#1 if not goto step 9

Step 9: ifc##1RRb]i^##404- is true theni- b]i^#414! c#0 if not goto Step *

Step *: ifc##1RRb]i^##414- then b]i^#404!c#1Step ;: displaed b]l^#4Q04Step +: print b and return to main program

89


96/259

lo3 chart/

Start

nitialiIe ! a] ^

Bead a

# 0 i((

A]i^'#4Q04

true

f a]i^'#404RRa]i^'#414

umber isnot valid

true

false

Call sub programComplementa-

false

Stop

8*


97/259

Complement-

nitialiIe

!!c#0!b] ^

1#strlena-

#i$1 $$

iM#0

true

fa]i^##404

H]i^#414

true

H]i^#^0^

false

#l$1 i$$

iM#0

true

f i##l$1

true

fb]i^##404

H]i^#414

true b]i^#404

C#1

false

f

c##1RRH]i^##404

H]i^#414 C # 0

A H

false

true

C

8;


98/259

Program/

Linclude ?stdio.hMLinclude?conio.hM

void complementchar 6a-Ovoid main-N

char a]1*^Oint iO

clrscr-OprintfPEnter the binar numberP-Ogetsa-Ofori#0Oa]i^'#\Q0\O i((-N

if a]i '#\0\ RR a]i^'#\1\-

N printfPhe number entered is not a binar number. Enter thecorrect numberP-O

e%it0-O

complementa-Ogetch-O

A

f c##1RRb]i^##1

H]i^#40^C#1

true

H

C

H]i^#4Q04

=rint 3b4

Beturn tomainprogram

8+


99/259

void complement char 6a-N int l! i! c#0O char b]1*^O

l#strlena-Ofor i#l$1O iM#0O i$$-N

if a]i^##\0\- b]i^#\1\O else b]i^#\0\O

fori#l$1O iM#0O i$$-N

ifi##l$1-

N if b]i^##\0\- b]i^#\1\O else

N b]i^#\0\O

c#1O

else

N ifc##1 RR b]i^##\0\- N b]i^#\1\O c#0O

else if c##1 RR b]i^##\1\-N

b]i^#\0\O c#1O

b]l^#\Q0\OprintfPhe "\s complement is 7sP! b-O

88


100/259

,utut/

1.Enter the binar number101010he "\s complement is 010110

Enter the binar number11111he "\s complement is 00001Enter the binar number""""he number entered is not a binar number. Enter the correct number4oncluion/the program is error free

5I5A 6789ATI,N9/

1- E%pand ASC @Ans: American standarad code for information interchange"-What is binar number @Ans: he number which contains onl 0 and 1 is called binar number.,- efine "[s complement @Ans: he given binar number is first covert the numbers 0 to1 and 1 to 0. And finalladd the 1 to the converted number. hen we will get the "4s complement number.

100


101/259

,bectie 2'/

o convert roman number to it4s decimal equivalentecrition/

n this program we have to ta5e the roman value. his value is converted into ait4s equivalent decimal number.

E%: _#10Algoritham/Step 1: StartStep ": read the roman numerical as stringStep ,: find length of roman numericalStep ): for each charcter in the string

i- ifchar#- then decimal#1ii- ifchar#T- then decimal#9iii- ifchar#_- then decimal#10iv- ifchar#G- then decimal#90v- ifchar#C- then decimal#100

vi- ifchar#- then decimal#900vii- ifchar#- then decimal#1000viii- otherwise invalid character

Step 9: repeat step ) until the length of the stringStep *: 5#char]length$1^Step ;: for each character of decimal string

i- ifdecimal]i^Mdec]i$1^- then 5#5$decimal]i$1^ii- else ifdecimal]i^#decimal]i$1 or decimal]i^?decimal]i$1- then

5#5(decimall]i$1^Step +: repate step ; until the length of decimal stringStep 8: print decimal valueStep 10: Stop

101


102/259

lo3 chart/

Start

Bead

roman value

Gen#strlenrom-

#0 i((

? len

from]i^##44

A]i^#1

rue

false

from]i^##4T4

A]i^#9

true

false

from]i^##4_4

A]i^#10

true

false

H

C

#a]len$1^

#len$1 i

M 0

true

A]i^Ma]i$

1^

true5#5$a]i$1^

F

A]i^##a]i$1^YYa]i^?a]i$1^

true

5#5(ai$1^

10"

