EXERCISE: 1 - cse.gecgudlavalleru.ac.incse.gecgudlavalleru.ac.in/pdf/manuals/clab.pdf · I B.Tech C...

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P a g e | 1 Department of Computer Science & Engineering GEC

EXERCISE: 1

a) Convert temperature from centigrade to Fahrenheit and Fahrenheit to centigrade.

DESCRIPTION:

To convert Fahrenheit to centigrade, subtract 32 and multiply by 0.56.

To convert centigrade to Fahrenheit, multiply by 1.8 and add 32 degrees.

ALGORITHM:

Step 1: start

Step2: read temperature in Fahrenheit

Step 3: calculate centigrade = (Fahrenheit -32)*0.56

Step 4: display centigrade value

Step 5: read temperature in centigrade

Step 6: calculate Fahrenheit = 32 + (centigrade * 1.8)

Step 7: display Fahrenheit value

Step 8: stop

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FLOW CHART:

OUTPUT:

Enter Temperature in Fahrenheit:52

Temperature in Celsius is: 11.200000

Enter Temperature in Celsius: 37

Temperature in Fahrenheit is: 98.600000

stop

Start

Read c

fa = 32+(c*1.8)

Display c

Read f

c=(fa-32)*0.56

Display fa

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C Programming Lab


b)To find Student grading.

DESCRIPTION:

Find the grade of a student by reading marks.

ALGORITHM:

Step 1 : start

Step 2 : read 6 subjects marks

Step 3 : compute total and average of marks

Step 4 : if avg >= 80 then grade =A go to step 8

Ste p 5 : if avg >= 60 and avg <=80 then grade = B go to step 8

Step 6 : if avg >=40 and avg <=60 then grade = C go to step 8

Step 7 : if avg <40 then display failed

Step 8 : display grade.

Step 9 : stop

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FLOW CHART:

yes

no no avg >80 avg >=60

and <=80

avg >=40

and <=60 Display fail

Display

grade =A

Display

grade =B

Display

grade = C

stop

no

start

Read marks

total=m1+m2+m3+m4

+m5+m6;

avg=total/6;

yes yes

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OUTPUT:

Enter your marks 36

36

36

36

36

36

You Failed in this exam

Enter your marks 60

70

35

69

35

69

You got C grade

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C Programming Lab


c) Program for income tax for calculation.

ALGORITHM:

Step 1: start

Step 2: read income

Step 3: if income >0 and <200000 display “No tax” otherwise go to step 4

Step 4: if income > 200001 and < 500000 then tax = income *0.1 go to step 7 otherwise go to step 5

Step 5: if income > 500001 and < 100000 then tax = income*0.2 go to step 7 otherwise go to step 6

Step6: tax=income*0.3

Step 7: display tax

Step 8: stop

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C Programming Lab


FLOWCHART:

OUTPUT:

Enter Income: 500000

INCOME TAX AMOUNT= 50000.000000

REMAINING INCOME= 450000.000000

Enter Income: 225000

INCOME TAX AMOUNT= 22500.000000

REMAINING INCOME= 202500.000000

no

yes yes

yes

no no

tax = income *

0.3

start

Read income

income >

0 and <

200000

income >

200001 and

<500000

income >

500001 and

< 1000000

Display

No tax

Display tax

stop

tax = income*0.1

tax = income*0.2

Display tax Display tax

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C Programming Lab


EXERCISE: 2

To convert the given binary number to 2’s complement.

DESCRIPTION:

In this program the given binary number is first covert the numbers 0 to1 and 1 to 0.

And finally add the 1 to the converted number. Then we will get the 2‟s complement

number.

ALGORITHM:

Step1: Start

Step2: Read a binary number b

Step3: store length of the binary number-1 value in n

Step4: initialize i=n

Step5: if i>=0 otherwise go to step7

(i) if b[i]==1 is true, otherwise go to step6

(A) initialize j=i-1

(B) if j>=0 otherwise break

(a) if b[j]==1 otherwise b[j]=1 go to step5(i)(B)(b)

b[j]=0

(b) decrement j go to step5(i)(B)

Step6: decrement i, go to step5

Step7: print b

Step8: stop

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FLOWCHART:

OUTPUT:

Enter Binary Number: 101010

Its 2's Complement Is: 010110

Enter Binary Number: 11111

Its 2's Complement Is: 00001

true

true

false

true

false

false

true

false

i = n i--

i >=0

j = i-1 j>=0

j--

Start

n= strlen(b)-1

Read a binary

number

If

b[i]==‟1‟

Stop

if b[j]==‟1‟

b[j] = „0‟ b[j] = „1‟

Display 2‟s

complement

of number

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C Programming Lab


EXERCISE 3

a) To find the sum of individual digits of a given number

DESCRIPTION:

Sum of the individual digits means adding all the digits of a number

Ex: 123 sum of digits is 1+2+3=6

ALGORITHM:

Step 1: start

Step 2: initialize the s=0

Step 3: read n

Step 4: if n<0 go to Step 7

Step 5: if n!=0 go to Step 6 else go to step 7

Step 6: store n%10 value in p

Add p value to s

Assign n/10 value to n

Go to Step 5

Step 7: print s

Step 8: stop

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FLOW CHART:

OUTPUT:

enter the value for n:

123

sum of individual digits is 6

enter the value for n:

-11

The given number is not valid

FALSE TRUE

TRUE

START

IF

N! =0

s=0

Read N

IF N < 0

Print S

STOP P=N%10

S=S+P

N=N/10

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b) To print the Fibonacci series for 1 to n value

DESCRIPTION:

A Fibonacci series is defined as follows

The first term in the sequence is 0

The second term in the sequence is 1

The sub sequent terms 1 found by adding the preceding two terms in the sequence

Formula: let t1,t2,…………tn be terms in Fibonacci sequence

t1=0, t2=1

tn=tn-2+tn-1……where n>2

ALGORITHM:

Step 1: start

Step 2: initialize the a=0, b=1

Step 3: read n

Step 4: if n== 1 print a go to step 7. else goto step 5

Step 5: if n== 2 print a, b go to step 7 else print a,b

Step 6: initialize i=3

i) if i<= n do as follows. If not goto step 7

c=a+b

print c

a=b

b=c

increment i value

goto step 6(i)

Step 7: stop

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FLOWCHART:

OUTPUT:

enter n value5

0 1 1 2 3

enter n value10

0 1 1 2 3 5 8 13 21 34

True False

False True

False

Start

Initialize

a=0,b=1

Read n

If n==1

Output a If n ==2

Output a,b Output a,b

i=3 i++

i<=n

c =a + b

Output c

a = b

b = c

Stop

true

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C Programming Lab


c) To print prime numbers up to 1 to n

DESCRIPTION:

Prime number is a number which is exactly divisible by one and itself only

Ex: 2, 3,5,7,………;

ALGORITHM:

Step 1: start

Step 2: read n

Step 3: initialize i=1,c=0

Step 4:if i<=n

Initialize c=0 go to step 5

If not go to step 10

Step 5: initialize j=1

Step 6: if j<=i do the following. If no goto step 7

i)if i%j==0 increment c otherwise increment j go to step 6

ii) increment j

iii) goto Step 6

Step 7: if c== 2 print i

Step 8: increment i

Step 9: goto step 4

Step 10: stop

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true

false

false

true

true

false

j = 1 j++

j<=i

if c==2

if i % j == 0

c ++

Display i

Start

Read n

i = 1 i++

i<=n

stop

c=0

false

true

FLOW CHART:

OUTPUT:

Enter Value of n:5

Prime Numbers upto 5 are:

2 3 5

Enter Value of n:9

Prime Numbers up to 9 are:2 3 5 7

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C Programming Lab


c) To check a given integer is Fibonacci number or not.

