Hey below I have given you the CCP lab manual preferred by VTU so you can have a look

VTU CCP Lab Programs

PART – A

1. Design, develop and execute a program in C to find and output all the roots of a given
quadratic equation, for non-zero coefficients.

2. Design, develop and execute a program in C to implement Euclid’s algorithm to find the
GCD and LCM of two integers and to output the results along with the given integers.

3. Design, develop and execute a program in C to reverse a given four digit integer number
and check whether it is a palindrome or not. Output the given number with suitable
message.

4. Design, develop and execute a program in C to evaluate the given polynomial
f(x) = a4x4 + a3x3 + a2x2 + a1x1 + a0 for given value of x and the coefficients using Horner’s
method.

5. Design, develop and execute a program in C to copy its input to its output, replacing each
string of one or more blanks by a single blank.

6. Design, develop and execute a program in C to input N integer numbers in ascending order
into a single dimension array, and then to perform a binary search for a given key integer
number and report success or failure in the form of a suitable message.

7. Design, develop and execute a program in C to input N integer numbers into a single
dimension array, sort them in to ascending order using bubble sort technique, and then to
print both the given array and the sorted array with suitable headings.
8. Design, develop and execute a program in C to compute and print the word length on the
host machine.
PART –B
9. Design, develop and execute a program in C to calculate the approximate value of exp (0.5)
using the Taylor Series expansion for the exponential function. Use the terms in the
expansion until the last term is less than the machine epsilon defines as FLT_EPSILON in the
header file <float.h>. Print the value returned by the Mathematical function exp ( ) also.

10. Design, develop and execute a program in C to read two matrices A (M x N) and B (P x Q)
and to compute the product of A and B if the matrices are compatible for multiplication.
The program is to print the input matrices and the resultant matrix with suitable headings
and format if the matrices are compatible for multiplication, otherwise the program must
print a suitable message.(For the purpose of demonstration, the array sizes M, N, P, and Q
can all be less than or equal to 3)



11. Design, develop and execute a parallel program in C to add, element-wise, two onedimensional
arrays A and B of N integer elements and to store the result in another onedimensional
array C of N integer element.

12. Design and develop a function rightrot(x, n) in C that returns the value of the integer x
rotated to the right by n bit positions as an unsigned integer. Invoke the function from the
main with different values for x and n and print the results with suitable headings.

13. Design and develop a function isprime(x) that accepts an integer argument and returns 1 if
the argument is prime and 0 otherwise. The function is to use plain division checking
approach to determine if a given number is prime. Invoke this function from the main with
different values obtained from the user and print appropriate messages.

14. Design, develop and execute a parallel program in C to determine and print the prime
numbers which are less than 100 making use of algorithm of the Sieve of Eratosthenes.

15. Design and develop a function reverses(s) in C to reverse the string s in place. Invoke this
function from the main for different strings and print the original and reversed strings.

16. Design and develop a function matchany (s1,s2) which returns the first location in the string
s1 where any character from the string s2 occurs, or – 1 if s1 contains no character from s2.
Do not use the standard library function which does a similar job! Invoke the function
matchany (s1. s2) from the main for different strings and print both the strings and the
return value from the function matchany (s1, s2).
1. Note: In the practical examination the student has to answer two questions. One
question from Part A and one question from Part B, will be selected by the student by
lots. All the questions listed in the syllabus have to be included in the lots. The change of
question (Part A only / Part B only / Both Part A & Part B) has to be considered, provided
the request is made for the same, within half an hour from the start of the examination.
The allotment of marks is as detailed below:
Sl. # Activity Max. Marks
1 Procedure
Writing program & procedure for the assigned problems along with
algorithms / flowchart
Part A 5*
Part B 5*
2 Conduction
Execution of the program and showing the results in proper format
Part A 10
Part B 20
3 Viva-voce** 10
Total Max. Marks 50
Minimum Passing Marks (40% of Max. Marks) 20

PART – A

1) Design, develop and execute a program in C to find and output all the roots of
a given quadratic equation,for non-zero co-efficient
Algorithm
1. Start.
2. Input co-efficient of equation a, b, c .
3. IF any or all the coefficients are zero
Print Invalid input
ELSE
d ← b2- 4ac
r ←√ |d|
IF d > 0
r1 ← (-b +r)/ (2a)
r2 ← (-b -r)/ (2a)
Print “Roots are REAL and DISTINCT”
Print r1, r2
ELSE IF d < 0
r1 ←-b/ (2a)
r2 ←r/ (2a)
Print “Roots are COMPLEX”
Print r1 “+i” r2, r1 “- i” r2
ELSE
r1←-b/(2a)
Print “Roots are EQUAL”
Print r1, r1
END IF
END IF
END IF.
4. Stop