A


103/259

H

from]i^##4G4

A]i^#90

true

false

from]i^##4C4

A]i^#100

true

false

from]i^##44

A]i^#900

true

from]i^##4C4

A]i^#1000

false

true

C

=rintinvalid

A

=rint 5

Stop

10,


104/259

Program/


Linclude?string.hMLinclude?stdlib.hM

void main-N

int 6a!len!i!U!5Ochar 6romO

clrscr-O

printfPEnter the Boman umeral:P-OscanfP7sP!rom-O

len#strlenrom-O

fori#0Oi?lenOi((- && loop will continue until is not graterthan length.N ifrom]i^##\\- a]i^#1O else ifrom]i^##\T\- a]i^#9O else ifrom]i^##\_\-

a]i^#10O else ifrom]i^##\G\- a]i^#90O

else ifrom]i^##\C\- a]i^#100O

else ifrom]i^##\\- a]i^#900O

else ifrom]i^##\\- a]i^#1000O else

N printfPQnnvalid TalueP-O getch-O e%it0-O 5#a]len$1^Ofori#len$1OiM0Oi$$- && loop will continue until lessthan Iero

10)


105/259

N ifa]i^Ma]i$1^- && chec5 the condition 5#5$a]i$1^O

else ifa]i ##a]i$1^ YY a]i^?a]i$1^- 5#5(a]i$1^O

printfPQnts ecimal Equivalent is:P-OprintfP7dP!5-Ogetch-O

,utut/

Enter the Boman umeral:

ts ecimal Equivalent is:900

Enter the Boman umeral:_

ts ecimal Equivalent is:10Enter the Boman umeral:",

nvalid Talue4oncluion/he program is error free

5I5A 6789ATI,N9/

1- What is difference between structure and unions @Ans : he amount of memor required to store a structure variable is the sum of siIe allthe members in addition to the padding btes that ma be provided b the compiler. ncase of a union the amount of memor required is the same as that required b its largestmember."- What are various operations performed on union @Ans: i-An union variable can be assigned to another union variable ii- A union variable can be passed to a function as a parameter iii- he address of the union variable can be e%tracted b using the address of

operator R-.

109


106/259

,bectie 2(/

o read the two comple% numbers and perform the addition and multiplicationof these two numbers.

ecrition/

n this program the comple% number means it contains the two parts . first one is

real part and second one is imaginar part"(,i-.b ta5ing these two comple%numbers we can perform the addition and multiplication operation.Algorithm/

Step 1: StartStep ": declare structure for comple% numbersStep ,: read the comple% numberStep ): read choiceStep 9: if choice#1 then addition operation will perform and it contains following

steps i- w.realpart # w1.realpart(w".realpartO

ii- w.imgpart # w1.imgpart(w".imgpartO goto step )

Step *: if choice#" then multiplication operation will perform and it containsfollowing steps i- w.realpart#w1.realpart6w".realpart-$w1.imgpart6w".imgpart-O

ii- w.imgpart#w1.realpart6w".imgpart-(w1.imgpart6w".realpart-O gotostep )

Step ;: if choice#0 then e%it operation will perform

Step +:if w.imgpartM0 then print realpart(imgpart else =rint realpart.

Step 8: Stop

10*


107/259

Start

eclare structure

Beadoption

foption#addor mul-

truefalse Bead real part of first numberimage part of first numberBead real part of second numberimage part of second number

foption#add

false

w.realpart#w1.realpart(w".realpartw.imgpart#w1.imgpart(w".imgpart f

option#mul

w.realpart#w1.realpart6w".realpart-w.imgpart#w1.imgpart6w".imgpart-f

w.imgpartM0

true

=rintw.realpart(w.img

part i

false

=rint realpart

Stop

lo3 chart/

10;