DESCRIPTION :

Find the given number is Fibonacci or not. It means if the given number is present in the Fibonacci

series it satisfies the condition otherwise it fails.

ALGORITHM:

Step 1: start

Step 2: initialize a=0 , b=1

Step 3 : compute c=a+b

Step 4 : read n

Step 5 : if c<=n go to step 6 otherwise goto step 12

Step 6 : compute c = a+b

Step 7 : a=b

Step 8 : b = c goto step 5

Step 9 : if n==c or n==0 or n==1 goto step 10 otherwise goto step 11

Step 10 : display “ given number is Fibonacci “ n

Step 11 : display “ given number is not Fibonacci “ n

Step 12 : stop

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FLOWCHART:

OUTPUT:

Enter no to find: 21

21 is Fibonacci

Enter no to find: 25

25 is not Fibonacci

START

a=0,b=1

Read n

if c< n if n ==0 or

n==1 or

n ==c

Display n is

Fibonacci

number

Stop

c = a+b

c= a+b

a=b, b=c

Display n is not

a Fibonacci

number

no no

yes

yes

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C Programming Lab


EXERCISE: 4

a) To calculate the sum. Sum=1-x2/2! + x

4/4! - x

6/6! + x

8/8! - x

10/10!

ALGORITHM:

Step 1: start

Step 2: declare sum=0.0,p=0

Step 3: read x,n value


Step5: if i<=n initialize f=1 otherwise go to step 11

Step6: initialize j=1

Step7: if j<=p

assign f*j to f

increment j

otherwise go to step 8

Step8: if i%2==0 sum=sum-(pow(x,p)/f) otherwise sum=sum+(pow(x,p)/f)

Step9: compute p=p+2

Step 10: increment i

Step 11: print sum value

Step 12: stop

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true

false

false

Start

Read x,n

f= 1

i=1, i<=n

i++

Display sum

Stop

Initialize Sum = 0.0,

p=0

j=1, j<=p

j++

f= f*j

if i%2

== 0

sum=sum-

(pow(x,p)/f)

sum=sum+

(pow(x,p)/f)

p = p+2

true

true

false

FLOW CHART:

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OUTPUT:

Enter Number of Terms in the Series:3

Enter x value:1

The sum of the Series upto 3 terms: 0.541667

Enter Number of Terms in the Series:3

Enter x value:2

The sum of the Series upto 3 terms: -0.333333

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b) To find the roots of the quadratic equation

DESCRIPTION:

Nature of roots of quadratic equation can be known from the quadrant = b2-4ac

If b2-4ac >0 then roots are real and unequal

If b2-4ac =0 then roots are real and equal

If b2-4ac <0 then roots are imaginary

ALGORITHM:

Step 1: start

Step 2: read the a,b,c value

Step3: assign pow((b*b-4*a*c),0.5) value to d

Step4: if d==0 otherwise go to step5

Compute r1=-b/(2*a)

r2=-b/(2*a)

go to step 7

Step5: if d<0 print roots are imaginary otherwise go to step6

Step6: compute r1 = ((-b+d) / (2*a))

r2 = ((-b-d) / (2*a))

go to step 7

Step7: print r1, r2

Step8: stop

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true

false

false

Start

Read a,b,c

d = pow((b*b-4*a*c),0.5)

If d < 0

r1 = ((-b+d) / (2*a))

r2 = ((-b-d) / (2*a))

Display

r1, r2

Stop

If d== 0

r1=-b / (2 * a )

r2= -b / (2 * a)

Display

roots are

imaginary

true

FLOWCHART:

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OUTPUT:

Enter the values of a,b,c:1 6 9

Roots are Equal:-3.000000,-3.000000


Roots are:-6.750000,-7.250000


Roots are Imaginary

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EXERCISE: 5

a) The total distance travelled by vehicle in 't' seconds is given by distance = ut+1/2at2

where 'u' and 'a' are the initial velocity (m/sec.) and acceleration (m/sec2). Write C

program to find the distance travelled at regular intervals of time given the values of 'u'

and 'a'. The program should provide the flexibility to the user to select his own time

intervals and repeat the calculations for different values of 'u' and 'a'.

DESCRIPTION:

The total distance travelled by vehicle in 't' seconds is given by distance = ut+1/2at2 where

'u' and 'a' are the initial velocity (m/sec.) and acceleration (m/sec2).

ALGORITHM:

Step 1:Start

Step2 : Read u,a,t

Step 3: compute s=(u*t)+(0.5*a*t*t)

Step 4:print d value

Step 5: read ch

Step 6: if ch==1 go to step 2 otherwise go to step 7

Step 7: stop

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FLOWCHART:

true

false

start

Read u, a, t

d= (u*t)+(0.5*a*t*t)

Display distance as d

Read ch

If

ch==1

stop

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OUTPUT:

Enter the value of initial velocity:12

Enter the value of acceleration:14

Enter time interval:2

The distance is: 52.000000

Press 0 to quit ---- 1 to continue

1

Enter the value of initial velocity:16

Enter the value of acceleration:15

Enter time interval:4

The distance is: 184.000000

Press 0 to quit ---- 1 to continue

0

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b) Two integer operands and one operator form user, performs the operation and then prints the

result. (Consider the operators +,-,*, /, % and use Switch Statement)

Description:

To take the two integer operands and one operator from user to perform the some arithmetic

operations by using the following operators like +,-,*, /, %

Ex: 2+3=5

Algorithm:

Step 1: Start

Step 2: Read the values of op1,op2 and operator op

Step 3: if the operator is „+‟ then

R=op1+op2

Go to step 8

Break

Step 4: Else if the operator is „-„ then

R=op1-op2

Go to step 8

Step 5: Else if the operator is „*„ then

R=op1*op2

Go to step 8

Step 6: Else if the operator is „/„ then

R=op1/op2

Go to step 8

Step 7: Else if the operator is „%„ then

R=op1%op2

Go to step 8

Step 8: print R

Step 9: stop

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FLOWCHART:

false

false

false

false

false

true

true

true

true

true

start

Read

op1,op2, op

If op

==‟+‟

If op

==‟-‟

r=op1+op2

r=op1-op2

If op

==‟*‟

If op

==‟/‟

If op

==‟%‟

r=op1*op2

r=op1/op2

r=op1%op2

Display r

stop

Display

invalid

operation

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C Programming Lab


OUTPUT:

Operations Performed are: + - * / %

Enter 2 Integers & an Operator:2

4

*

MULTIPLICATION -- Product is 8



4

+

ADDITION -- Sum is 6



4

-

SUBTRACTION -- Difference is -2



4

/

DIVISION -- Quotient is 0



4

%

MODULUS -- Remainder is 2

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C Programming Lab


EXERCISE: 6

a) String functions (predefined) examples.