Program
#include <stdio.h>
#include <math.h>
void main()
{
int a,b,c;
float d,x1,x2,r;
printf("Enter the three co-efficient :\n");
scanf("%d%d%d",&a,&b,&c);
if (a* b* c == 0)
{
printf("\n Invalid Input ");
}
else
{
d = b * b - 4 * a * c;
r=sqrt(fabs(d));
if (d > 0)
{
x1 = (-b +r) / (2.0*a);
x2 = (-b -r) / (2.0*a);
printf("\n The roots are real and distinct\n");
printf("\n The roots are \n 1) x1=%f\t\t \n 2) x2=%f",x1,x2);
}
else if (d == 0)
{
x1 = x2 = -b/(2.0*a);
printf("\n The roots are real and equal\n");
printf("\n The roots are: \n 1) x1=x2=%f",x1);
}
else
{
x1 = -b / (2.0 * a);
x2 = r / (2.0*a);
printf("\n The roots are real and imaginary\n");
printf("\n The roots are:\n 1) %f +i %f \t\t\n 2) %f –i %f ",x1,x2,x1,x2);
}
}
}

Sample Output
1. Enter the three co-efficient :
1 4 4
The roots are real and equal
The roots are:
X1=X2=2.0000
2. Enter the three co-efficient :
1 - 5 6
The roots are real and distinct
The roots are:
X1=3.0000
X2=2.0000
3. Enter the three co-efficient :
2 3 4
The roots are real and imaginary
The roots are:
1) -0.750000 +i 1.198958
2) -0.750000 - i 1.198958
4. Enter the three co-efficient :
1 0 5
Invalid Input

2) Design, develop and execute a program in C to implement Euclid’s algorithm to
find the GCD and LCM of two integers and to output the results along with the
given integers.

Algorithm
1. Start.
2. Input m , n.
3. Initialize p ← m , q ← n.
4. Until n <> 0
rem ← m mod n
m ← n
n ← rem
END until
5. GCD ← m
6. LCM ← p * q / GCD.
7. Print p, q, GCD, LCM.
8. Stop



Program
#include<stdio.h>
void main()
{
int m,n,p,q,gcd,lcm,rem;
printf("Enter two numbers : ");
scanf("%d%d",&m,&n);
p = m;
q = n;
while(n!=0)
{
rem=m%n;
m=n;
n=rem;
}
gcd = m;
lcm = (p * q) / gcd;
printf("\n The LCM of %d and %d = %d",p,q,lcm);
printf("\n The GCD of %d and %d = %d",p,q,gcd);
}
Sample Output
Enter two numbers:
34
66
The LCM = 1122
The GCD = 2

3) Design, develop and execute a program in C to reverse a given four digit
integer number and check whether it is a palindrome or not. Output the given
number with suitable message.
Algorithm
1. Start.
2. Input n.
3. Initialize a ← n, rev ← 0, rem ← 0.
4. IF n <= 999 OR n>9999
Print “Not a 4 digit number”
Goto step 7.
5. Until n <>0
rem ← n % 10
rev ← rev *10 + rem.
n ← n / 10.
END until
6. IF a EQUAL TO rev
Print “Palindrome”.
ELSE
Print “Not a Palindrome”.
7. Stop.

Program
#include <stdio.h>
void main()
{
int n,rev=0,rem,a;
printf("Enter a number : ");
scanf("%d",&n);
a = n;
if(n<=999 || n>9999)
{
printf(“ Not a 4 digit number\n”);
exit(0);
}
while(n != 0)
{
rem=n%10;
rev= rev*10+rem;
n = n / 10;
}
if(a==rev)
printf("\n The given Number %d is Palindrome",a);
else
printf("\n The given Number %d is not Palindrome",a);
}
Sample Output
1. Enter a number:
201
Not a four digit number
2. Enter a number:
5642
The Number is not Palindrome
3. Enter a number:
8118
The Number is Palindrome

4) Design, develop and execute a program in C to evaluate the given polynomial
f(x)= a4x4 + a3x3 + a2x2 + a1x1 + a0 for given value of x and the coefficients using
Horner’s method.
Algorithm
1. Start.
2. Read n.
3. FOR i ← 0 to n in steps of 1
Read a[i]
END FOR
4. Read x.
5. poly ← a[0]
6. FOR i ← 1 to n in steps of 1
poly ← poly * x + a[i]
END FOR.
7. Print poly.
8. Stop.