108/259

Program/

Linclude?stdio.hMLinclude?math.hMvoid arithmeticint opern-O

struct compNdouble realpartOdouble imgpartO

Ovoid main-Nint opernOclrscr-OprintfPQnQn QtQtQt66666 A EJ 66666P-OprintfPQnQn Select our option: Qn 1 : AQn " : JG=GXQn 0 : E_ QnQnQtQt Enter

our Kption ] ^QbQbP-OscanfP7dP!Ropern-OifopernM"-

N printfPinvalid optionP-O

else N switchopern- N

case 0: e%it0-O case 1: case ": arithmeticopern-O

default: main-O

getch-Ovoid arithmeticint opern-Nstruct comp w1! w"! wOprintfPQn Enter two Comple% umbers %(i-:Qn Beal =art of First umber:P-OscanfP7lfP!Rw1.realpart-OprintfPQn maginar =art of First umber:P-OscanfP7lfP!Rw1.imgpart-OprintfPQn Beal =art of Second umber:P-OscanfP7lfP!Rw".realpart-O

10+


109/259

printfPQn maginar =art of Second umber:P-OscanfP7lfP!Rw".imgpart-Oswitchopern-N &6addition of comple% number6&

case 1: w.realpart # w1.realpart(w".realpartO w.imgpart # w1.imgpart(w".imgpartO brea5O

&6multiplication of comple% number6& case ":

w.realpart#w1.realpart6w".realpart-$w1.imgpart6w".imgpart-O w.imgpart#w1.realpart6w".imgpart-(w1.imgpart6w".realpart-O brea5O if w.imgpartM0-

printfPQn Answer # 7lf(7lfiP!w.realpart!w.imgpart-OelseprintfPQn Answer # 7lf7lfiP!w.realpart!w.imgpart-O

getch-Omain-O

,utut/

66666 A EJ 66666

Select our option:1 : A" : JG=GX0 : E_ Enter our Kption ] 1^

Enter two Comple% umbers %(i-:Beal =art of First umber:"

maginar =art of First umber:"

Beal =art of Second umber:"

maginar =art of Second umber:"

Answer # ).000000().000000i

66666 A EJ 66666

108


110/259

Select our option:1 : A" : JG=GX0 : E_

Enter our Kption ] "^

Enter two Comple% umbers %(i-:Beal =art of First umber:"

maginar =art of First umber:"

Beal =art of Second umber:"

maginar =art of Second umber:"

Answer # 0.000000(+.000000i 66666 A EJ 66666


Enter our Kption ] ,^invalid option

66666 A EJ 66666


Enter our Kption ] 0^

4oncluion/he program is error free5I5A 6789ATI,N9/

1- efine structure @Ans: Structure is amethod for pac5ing data of different tpes. Structure help to organiIecomple% data in a more meaninigful wa.

"- What is use of ?math.hM header file @Ans: t is used to access the mathematical functions in programs.

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,bectie 2)/

=rogram which copies one file to anotherecrition/

n this program we have to use the file functions to perform the copoperation from one file to another file.

Algorithm/Step 1: StartStep ": read command line argumentsStep ,: chec5 if no of arguments #, or not. f not print invalid no of argumentsStep ): open source file in read modeStep 9: if JGG pointer! then print source file can not be openStep *: open destination file in write modeStep ;: if JGG pointer! then print destination file can not be openStep + : read a character from source file and write to destination file until EKFStep 8: Close source file and destination fileStep 10: Stop

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Start

Bead arguments

from command line

farg'

#,

=rint invalid noof arguments

truefalse

Fs#fopenarg]1^![r[

ffs#JG

G

true

=rint sourcefile can notbe opened

false

Ft#fopenarg]"^![w[

fft##JGG

=rint target filecan not be opened

Fclosefs-false

Ch#fgetcfs-

fch##EKF

Fclosefs-Fcloseft-

true

false

Fputcch!ft-

Stop

lo3 chart/

11"


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Program/

Linclude ?stdio.hMLinclude ?conio.hM

Linclude ?process.hM

void mainint argc! char 6argv]^-NFGE 6fs!6ftOchar chOclrscr-Oifargc'#,- N putsPnvalid number of arguments.P-O e%it0-O

fs # fopenargv]1^!PrP-Oiffs##JGG- N

putsPSource file cannotbe opened.P-Oe%it0-O

ft # fopenargv]"^!PwP-Oif ft##JGG- && chec5 the condition if the file pointer is JGG or not

NputsParget file cannot be opened.P-Ofclosefs-Oe%it0-O

while1-N ch#fgetcfs-O if ch##EKF- && chec5 the condition if the file is end or not

brea5O else fputcch!ft-O fclosefs-O fcloseft-O getch-O

11,


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,utut/

source.c this is source te%touput.c

Command line argumentssource.c ouput.csource.c this is source te%touput.c

this is source te%t

Command line argumentssource.cnvalid number of arguments.