DESCRIPTION:

Apply some of the predefined functions on given strings. These are included in string.h header file.

Algorithm

Step 1 : start

Step 2: read s1,s2,s3

Step 3: compute l1= strlen(s1)

Step 4: print l1

Step 5: compute s=strcpy(s3,s1)

Step 6: print s

Step7: compute e= strcmp(s1,s2)

Step 8: if e==0 go to step 9 otherwise go to step 10

Step 9: display “strings are equal”

Step 10: display “Strings are not equal “

Step 11: display “reverse of 1st string is “, strrev(s1)

Step 12: display “after concatenation of two strings s1 and s2 is “ strcat(s1,s2)

Step 13: stop

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Flowchart:

true

start

Read strings

s1,s2,s3

l1=strlen(s1)

Display l1

s=strcpy(s3,s1)

Display s1,s2

are equal

e=strcmp(s1,s2)

If

e==0

Display s

Display s1,s2 are not equal

r=strrev(s1)

Display r

c=strcat(s1,s2)

Display c

stop

false

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OUTPUT:

Enter three strings sagf

gec

gudlavalleru

length of s1 is 4

After copying 1st string into 3rd string is sagf

The two Strings s1 and s2 are not equal

Reverse of first string is fgas

After adding first two strings fgasgec

Enter three strings gec

gec

gudlavalleru

length of s1 is 3

After copying 1st string into 3rd string is gec

The two strings s1 and s2 are equal

Reverse of first string is ceg

After adding first two strings ceggec

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b) Verifying a string for its palindrome property.

DESCRIPTION:

Read a string, compare first and last characters of a string, if equal continue up to middle of the

string. If the comparison fails at any character the string is not a palindrome otherwise palindrome

property satisfies.

Algorithm:

Step 1: start

Step 2: read the string s1

Step 3: store reverse of the given string s1 in a temporary string s2

Step 4: compare the two strings s1 and s2

Step 5: if both are equal then print the given string is palindrome

Step 6: otherwise print the given string is not palindrome

Step 7: stop

FLOWCHART:

true

false

start

Read string s1

strcpy(s2,s1)

strrev(s2)

p=strcmp(s1,s2)

if p

== 0

The given

string s1 is

palindrome

The given

string s1 is not

palindrome

stop

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OUTPUT:

Enter a string: madam

The given string is palindrome

Enter a string : manual

The given string is not palindrome

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c) copy one string to another string

DESCRIPTION:

Copying one string to another string

ALGORITHM:

step1: start

step2: read a string s1

step3: initialize i=0

step4: if s1[i]!=‟\0‟ otherwise s2[i]=‟\0‟

s2[i]=s1[i]

increment i

step5: print s2

step6: stop

FLOWCHART:

false

true

start

Read a string

s1

i=0 s1[i]!=‟\0‟

i++

s2[i]=s1[i]

s2[i]=‟\0‟

Display s2

stop

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OUTPUT:

Enter a string : malayalam

The copied string is malayalam

Enter a string : haihello

The copied string is haihello

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d) calculate the length of the given string

DESCRIPTION:

Calculating the length of the given string

ALGORITHM:

Step 1:start

Step 2: initialize sum=0

Step3: read the string s1

Step 4:initialize i=0

Step5: if s1[i]!=‟\0‟ otherwise go to step6

increment sum

increment i

step6: print sum

step7: stop

FLOWCHART:

true

Start

Initialize sum = 0

Read

string s1

i=0 s1[i]!=‟\0‟

i++

sum ++

stop

Display

sum

false

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OUTPUT:

Enter a string : gudlavalleruengineeringcollege

30

Enter a string : string

6

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e) To reverse the given string

DESCRIPTION:

Reverse a given string

ALGORITHM:

Step1: start

Step2: initialize sum=0

Step3: read string s1

Step4: initialize i=0

Step5: if s1[i]!=‟\0‟ otherwise go to step6

(i) increment sum

(ii) increment i

Step6: initialize i=0,j=sum-1

Step7: if s1[i]!=‟\0‟and j>=0 otherwise go to step8

(i) rev[j]=s[i];

(ii) increment i

(iii) decrement j

Step8: rev[j]=‟\0‟

Step9: stop

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FLOW CHART:

false

true

start

Initialize sum = 0

Read

string s1

i=0 s1[i]!=‟\0‟

i++

sum ++

stop

Display

rev

i=0, j=sum-1 s1[i]!=‟\0‟, j>=0

i++,j--

rev[j]=s[i]

rev[j]=‟\0‟

true

false

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OUTPUT:

Enter a string : sai

Reverse of the given string is ias

Enter a string : cprogramming

Reverse of the given string is gnimmargorpc

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C Programming Lab


EXERCISE: 7

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

DESCRIPTION:

In this program we need to insert a string into another string from a specified position.

ALGORITHM:

Step 1: start

Step 2: read main string and sub string

Step 3: find the length of main string(r)

Step 4: find length of sub string (n)

Step 5: copy main string into sub string

Step 6: read the position to insert the sub string (p)

Step 7: copy sub string into main string from position p-1

Step 8: copy temporary string into main string from position p+n-1

Step 9: print the strings

Step 10: stop

FLOWCHART:

Start

Read the strings S1,

S2, position i

Insert (s1,s2,i)

Start

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Yes

No

Yes

No

Yes

No

Insert (s1, s2, x)

i=0 i++,c++

s1[ i]!=\0

if i==x

j=0 j++,c++

s2 [j]! =\0

s[c] =s2[j]

s[c]='\0'

Print

string s

Return to main()

s[c]=s1[i]

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OUTPUT:

Enter the Original String: computer

Enter the subString: gec

Enter the index where you want to insert the subString: 3

The String now is:

------------------

comgecputer

Enter the Original String: gudlavalleru

Enter the subString: enggcollege

Enter the index where you want to insert the subString: 5

The String now is:

------------------

gudlaenggcollegevalleru

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b) To delete n characters from a given position in a given string

DESCRIPTION:

In this program we need to delete a string from the given string at a specified position.

ALGORITHM:

Step 1: start

Step 2: read string

Step 3: find the length of the string

Step 4: read the value of number of characters to be deleted and positioned

Step 5: string copy part of string from position to end, and (position+number of

characters to end)

Step 6: stop

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OUTPUT:

1. enter the string

nagaraju

Enter the position from where to delete:4

Enter the number of charcters to be deleted3

nagju

2. enter the string

kaliraju

Enter the position from where to delete:0

Enter the number of charcters to be deleted4

Raju

Start

Read string

Read position, no of

characters

delchar( string, n, pos)

Stop

delchar( )

if ((a+b-1)<=

strlen(x))

Strcpy(&x[b-1],&x[a+b-1])

Print x

Return to

main()

FLOW CHART:

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C Programming Lab


c) To replace a character of string either from beginning or ending or at a specified location.

DESCRIPTION:

Replace a character of string either from beginning or ending or at a specified location.