Program
#include <stdio.h>
void main()
{
int n,i,x,a[10],poly=0;
printf("\n Enter the degree of the polynomial : ");
scanf("%d",&n);
printf("\n Enter the %d coefficients\n",n+1);
for(i = 0 ; i <= n ; i++)
{
scanf("%d",&a[i]);
}
printf("\n Enter the value of x :");
scanf("%d",&x);
poly=a[0];
for(i = 1 ; i <= n ; i++)
{
poly = poly* x+a[i];
}
printf("\n The sum of polynomial = %d",poly);
}
Sample Output
Enter the degree of the polynomial : 4
Enter the 5 coefficient
1 2 3 4 5
Enter the value of x :
1
The sum of polynomial = 15

5) Design, develop and execute a program in C to copy its input to its output,
replacing each string of one or more blanks by a single blank.
Algorithm
1. Start
2. Read the text in Array c
3. FOR i← 0 to c [i] <> '\0' in steps of 1
IF c [i] = ' '
Print ' '
End If
Until c [i] = ‘\0'
Increment i
End Until
Print the character
End For
4. Stop

Program
#include <stdio.h>
#include<string.h>
void main()
{
char c[50];
int I;
printf("Enter the text");
scanf(“%[^\t\n]”, c);
for(i=0;c[i]!='\0';i++)
{
if (c[i]==' ')
printf(“%c”, c[i]);
while (c[i]==' ')
i++;
printf(“%c”, c[i]);
}
}
Sample output
Enter the text
welcome
welcome

6) Design, develop and execute a program in C to input N integer numbers in
ascending order into a single dimension array, and then to perform a binary
search for a given key integer number and report success or failure in the form of
a suitable message.
Algorithm
1. Start.
2. Initialize Flag ← 0
3. Read n.
4. FOR i← 0 to n-1 in steps of 1
Read numbers in ascending order in Array a[ ]
END FOR.
5. Read key.
6. Initialize low←0, high←n -1.
7. Until low <= high
mid←(low + high)/ 2.
8. IF key = a[mid]
Flag=1
Goto Step 10
END IF
9. IF key > a[mid]
low←mid + 1
ELSE
high←mid – 1
END IF
END Until
10. IF Flag = 1
Print “Successful Search”.
ELSE
Print “Unsuccessful Search”.
END IF
11. Stop.

program
#include <stdio.h>
void main()
{
int n,i,a[10],key,low,high,mid,Flag=0;
printf("\n Enter the no. of elements : ");
scanf("%d",&n);
printf("\n Enter %d elements in ascending order ",n);
for(i = 0 ; i < n ; i++)
scanf("%d",&a[i]);
printf("\n Enter the key element to search : ");
scanf("%d",&key);
low = 0;
high = n-1;
while(low <= high)
{
mid=(low+high)/2;
if(key==a[mid])
{
Flag=1;
break;
}
if(key > a[mid])
low=mid+1;
else
high = mid-1;
}
if(Flag==1)
printf("\n Successful Search ");
else
printf("\n Unsuccessful Search");
}

Sample Output
Enter the no. of elements: 5
Enter 5 elements in ascending order
10
20
30
40
50
Enter the key element to search: 30
Search Successful
Enter the no. of elements: 5
Enter 5 elements in ascending order
2345
6
Enter the key element to search : 7
unsuccessful Search

7) Design, develop and execute a program in C to input N integer numbers into a
single dimension array, sort them in to ascending order using bubble sort
technique, and then to print both the given array and the sorted array with
suitable headings.
Algorithm
1. Start.
2. Read n.
3. FOR i←0 to n-1 in steps of 1
Read integer numbers in array a [ ]
b[i]←a[i]
END FOR.
4. FOR i←0 to n-1 in steps of 1
FOR j←0 to n-i in steps of 1
IF a[j] > a[j + 1]
temp= a[j]
a[j] = a[j + 1]
a[j+1] = temp
END IF
END FOR
END FOR
5. Print original array b[i]
FOR i←0 to n-1 in steps of 1
Print b[i]
END FOR.
6. Print sorted array a[i]
FOR i←0 to n-1 in steps of 1
Print a[i]
END FOR.
7. Stop.

Program
#include <stdio.h>
void main()
{
int n,i,j,a[10],b[10],temp;
printf("\n Enter the no. of elements : ");
scanf("%d",&n);
printf("\n Enter %d elements ",n);
for(i = 0 ; i < n ; i++)
{
scanf("%d",&a[i]);
b[i]=a[i];
}f
or(i = 0 ; i < n-1 ; i++)
{
for(j = 0 ; j < n-i; j++)
{
if(a[j] > a[j+1])
{
temp = a[j];
a[j] = a[j+1];
a[j+1] = temp;
}
}
}
printf("\n The original elements are\n ");
for(i = 0 ; i < n ; i++)
printf("%d \n",b[i]);
printf("\n The Sorted elements are ");
for(i = 0 ; i < n ; i++)
printf("%d \n",a[i]);
}