5I5A 6789ATI,N9/

1- What is file @Ans: he collection of alphabets iscalled file

"- What are the various operations performed on the file @Ans: fopen-! fread-! fwrite-! fclose- etc..!

,- What is the use of file pointer @Ans: he file pointer must be used in subsequent operations on the file

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,bectie 2*/

o reverse the first n characters in afileecrition/

his program perform the reverse operation of n characters in the fileAlgorithm/

Step 1: StarStep ": read the command line argumentsStep ,: chec5 if arguments#, or not

f not print invalid no of argumentsStep ): open source file in read modeStep 9: if JGG pointer! then print file can not be openStep *: Store no of chars to reverse in 5

# 6argv]"^$)+Step ;: read the item from file stream using freadStep +: Store chars from last position to initial position in another stringtemp-Step 8: print the temp string

Step 10: Stop

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Start

Bead command

line args

fargc'#,

false

Fp#fopenargv]1^![r[

=rint invalidno of args

ffp##JG

G

=rint file cannot be

opened

true

false

5#6argv]"^$)+ n # freada!1!5!fp- a]n^#\Q0\ len#strlena-

# len$1 i

M#0

true

S]U^#a]i^

=rint s]U^

V((

false

S]U(1^#4Q04

Stop

lo3 chart/

11*


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Program/


Linclude ?process.hM

void mainint argc! char 6argv]^-N char a]19Ô char s]"0Ô char nO int 5O int U#0O int iO int lenO

FGE 6fpO

ifargc'#,- N putsPmproper number of arguments.P-O e%it0-O fp # fopenargv]1^!PrP-O iffp ## JGG- N

putsPFile cannot be opened.P-O e%it0-O

5#6argv]"^$)+O n # freada!1!5!fp-O a]n^#\Q0\O len#strlena-O fori#len$1OiM#0Oi$$- N s]U^#a]iÔ printfP7cP!s]U^-O U#U(1Os]U(1^#\Q0\Ogetch-O

11;


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,utut/

source.c this is sourceouput.c

Command line argumentssource.c ouput.csource.c this is source

ecruos si siht

Command line argumentssource.cnvalid number of arguments.


5I5A 6789ATI,N9/

1- Gist out the file handling functions @Ans: fopen-! fprintf-!fclose-!fscanf-!fgetc-!fputc-! etc..!

"- What is the use of fsee5- function @Ans: he function fsee5 sets the file pointer associated with a stream to a new position

,- What is use of the fflush- function @Ans: f the given stream has a buffered output! fflush writes the output of the stream tothe associate file.

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,bectie 2+/

=rogram that uses functions to perform the following operations on single lin5edlists. i- creation ii- insertion iii- deletion iv- traversalecrition/

n this program we have to create a single lin5ed list! insert the elements into that

list !delete the some elements from that list and then perform the sorting operationand traversal operation on that created lin5edlist Algorithm /

Step 1: Start

Step ": eclare a structure named lin5ed$list

Step ,: eclare the pointers ne%t! first! fresh! ptr

Step ): =rint main menu

Step 9: Bead choice

Step *: Switchchoice-

Step ;: fchoice##1-;.1 Assign fresh#mallocsiIe of node--;." Bead the element fresh$Mdata;., Bead the choice where to insert;.):Switchchoice-

;.).1: f choice##1;..).": Call the function Hegin-

;.).,: f choice##";.).): Call the function end-;.).9: f choice##,;.).*: Call the function middle-

Step +: fchoice##"-+.1: Bead the position to delete+.": Switchchoice-

+.".1: f choice##1+..".": Call the function Hegin-

+.".,: f choice##"+.".): Call the function end-+.".9: f choice##,+.".*: Call the function middle-

Step 8: f choice##,8.1 Call function view

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Step 10: f choice##)10.1 E%it-

Step 11: Start insert function

Step 1": ffirst##null-

Step 1,: First$Mdata#e

Step 1): First$Mne%t#null

Step 19: Else declare new node

Step 1*:fresh$Mdata#e

Step 1;: f choice#1

Step 1+: frsh$Mne%t#first

Step 18: first#fresh

Step "0:if choice#"