ALGORITHM:

Step 1: start

Step 2: read string

Step 3: find the length of the string

Step 4: read the replace characters

Step 5: read the position to be placed like end,begin,pos

Step 6: char placed in a part of string from position to end, and (position+number

of characters to end)

Step 7: print string

Step 8: stop

FLOW CHART:

stop

false

switch

opt

case e case b

case p

s1[0] = ch; s1[l-1] = ch;

s1[pos]= ch;

pos >=0

and pos

<=l-1

Not match

Print string

true

start

Read string

Read replace

character

chara

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OUTPUT:

Enter string :cplab

Enter character to replace :n

Enter b-begin, e-ending , p-position b

after replacing string is nplab

Enter string :clab

Enter character to replace :w

Enter b-begin, e-ending , p-position e

after replacing string is claw

Enter string :cprogramming

Enter character to replace :l

Enter b-begin, e-ending , p-position p

Enter postion to replace 1

after replacing string is clrogramming

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C Programming Lab


EXERCISE : 8

Write a C program that uses functions to perform the following operations using Structure:

i) Reading a complex number ii) Writing a complex number

iii) Addition of two complex numbers iv) Multiplication of two complexnumbers

DESCRIPTION:

In this program the complex number means it contains the two parts . first one is real part and

second one is imaginary part(2+3i).by taking these two complex numbers we can perform the

addition and multiplication operation.

ALGORITHM:

Step 1: Start

Step 2: declare structure for complex numbers and perform the following

operations

Step 3: read choice

Step 4: if choice=1 then read the complex numbers

Step 5: if choice=2 then write the complex numbers

Step 6: if choice=3 then addition operation will perform and it contains following

steps

i) c3.real = c1.real+c2.real;

ii) c3.img = c1.img+c2.img;

print the result by calling write() function then goto step 3

Step 6: if choice=4 then multiplication operation will perform and it contains

following steps

i) c4.real=(c1.real*c2.real)-(c1.img*c2.img);

ii) c4.img=(c1.real*c2.img)+(c1.img*c2.real);

print the result by calling write() function then goto step 3

Step 7: if choice=5 then exit

Step 9: Stop

I B.Tech

C Programming Lab


case exit

Start

Declare structure

Read complex

numbers

switch

choice

case read

c3.real=c1.real+c2.real

c4.img=c1.img+c2.img

c4.real=(c1.real*c2.real)-(c1.img*c2.img)

c4.img=(c1.real*c2.img)+(c1.img*c2.real)

Stop

Read choice

Display the

result

case mul case add

case write

FLOW CHART:

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C Programming Lab


OUTPUT:

1:Reading 2:Writing 3:Addition 4:Multiplication 5:Exit

enter your choice:1

enter real and imaginary parts of first complex number: 2 3

enter real and imaginary parts of second complex number: 4 5


enter your choice:2

answer=2.000000+3.000000i

answer=4.000000+5.000000i


enter your choice:3

answer=6.000000+8.000000i


enter your choice:4

answer= -7.000000+22.000000i


enter your choice:5

.

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C Programming Lab


EXERCISE : 9

a) To perform the addition of two matrices

Description:

The program takes the two matrixes of same size and performs the addition

Algorithm:

Step 1: start

Step 2: read the size of matrices A,B – m,n

Step 3: read the elements of matrix A

Step 4: read the elements of matrix B

Step 5: perform the addition operation

Step 6: print sum of matrices A and B

Step 7: Stop

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C Programming Lab


FLOWCHART:

true

true

false

false

true

false

i=0 i++

i<r1

j=0 j<c1

j++

start

Read matrices A and B

stop

Read number of rows

r1 and number of

columns c1 of matrix 1

Read number of rows

r2 and number of

columns c2 of matrix 2

If r1!=r2

and c1!=c2

c[i][j]=a[i][j]+b[i][j];

Print result matrix

Display

invalid order

of matrices

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C Programming Lab


OUTPUT:

Enter Number of rows & columns for Matrix1:2

2


2

Enter elements for Mat1:2

2

2

2


2

2

2

Matrix Addition Result:

4 4

4 4


2


3

Invalid order of Matrices

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C Programming Lab


b) Calculating transpose of a matrix in-place manner.

DESCRIPTION:

The transpose of a matrix is obtained by interchanging the row and column values of the matrix,

hence the order of the resultant matrix changes.

ALGORITHM:

Step 1: start

Step 2: read the size of matrix A


Step 4: perform the transpose operation by interchanging the row and

column values, the order of the resultant matrix

number of rows in transpose matrix=number of columns in the given matrix

number of columns in transpose matrix=number of rows in the given matrix

Step 6: Transpose is obtained through

at[i][j]=a[i][j]

Step 7: print the resultant transpose matrix at.

Step 8: stop

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C Programming Lab


FLOWCHART:

OUTPUT:

Enter Number of Rows: 2

Enter Number of Columns: 3

Enter 6 Elements: 2

3

4

5

6

7

false

true

true

i=0 i++

i<c

Read elements of

matrix

Read rows and

columns of matrix

j=0 j<r

j++

at[i][j]=a[j][i]

stop

Print resultant transpose matrix

start

false

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C Programming Lab


After Transpose:

2 5

3 6

4 7

Enter Number of Rows: 3

Enter Number of Columns: 4

Enter 12 Elements: 1

2

3

4

5

6

7

8

9

10

11

12

After Transpose:

1 5 9

2 6 10

3 7 11

4 8 12

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C Programming Lab


c) Matrix multiplication by checking compatibility

DESCRIPTION:

Takes the two matrixes of different sizes and checks for possibility of multiplication and perform

multiplication if possible.

ALGORITHM:

Step 1: start

Step 2: read the size of matrices A,B

Step 3: check compatibility of matrices for multiplication i.e, number of columns

in the first matrix should be equal to number of rows in the second matrix.


Step 5: read the elements of matrix B

Step 6: perform the multiplication operation by storing the resulting values into

matrix C.

Step 7: print the resultant matrix C.

Step 8: Stop

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C Programming Lab


FLOWCHART:

true

true

false

true

false

false

true

i=0 i++

i<r1

j=0 j<c2

j++

start

Read matrices A and B

stop

Read number of rows

r1 and number of

columns c1 of matrix

A

Read number of rows

r2 and number of

columns c2 of matrix B

if c1!=r2

c[i][j]=c[i][j]+(a[i][k]*b[k

][j])

Print result matrix

Display

invalid order

of matrices

c[i][j]=0

k=0 k<c1

k++

false

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C Programming Lab


OUTPUT:


2


2


2

2

2


2

2

2

Matrix Multiplication Result:

8 8

8 8

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C Programming Lab


EXERCISE : 10

a) (i) Programs that use recursive function to find the factorial of a given integer.

DESCRIPTION:

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

ALGORITHM: main program

Step 1: start

Step 2: read n

Step 3: call sub program

Step4: f=fact(n)

Step 5: print f value

Step 6: stop

Sub program:

Step 1: initialize the f

Step 2: if n= = 0 or n == 1 return 1 to main program if not goto step 3

Step 3: return n*fact(n-1) go to main program

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C Programming Lab


true

Start

Read n

Call subprogram

print f

Stop

Sub program

fact()

If n=0 || n=1

Return

n*fact(n-1)

Return to main

program

false

f = fact(n)

Return 1

FLOWCHART:

OUTPUT:

enter the number :7

factorial of number 7 is 5040

enter the number :5

factorial of number 5 is 120

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C Programming Lab


a) ii) Program that use non recursive function to find the factorial of a given integer.

DESCRIPTION:

Factorial of a number is nothing but the multiplication of numbers from a given number to 1 Ex: 5!

=5*4*3*2*1= 120

ALGORITHM:

main program

Step 1: start

Step 2: read n

Step 3: call the sub program fact(n)

Step 4: print the f value

Step 5: stop

Sub program:

Step 1: initialize the f=1

Step 2: if n==0 or n=1 return 1 to main program. If not goto step 3

Step 3: perform the looping operation as follows

For i=1 i<=n; i++

Step 4: f=f*i

Step 5: return f value to the main program

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C Programming Lab


false

i= 1 i++

i<=n

start

Read n

Call subprogram

Fact(n)

Print F value

Stop

Fact (n)

Initialize F = 1

if n == 0 ||

n == 1

Return to main program

F = F * i

Return F

Return 1

true

Subprogram Flowchart

FLOWCHART:

OUTPUT:

enter the number :5

factoria of 5 is 120

enter the number :7

factoria of 7 is 5040

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C Programming Lab


b) To find the GCD of two given integers by using the recursive function

Description:

GCD means Greatest Common Divisor. i.e the highest number which divides the given number

Ex: GCD (12,24) is 12

Formula: GCD= product of numbers/ LCM of numbers

ALGORITHM:

main program

Step 1: start

Step 2: read a,b

Step 3: call the sub program GCD(a,b) and then print the value

Step 4: stop

Sub program:

Step 1: if n==0 return m else goto step 2

Step 2: return GCD (n,m%n)

Step 3: return to main program

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C Programming Lab


FLOW CHART:

OUTPUT:

enter the two numbers whose gcd is to be found:36 63

GCD of a,b is 9


GCD of a,b is 5

true false

gcdrec( )

If n==0

Return m Call the same function

Return gcdrec(n,m%n)

Return to main()

Start

Read a,b

Call sub program

G=gcdrec(a,b)

Print gcdvalue

Stop


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C Programming Lab


b) ii)To find the GCD of two given integers by using the non recursive function

Description:

GCD means Greatest Common Divisor. i.e the highest number which divides the given number

Ex: GCD(12,24) is 12

Formula: GCD= product of numbers/ LCM of numbers

ALGORITHM:

Main program

Step 1: start

Step 2: read a,b

Step 3: call sub program g=GCD(a,b)

Step 4: print the g value

Step 5: stop

Sub program:

Step 1: initialize the p=1, q, remainder

Step 2: remainder=p-(p/q*q)

Step 3: remainder=0 return q else goto step 4

Step 4: GCD(q,remainder) return to main program

I B.Tech

C Programming Lab


false true

start

Read a, b

Call subprogram

gcdnonrec(a,b)

Output g

stop

gcdnonrec(m,n)

r=m%n

if r!=0

m=n,

n=r,

r=m%n

Return n

Return to main program


FLOWCHART:

OUTPUT:


GCD of a,b is 9


GCD of a,b is 5


GCD of a,b is 18

I B.Tech

C Programming Lab


c).To solve the towers of Hanoi problem by using the recursive function

DESCRIPTION:

Towers of Hanoi problem means we have three towers

Here source ,intermediate and destination are the three towers. We have to transfer all the disks

from source to destination towers. Here the restriction is not to place a big disk on smaller one . for

this we use intermediate tower. Finally the arrangements in the destination tower must be as same

as the disks in the source tower at first.

ALGORITHM:

main program

Step 1: start

Step 2: initialize the source=a, intermediate=c, destination = d

Step 3: read n

Step 4: call the sub program Hanoi recursion (n value,a ,b, c)

Step 5: stop

Sub program:

Step 1: if n== 1 call the sub program Hanoi recursion (num-1, a, c, b)

Step 2: print the output from a to b

Step 3: call the sub program Hanoi recursion(num-1, b, c, a)

Step 4: return to main program

Source intermediate destination

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C Programming Lab


false true

Print A,C Return to main

program

Hanoirec(num,n1,n2,n3)

if num==1

Print A,B

Call subprogram

Hanoirec(num-1,n3,n2,n1)

Call sbgroram

Hanoirec(num-1,n1,n3,n2)

Start

Assume Source = A

Intermediate = C

Destination = B

READ no

Call subprogram

Hanoirec(no,source,intermediate,destination)

Stop

Sub program

FLOWCHART:

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C Programming Lab


OUTPUT:

Enter the no. of disks to be transferred :3

recursive transfer with source needle A and destination needle B

Move top disk from needle A to needle B

Move top disk from needle A to needle C

Move top disk from needle B to needle C


Move top disk from needle C to needle A

Move top disk from needle C to needle B


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C Programming Lab


ii)To solve the towers of Hanoi problem by using the non recursive function

DESCRIPTION:

Towers of Hanoi problem means we have three towers

Here source ,intermediate and destination are the three towers. We have to transfer all the disks

from source to destination towers. Here the restriction is not to place a big disk on smaller one . for

this we use intermediate tower. Finally the arrangements in the destination tower must be as same

as the disks in the source tower at first.

ALGORITHM:

Step 1: start

Step 2: read no of disks

Step 3: initialize x=1

Step 4: if x < (1 << n) goto step 5 otherwise step 6

Step 5: print (x&x-1)%3, ((x|x-1)+1)%3 value

Step 6: stop

FLOW CHART:

Source intermediate destination

false x=1 x++

x < (1 << n)

Start

Read no of

disks

Print (x&x-1)%3,

((x|x-1)+1)%3

Stop

true

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C Programming Lab


Output:

1.Enter the no. of disks to be transferred:3

nonrecursive

Move top disk from tower A to tower C

Move top disk from tower A to tower B

Move top disk from tower C to tower B


Move top disk from tower B to tower A

Move top disk from tower B to tower C


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C Programming Lab


EXERCISE: 11

a) To find both the largest and smallest number in a list of integers

DESCRIPTION:

This program contains n number of elements, in these elements we can find the largest and smallest

numbers and display these two numbers

ALGORITHM:

Step 1: start

Step 2: read n


Step 4: if i<n do as follows. If not goto step 5

Read a[i]

Increment i

Goto step 4

Step 5: min=a[0], max=a[0]


Step 7: if i<n do as follows. If not goto step 8

If a[i]<min

Assign min=a[i]

Increment i goto Step 7

Step 8: print min,max

Step 9: stop

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C Programming Lab


false

true

true

true

false

true false

Read n

min = a[0]

max = a[0]

Print min,

max

i = 0 i++

i< n

if a[i]< min

a[i]>max min=a[i]

max=a[i]

i= 0

i<n i++

stop

Read a[i]

Start

FLOWCHART:

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C Programming Lab


OUTPUT:

enter the array size:5

Enter the elements of array50

40

30

20

10

maximum value is:50

minimum value is:10

enter the array size:8

Enter the elements of array21

-31

54

65

76

89

90

23

maximum value is:90

minimum value is:-31

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C Programming Lab


b) Program that displays the position or index in the string S where the string T begins , or -

1 if S doesn’t contain T

ALGORITHM:

Step 1: start

Step 2: read the string and then displayed

Step 3: read the string to be searched and then displayed

Step 4: searching the string T in string S and then perform the following steps

i. found=strstr(S,T)

ii. if found print the second string is found in the first string at the position.

otherwise goto step 5

Step 5: print -1

Step 6: stop

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C Programming Lab


FLOW CHART:

no yes

Start

Read first string s

Read string t to be searched

Found=strstr(s,t)

If found

Print -1 Print string is

found

Stop

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C Programming Lab


OUTPUT:

Enter the first string:

gudlavalleru

Enter the string to be searched:

lava

Second String is found in the First String at 3 position.