Step "1: ptr#first

Step "": ptr$Mne%t#fresh

Step ",: fresh$Mne%t#full

Step "): f choice #,

Step "9: Enter the position

Step "*:at p$1 node

Step ";: fresh$Mne%t# ptr$Mne%t

Step "+: ptr$Mne%t#fresh

Step "8: for delete function

Step ,0: f first'#null

Step ,1: Enter the position to delete

Step ,": f choice#1

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Step ,,: d#first$Mdata

Step ,): first#first$Mne%t

Step ,9: if choice#"

Step ,*: ptr#first

Step ,;: raverse to last node

Step ,+: d#ptr$Mne%t$Mdata

Step ,8: ptr $Mne%t#ptr$Mne%t$Mne%t

Step )0: =rint d value

Step )1: for function view

Step )": for ptr#first and ptr'#null and ptr#ptr$Mne%t

Step ),: =rint ptr$Mdata

Step )): End

1"1


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lo3chart/

1""


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Program/

L include?stdio.hML include?malloc.hM

int ch!i!n!U!p!itemO &6 TABAHGE ECGABAK 6&

&6 SAB KF SBJCJBE EFK 6&

struct lin5N

int dataOstruct lin5 6ne%tO

6start!6new!6l!6l1!6start1!6tO

&6 E KF SBJCJBE EFK 6&

&6 SAB KF A FJCK 6&

main-Nclrscr-Ostart#JGGOstart1#JGGOprintfP 6666 EJ 6666 P-OprintfPQn 1.nsertionQn ".eletionQn ,.raverseQn ).SearchQn 9.SortQn *.ergeQn;.BeverseQnP-Owhile1-N

printfPenter the choice:P-OscanfP7dP!Rch-Oswitchch-Ncase 1: insert-O

brea5Ocase ": delete-O

brea5Ocase ,: traverse-O

brea5Ocase ): search-O

brea5Ocase 9: sort-O

brea5Ocase *: merge-O

brea5Ocase ;: reverse-O

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brea5Ocase +:e%it-O

getch-O

&6 E KF A FJCK 6&

&6 SAB KF SEB FJCK 6&

insert-Nl#startOprintfPenter the item to be inserted:P-OscanfP7dP!Ritem-O

new#mallocsiIeofstruct lin5--Onew$Mdata#itemOifstart##JGG-N

new$Mne%t#JGGOstart#newO

elseN

printfP1.startQn".middleQn,.endQnP-OprintfPenter the place to place the item:P-O

scanfP7dP!Rch-Oifch##1-N

new$Mne%t#startOstart#newO

ifch##"-N

printfPenter the position to place item:P-OscanfP7dP!Rp-Ofori#1Oi?p$1Oi((-l#l$Mne%tOnew$Mne%t#l$Mne%tOl$Mne%t#newO

ifch##,-N

whilel$Mne%t'#JGG-l#l$Mne%tO

1")


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new$Mne%t#JGGOl$Mne%t#newO

&6 E KF SEB FJCK 6&

&6 SAB KF S=GAX FJCK 6&

traverse-N

ifstart##JGG-printfPGS S E=XQnP-OelseN

forl#startOl$Mne%t'#JGGOl#l$Mne%t-

ifl##start-printfPQnstart:7d$MP!l$Mdata-OelseprintfPQn7;d$MP!l$Mdata-Oifl$Mne%t##JGG-printfPQn last:7d$MQnP!l$Mdata-O

&6 E KF S=GAX FJCK 6&

&6 SAB KF EGEE FJCK 6&

delete-N

l#startOifstart##JGG-printfPK ES 2E GSQnP-OelseN

printfP1.startQn".middleQn,.endQnP-OprintfPenter the place to delete the item:P-OscanfP7dP!Rch-Oifch##1-N

item#start$MdataOprintfPdeleted item is:7dQnP!item-Ostart#start$Mne%tO

ifch##"-

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NprintfPenter the position to delete item:P-OscanfP7dP!Rp-Oifl$Mne%t##JGG-N

item#l$MdataOprintfPdeleted item is:7dQnP!item-Ol#start#JGGO

elseN

fori#1Oi?p$1Oi((-l#l$Mne%tOitem#l$Mne%t$MdataOprintfPdeleted item is:7dQnP!item-Ol$Mne%t#l$Mne%t$Mne%tO

ifch##,-N

ifl$Mne%t##JGG-Nitem#l$MdataOprintfPdeleted item is:7dQnP!item-Ol#start#JGGO

elseN

whilel$Mne%t$Mne%t'#JGG-l#l$Mne%tOitem#l$Mne%t$MdataOprintfPdeleted item is:7dQnP!item-Ol$Mne%t#JGGOl#l$Mne%tO