Enter the first string:

kangaroo

Enter the string to be searched:

rrr

-1

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C Programming Lab


c) To count the lines, words & characters in a given text

DESCRIPTION:

In this program we have to count the no of lines, no of words and no of characters in a given

program or given text by using the string function

ALGORITHM:

Step 1: Start

Step 2: Read the text until an empty line

Step 3: Compare each character with newline char „\n‟ to count no of lines

Step 4: Compare each character with tab char „\t\‟ or space char „ „ to count no

of words

Step 5: Compare first character with null char „\0‟ to find the end of text

Step 6: No of characters = length of each line of text

Step 7: Print no of lines, no of words, no of chars

Step 8: Stop

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C Programming Lab


false

true

true

false

false

false

true

Start

Initialize end=0,chars=0,words=0,lines=0

if

end!=0

c= 0

true

if

(ctr=getchar())==‟\n‟

line[c]=‟\0‟

line[c++]=ctr

if

line[0]=‟\0‟

Words ++ i = 0

i ++ line[i]!=‟\0‟

If line[i]==‟ „||

line[i]==‟\t‟

lines++

chars+=strlen(line)

Words ++

Print lines,

Words, chars

stop

true

FLOW CHART:

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C Programming Lab


OUTPUT:

hai

hello

how r u

Number of lines = 3

Number of words = 5

Number of characters = 15

Admiration is a very short-lived passion.

Admiration involves a glorious obliquity of vision.

Always we like those who admire us but we do not

like those whom we admire.

Fools admire, but men of sense approve.

Number of lines = 5

Number of words = 36

Number of characters = 205

I B.Tech

C Programming Lab


EXERCISE : 12

a) To generate Pascal’s triangle

DESCRIPTION:

Pascal‟s triangle which is used for a coefficient in the equation in polynomials.

ALOGRITHM:

Step 1: Start

Step 2: Read n


Step4: if i<n otherwise go to step 11


Step6: if j<=i otherwise go to step10

Step7: if j==0 or i==j otherwise p[i][j]=p[i-1][j-1] + p[i-1][j] go to step 9

Step8: p[i][j]=1

Step9: increment j go to step6

Step10: increment i go to step 4


Step12: if i<n otherwise go to step19

Step13: initialize k=0

Step14: if k<=n-i otherwise go to step16

Step15: print space ,Increment k, go to step14


Step17: if j<=i otherwise print new line

Step18: print p[i][j] ,increment j go to step17

Step19: stop

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C Programming Lab


OUTPUT:

i=0 i<n

i++

true

False

True

true

true

true

false

Start

Read n

i=0 i< n

i++

j=0 j ++

j<=i

true

false

false

p[i][j]=1

Print

p[i][j]

Print

space

Stop

if((j==0)

|| (i==j))

p[i][j]=p[i-1][j-1]+p[i-1][j]

k=0 k++

k<=n-i

j=0 j++

j<=i

Print

newline

false

false

FLOW CHART:

I B.Tech

C Programming Lab


Enter no of lines to print: 5

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

Enter no of lines to print: 8

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

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C Programming Lab


b) To construct a pyramid of numbers

DESCRIPTION:

In this program the we have to construct output in the pyramid shape manner

ALGORITHM:

Step 1: start

Step 2: read no of lines n


Step 4: if i<=n otherwise go to step 9


Step6: if j<=n-1 then print “ “ and increment j otherwise go to step 7

Step 7: initialize k=1

Step8: if k<=i then print k and increment k otherwise print new line, increment i, go to

step 4

Step 9: stop

I B.Tech

C Programming Lab


FLOW CHART:

true

true

false

start

Read n

i=1 i++

i<=n

Print space

Print k

k=1 k++

k<=i

stop

j=0 j++

j<=n-i

Print new line

false

false

true

I B.Tech

C Programming Lab


OUTPUT:

Enter number of lines to print :6

1

1 2

1 2 3

1 2 3 4

1 2 3 4 5

1 2 3 4 5 6

Enter number of lines to print :8

1

1 2

1 2 3

1 2 3 4

1 2 3 4 5

1 2 3 4 5 6

1 2 3 4 5 6 7

1 2 3 4 5 6 7 8

I B.Tech

C Programming Lab


EXERCISE : 13

a) To read in two numbers x and n and then compute the sum of this geometric progression

1+x+x2+

x3+……….+x

n

DESCRIPTION:

In this program we have to read the two numbers and then calculate the sum of this geometric

progression in above mention.

ALGORITHM:

Step 1: Start

Step2: initialize sum=1

Step 3: read values of x and n

Step 4: if n<=0 || x<=0 otherwise go to step5

i) print values are not valid go to step8

Step 5: print value is valid


Step7: if i<n otherwise print sum go to step8

i) sum+=pow(x,n) and increment i go to step7

Step8: Stop

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C Programming Lab


FLOWCHART:

false

Start

Initialize

sum=1

Read x,n

if

n<=0||x<=0

Print not

valid i=1 i++

i<n

sum+=pow(x,i)

Print sum

Stop

true false

true

Print value

is valid

I B.Tech

C Programming Lab


Output:

Enter the values for x and n:4

4

Value is valid

Sum of series= 341

Enter the values for x and n:3

9

Value is valid

Sum of series= 29524

I B.Tech

C Programming Lab


a)Write a C function to read in two numbers, x and n(no. of terms), and then compute sin(x)

and cos(x).

DESCRIPTION:

In this we calculate sin and cos values by reading x value and no.of terms „n‟.

ALGORITHM:

Step 1: START

Step 2: initialize sum=o,term=0

Step 3: while true

Step 4: Read ch

Step 5: if ch!=3

Step 6: Read x,n

Step 7: compute x=(x*3.14)/180

Step 8: if ch==1 otherwise go to step 9

` i) term=sum=x

ii) initialize i=1

iii) if i<=n otherwise go to step8 (iv)

i) term= ( (-term)*(x*x) ) / ((2*i)*(2*i+1));

ii) sum=sum+term, increment i

iv) Prnt Sin series sum

v) break go to step3

Step 9: if ch==2 otherwise go to step 10

i) term=sum=1

ii) initialize i=1

iii) if i<=n otherwise go to step9 (iv)

i) term= ( (-term)*(x*x) ) / ((2*i)*(2*i-1));

ii) sum=sum+term, increment i

iv) Prnt Cos series sum

v) break go to step3

Step 10: if ch==3 exit

Step 11: STOP

I B.Tech

C Programming Lab


true

true

false

Start

Initialize

sum=0,term=0

Read x,n

If true

1.sin 2.cos 3.exit

If(ch!=3)

If

ch==1

true

i=1 i++

i<=n

Read ch

x=(x*3.14)/180

term=sum=x

term= ( (-term)*(x*x) ) / ((2*i)*(2*i+1));

sum=sum+term

B

A

C D

Flow chart:

Print sin

series sum

A

I B.Tech

C Programming Lab


true

true

false

false

i=1 i++

i<=n

if

ch==2

term=sum=1

term= ( (-term)*(x*x) ) / ((2*i)*(2*i-1));

sum=sum+term

Print cos

series sum

B

D

stop

if

ch==3

exit

I B.Tech

C Programming Lab


OUTPUT:

1.sin 2.cos 3.exit

1

Enter any number : 45

Enter no.of terms 5

The sin is 0.706825

1.sin 2.cos 3.exit

1


Enter no.of terms 5

The sin is 1.000003

1.sin 2.cos 3.exit

2


Enter no.of terms 5

The cos is 0.707388

1.sin 2.cos 3.exit

2


Enter no.of terms 5

The cos is 0.000821

1.sin 2.cos 3.exit

3

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C Programming Lab


EXERCISE : 14

a) Pointer based Function to exchange value of two integers using passing by address.

DESCRIPTION:

This program interchange values of two variable using functions and sending addresses.

ALGORITHM:

Step 1 : Start

Step 2 : Read num1,num2.

Step 3 : Display before swapping num1,num2.

Step 4 : Calling swap function.

Step 5 : Display after swapping.

Step 6 : Stop

Algorithm for swap:

Step 1 : start

Step 2 : assign t = *a;

Step 3 : assign *a =*b

Step 4 : assign *b = t

Step 5 : stop

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C Programming Lab


FLOWCHART:

Subprogram:

void swap()

temp=*a

*a=*b,

*b=temp

main()

start

After Swaping Print

num1,num2

Read num1, num2

Swap()

Stop

Before Swaping Print

num1,num2

I B.Tech

C Programming Lab


OUTPUT:

Enter 2 Numbers to Swap:10 20

Before Swapping:

Two Numbers are:Num1=10 Num2=20

After Swapping:


Enter 2 Numbers to Swap:55 66

Before Swapping:


After Swapping:


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C Programming Lab


EXERCISE : 14

a)Program which explains the use of dynamic arrays.

ALGORITHM

Step 1 : start

Step 2 : read n

Step 3 : allocate memory for n values of given type using malloc

Step 4 : assign 0 to i

Step 5 : if(i<n) go to step 6 else go to step 8

Step 6 : read a+i value

Step 7 : compute i = i+1 goto step 5


Step 9: if(i<n) go to step 10 else go to step 12

Step 10 : display *(a+i), (a+i)

Step 11 : increment i go to step9

Step 12: free the memory

Step13: allocate memory for n values of given type using calloc

Step 14: assign 0 to i

Step 15 : if(i<n) go to step16 else go to step 18

Step 16 : read a+i value



Step 19: if(i<n) go to step 20 else go to step 22

Step 20 : display *(a+i), (a+i)

Step 21 : increment i go to step19

Step 22: free the memory

Step 23 : Stop

I B.Tech

C Programming Lab


FLOWCHART:

true

true

true

false

false

false

i=0 i++

i<n

i=0 i++

i<n

Start

Read n

a=(int *) malloc(n*sizeof(int))

Read (a+i)

Print *(a+i), (a+i)

free(a)

Read (a+i)

a=(int *) calloc(n,sizeof(int))

A

i=0 i++

i<n

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C Programming Lab


OUTPUT:

Enter no.of values :5

enter 5 elements :50

40

30

20

10

The given elements are :

50 07B0

40 07B2

30 07B4

20 07B6

10 07B8

enter 5 elements :5

4

3

2

1

The given elements are :

5 07B0

4 07B2

3 07B4

2 07B6

1 07B8

true

false

Print *(a+i), (a+i)

i=0 i++

i<n

A

free(a)

Stop

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C Programming Lab


EXERCISE : 15

a) Program to display students information using structures.

DESCRIPTION:

This program demonstrates about structures.

ALGORITHM:

Step 1 : start

Step 2 : read n

Step 3 : initialize I =0

Step 4 : if(i<n) go to step 5 otherwise go to step 9

Step 5 : initialize total to 0

Step 6 : read student rollno, name, class, marks in 6 subjects

Step 7 : calculate total = sum of 6 subjects


Step 9 : initialize i =0

Step 10 : if(i<n) go to step 11 otherwise go to step 13

Step 11 : display “Name “ student name

Display “Roll no “ student rollno

Display “class “ student class

Display “total “ student total

Step 12 : compute i = i+1 , go to step 10

Step 13 : stop

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C Programming Lab


FLOWCHART:

true

true

true

false

false

START

STSTA

RT

Initialize

s[i].total=0

Read number of

students n

i=0 i++

i<n

Read roll num, name, class,

6 subjects marks

j=0 j<6

j++

s[i].total + = s[i].m[j]

Display student

details

i=0 i++

i<n

STOP

false

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C Programming Lab


OUTPUT:

Enter number of students :2

Enter 1 student details :

Enter rollno,name, class, marks in 6 subjects

2

ramu

first

20

21

22

23

24

25

Enter 2 student details :

Enter rollno,name, class, marks in 6 subjects

2

somu

second

31

32

33

34

35

36

The student‟s details:

NAME ROLLNO CLASS TOTALMARKS

ramu 1 first 135

somu 2 second 201

I B.Tech

C Programming Lab


b) Program to demonstrate the concept of unions.

DESCRIPTION

This program demonstrates the concept of unions clearly by nesting the union withing the structure

and showing the difference between structures and unions.

ALGORITHM

Step 1 : start

Step 2 : declare a structure.

Step 3 : Now declare a union nested within the structure.

Step 4 : create objects part1,part2 for the union and object obj for the structure.

Step 5 : assign the values for the members of both union and structure.

Step 6 : display the size of the structure and union inorder to see the difference between

the memory allocations of union and structure.

Step 7 : print the values of the members of union and structure using dot operator

obj.member


Step 9 : stop

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C Programming Lab


FLOWCHART:

OUTPUT:

size of structure=11

size of union=4

Item=123 Cost=45.67

Code=s Size=4

Assign values for

members

obj.part1.item=123;

obj.part2.cost=45.67;

obj.code='s';

obj.size=4;

print sizeof(obj)

sizeof(obj.part1)

obj.part1.item,obj.part2.cost,

obj.code, obj.size values

Stop

Start

Declare a union with

objects part1, part2

within a structure with

object obj

I B.Tech

C Programming Lab


EXERCISE: 16

a) Program which copies one file to another

DESCRIPTION:

In this program we have to use the file functions to perform the copy operation from one file to

another file.