&6 E KF EGEE FJCK 6&

&6 SAB KF SEABC2 FJCK 6&

search-Nint f#0O

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printfPenter the search item:P-OscanfP7dP!Ritem-Oifstart##JGG-printfPGS S E=XP-Oelse

N forl#start!i#1Ol'#JGGOl#l$Mne%t!i((-ifl$Mdata##item-N

f#1Obrea5O

iff##1-printfPitem 7d found at position :7dQnP!item!i-OelseprintfPitem 7d not foundQnP!item-O

&6 E KF SEABC2 FJCK 6&

&6 SAB KF SKB FJCK 6&

sort-Nint tOifstart##JGG-printfPGS S E=XP-OelseN

forl1#startOl1$Mne%t'#JGGOl1#l1$Mne%t-N

forl#startOl$Mne%t'#JGGOl#l$Mne%t-ifl$Mdata M l$Mne%t$Mdata-Nt#l$MdataOl$Mdata#l$Mne%t$MdataOl$Mne%t$Mdata#tO

printfP2E SKBE KBEB S:P-Oforl#startOl'#JGGOl#l$Mne%t-printfP7,dP!l$Mdata-OprintfPQnP-O

1";


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&6 E KF SKB FJCK 6&

&6 SAB KF EB/E FJCK 6&

merge-NprintfPenter no of elements to be inserted in second list :P-OscanfP7dP!Rn-OforU#1OU?#nOU((-N

l1#start1OprintfPenter the item to be inserted:P-OscanfP7dP!Ritem-Onew#mallocsiIeofstruct lin5--Onew$Mdata#itemO

new$Mne%t#JGGOifstart1##JGG-start1#newOelseN

printfP1.startQn".middleQn,.endQnP-OprintfPenter the place to place the item:P-OscanfP7dP!Rch-Oifch##1-

Nnew$Mne%t#start1Ostart1#newO

ifch##"-

NprintfPenter the position to place item:P-OscanfP7dP!Rp-Ofori#1Oi?p$1Oi((-l1#l1$Mne%tOnew$Mne%t#l1$Mne%tOl1$Mne%t#newO

ifch##,-

Nwhilel1$Mne%t'#JGG-l1#l1$Mne%tOl1$Mne%t#newO

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ifstart##JGG-start#start1OelseN

l#startO

whilel$Mne%t'#JGG-l#l$Mne%tOforl1#start1Ol1$Mne%t'#JGGOl1#l1$Mne%t-N

l$Mne%t#l1Ol#l$Mne%tO

printfP 666 GS S EB/E 666 QnP-O

&6 E KF EB/E FJCK 6&

&6 SAB KF BETEBSE FJCK 6&

reverse-Nifstart##JGG-printfPGS S E=XQnP-OelseN

l#startOl1#t#JGGOwhilel'#JGG-N

l1#tOt#lOl#l$Mne%tOt$Mne%t#l1O

start#tOprintfP 666 GS S BETEBSE 666QnP-O

&6 E KF BETEBSE FJCK 6&

66666 KJ=J 666666666 EJ 66661.nsertion

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".eletion,.raverse).Search9.Sort*.erge

;.Beverseenter the choice:1enter the item to be inserted:1enter the choice:1enter the item to be inserted:"1.start".middle,.endenter the place to place the item:1enter the choice:1enter the item to be inserted:,

1.start".middle,.endenter the place to place the item:,enter the choice:1enter the item to be inserted:)1.start".middle,.endenter the place to place the item:"enter the position to place item:,enter the choice:,

start:"$M 1$M )$Mlast:,$Menter the choice:)enter the search item:)item ) found at position :,enter the choice:*enter no of elements to be inserted in second list :,enter the item to be inserted:9enter the item to be inserted:*1.start".middle,.endenter the place to place the item:1enter the item to be inserted:;1.start

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".middle,.endenter the place to place the item:"enter the position to place item:"666 GS S EB/E 666

enter the choice:,

start:"$M 1$M )$M ,$M *$M ;$Mlast:9$Menter the choi

c and ds manuaL

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