ALGORITHM:

Step 1: Start

Step 2: read source file fname1, destination file fname2

Step 3: open source file in read mode

Step 4: open destination file in write mode

Step 5: if NULL pointer, then print unable to open files, go to step 8 otherwise go

to step6

Step 6: read a character from source file and write to destination file until EOF

Step 7: Close source file and destination file and print copied successfully

Step 8: Stop

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C Programming Lab


true

Start

Read source file

fname1

true

fs=fopen(fname1,"r");

ft=fopen(fname2,"w");

if

fs==NULL ||

ft==NULL

false

Print

Unable to open

false

ch=fgetc(fs);

fputc(ch,ft);

if ch!=EOF

fcloseall()

Print File copy

operation performed

successfully

Stop

Read destination

file fname2

exit

FLOW CHART:

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C Programming Lab


OUTPUT:

enter sourse file name1.c

enter dest file name2.c

File copy operation performed successfully

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C Programming Lab


b) To reverse the first n characters in a file

DESCRIPTION:

This program perform the reverse operation of n characters in the file

ALGORITHM:

Step 1: Start

Step 2: read the command line arguments

Step 3: check if arguments=3 or not

If not print invalid no of arguments

Step 4: open source file in read mode

Step 5: if NULL pointer, then print file can not be open

Step 6: Store no. of chars to reverse in k

k= *argv[2]-48

Step 7: read the item from file stream using fread

Step 8: Store chars from last position to initial position in another string(temp)

Step 9: print the temp string

Step 10: Stop

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C Programming Lab


true

true

false

false

true

Start

Read command

line args

If

argc!=3 fp=fopen(argv[1],”r” Print invalid

no of args

If

fp==NULL

Print file can

not be opened

false

k=atoi(argv[2])

n = fread(a,1,k,fp)

a[n]='\0'

len=strlen(a)

i = len-1 i--

i >=0

s[j]=a[i]

Print s[j]

j++

s[j+1]=‟\0‟

FLOW CHART:

Stop

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C Programming Lab


OUTPUT:

C:\TC\PROGRAMS>reverse source.c 8

ollehiah

C:\TC\PROGRAMS>reverse source.c

Invalid number of arguments.

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C Programming Lab


ADDITIONAL PROGRAMS:

EXCERCISE 1:

Program to find whether the given matrix is symmetric or not.

DESCRIPTION:

The transpose of a matrix is obtained by interchanging the row and column values of the matrix,

hence the order of the resultant matrix changes and therefore symmetry

ALGORITHM:

Step 1: start

Step 2: read the size of matrix

Step 3: read the elements of matrix

Step 4: perform the transpose operation by interchanging the row and

column values, the order of the resultant matrix

number of rows in transpose matrix=number of columns in the given matrix

number of columns in transpose matrix=number of rows in the given matrix

Step 6: Transpose is obtained through

at[i][j]=a[i][j]

Step 7: print the resultant transpose matrix at.

Step 8: check whether the given matrix is symmetric or not by comparing the

elements in the transpose matrix and the given matrix, if equal then goto

step 9 otherwise goto step 10.

Step 9: print the given matrix is symmetric

Step 10: print given matrix is not symmetric

Step 11: stop

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C Programming Lab


FLOWCHART:

false true if

s==1

Stop

Print

symmetric

Print not

symmetric

Read elements of

matrix

start

Read matrix size

s=symmetric(a,n)

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C Programming Lab


true

false

true

false

true

symmetric(a,n)

true

true

false

false

at[i][j]=a[j][i]

Print resultant

transpose matrix

i=0 i++

i<n

j=0 j<n

j++

false

j=0 j<n

j++

if a[i][j]!=

at[i][j]

Return 0

c++

Return to

main()

Return 1

i=0 i++

i<n

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C Programming Lab


OUTPUT:

Enter matrix size :2

Enter 4 elements :1

2

2

1

After Transpose :12

21

Symmetric

Enter matrix size :3

Enter 9 elements :1

2

3

4

5

6

7

8

9

After Transpose :147

258

369

Not symmetric

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C Programming Lab


EXCERCISE 2:

Write a C program to arrange the given numbers in the ascending order.

DESCRIPTION:

To arrange the given numbers in the ascending order i.e, sorting the given numbers in the ascending

order.

ALGORITHM:

Step1: Start

Step2: Read n


Step4: if i<n go to step5, otherwise go to step6

Step5: Read a[i], go to step4




Step9: if j<n go to step10, otherwise go to step12

Step10: if a[i]>a[j], otherwise go to step 11

i) temp=a[i];

ii) a[i]=a[j];

iii) a[j]=temp; go to step11

Step11: increment j, go to step9

Step12: increment i, go to step7



Step15: Print a[i], go to step14

Step16: Stop

I B.Tech

C Programming Lab


FLOWCHART:

true

false

false

false

true

true

true

Start

Read n

i=0 i++

i<n

i=0 i++

i<n

j=0 j<n

j++

If

a[i]>a[j

]

temp=a[i]

a[i]=a[j]

a[j]=temp

Read

a[i]

i=0 i++

i<n

Print a[i]

Stop

false

true

false

I B.Tech

C Programming Lab


OUTPUT:

Enter number of elements :5

5

4

3

2

1

The 5 numbers in Ascending order :

1

2

3

4

5

Enter number of elements :8

33

44

22

11

66

55

17

88

The 8 numbers in Ascending order :

11

17

22

33

44

55

66

88

I B.Tech

C Programming Lab


EXCERCISE 3:

Write a c program to merge the contents of two files into one file.

DESCRIPTION:

To merge the contents of two files into the one file. This program uses the standard predefined

functions to perform this operation.

ALGORITHM:

Step1: Start

Step2: Read two files names

Step3: Read another file name to store content of two files

Step4: open two files in read mode

Step5: if fs1 == NULL || fs2 == NULL go to step 6, otherwise go to step 7

Step6: Print Error ,go to step 15

Step7: Read third file in write mode

Step8: if ft==NULL print Error go to step 15, otherwise go to step 9

Step9: if (ch = fgetc(fs1) ) != EOF ,go to step10 otherwise go to step11

Step10: fputc(ch,ft) go to step 9

Step11: if ( ch = fgetc(fs2) ) != EOF, go to step12 otherwise go to step13

Step12: fputc(ch,ft) go to step 11

Step13: Print two files are merged successfully

Step14: close three files

Step15: Stop

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C Programming Lab


FLOWCHART:

true false

false

Start

Read another File

to store content of

two files

Read two

File names

fs1 = fopen(file1,"r");

fs2 = fopen(file2,"r");

if

fs1 == NULL ||

fs2 == NULL

Print Error

ft = fopen(file3,"w")

if

ft == NULL Print Error

A B C

true

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C Programming Lab


true false

false if

( ch = fgetc(fs1) )

!= EOF

fputc(ch,ft)

if

( ch = fgetc(fs2) )

!= EOF

fputc(ch,ft)

Two files are

merged

successfully

fclose(fs1)

fclose(fs2)

fclose(ft)

Stop

C

true

A B

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C Programming Lab


OUTPUT:

Enter name of first file 1.c

Enter name of second file 2.c

Enter name of file which will store contents of two files 4.c

Two files were merged into 4.c file successfully.

EXERCISE: 1 - cse.gecgudlavalleru.ac.incse.gecgudlavalleru.ac.in/pdf/manuals/clab.pdf · I B.Tech C...

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