Sample Output
Enter the no. of elements : 5
Enter 5 elements
91
15
2
31
4
The original elements are
91
15
2
31
4
The Sorted elements are
131
4
52
91

8) Design, develop and execute a program in C to compute and print the word
length on the host machine.
Algorithm
1. Start
2. Initialize
Var←-1
Wordlen ←0
3. Until (var)
Wordlen++
Var<<←1
End Until
4. Print The word length of host machine
5. Stop

Program
#include<stdio.h>
void main()
{
int var = -1,wordlen=0;
while (var)
{
wordlen++;
var <<= 1;
}
printf(" The Word length of this Host Machine is %d Bits", wordlen);
}
Sample output
The Word length of this Host Machine is 32 Bits

PART-B

9) Design, develop and execute a program in C to calculate the approximate value
of exp (0.5) using the Taylor Series expansion for the exponential function. Use
the terms in the expansion until the last term is less than the machine epsilon
defines as FLT_EPSILON in the header file <float.h>.Print the value returned by
the mathematical function exp( ) also.
Algorithm
1. Start
2. Read X
3. Initialize sum←0,term←1,fact←1
4. for i←1; term >= FLT_EPSILON; i++
fact ← fact*i
sum ← sum+term
term ← pow(x,i)/fact
5. Print calculated value
6. Print Library function value.
7. Stop

Program
#include<stdio.h>
#include <float.h>
#include <math.h>
void main()
{
int i;
float x,sum,fact,term;
printf("\nYou have this series : 1+x/1!+x^2/2!+ x^3/3! + x^4/4!+……………….x^n/n!");
printf("\n\nEnter the value for X : ");
scanf("%f",&x);
sum = 0;
term = 1;
fact = 1;
for(i=1;term >= FLT_EPSILON;i++)
{
fact =fact * i;
sum = sum + term;
term = pow(x,i)/fact;
}
printf("\n\nThe Calculated value of e^% .3f = %f",x,sum);
printf("\n\nThe Library Function Value of e^%.3f = %f",x,exp(x));
}
Sample Output
You have this series: 1+x/1!+x^2/2!+x^3/3!+x^4/4!+…….x^n/n!
Enter the value for X : 0.5
The Calculated value of e^ 0.500 = 1.648721
The Library Function Value of e^ 0.500 = 1.648721

10) Design, develop and execute a program in C to read two matrices A(M*N) and
B(P*Q) and to compute the product of A and B if the matrices are compatible for
multiplication. The program is to print the input matrices and resultant matrix
with suitable headings and format if the matrices are compatible for
Multiplication. Otherwise the program must print a suitable message.(For the
purpose of demonstration, the array sizes M,N,P and Q can all be less than or
equal to 3).
Algorithm
1. Start.
2. Read order m, n
3. Read order p, q
4. IF n==p THEN
FOR i ← 0 to m in steps of 1
FOR j ← 0 to n in steps of 1
Read a[i][j]
END FOR
END FOR
5. FOR i ← 0 to p in steps of 1
FOR j ← 0 to q in steps of 1
Read b[i][j]
END FOR
END FOR
6. Matrix multiplication
FOR i ← 0 to m in steps of 1
FOR j ← 0 to q in steps of 1
c[i][j] ← 0
FOR k ← 0 to n in steps of 1
c[i][j] ← c[i][j] + a[i][k] * b[k][j]
END FOR
END FOR
END FOR
End if
7. Print matrix A
FOR i ← 0 to m in steps of 1
FOR j ← 0 to n in steps of 1
Print a[i][j]
END FOR
END FOR
8. Print matrix B
FOR i ← 0 to p in steps of 1
FOR j ← 0 to q in steps of 1
Print b[i][j]
END FOR
END FOR
9. Print matrix C
sample out put
FOR i ← 0 to m in steps of 1
FOR j ← 0 to q in steps of 1
Read c[i][j]
END FOR
END FOR
ELSE
Print “Multiplication not possible”
END IF
10. Stop.

Program
#include <stdio.h>
void main()
{
int a[10][10],b[10][10],c[10][10];
int m,n,p,q,i,j,k;
printf("\n Enter the order of the matrix A :");
scanf("%d%d",&m,&n);
printf("\n Enter the order of the matrix B :");
scanf("%d%d",&p,&q);
if(n==p)
{
printf("\n Enter the elements of matrix A \n");
for(i = 0 ; i < m ; i++)
{
for(j = 0 ; j < n ; j++)
scanf("%d",&a[i][j]);
}
printf("\n Enter the elements of matrix B \n");
for(i = 0 ; i < p ; i++)
{
for(j = 0 ; j < q ; j++)
scanf("%d",&b[i][j]);
}f
or(i = 0 ; i < m ; i++)
{
for(j = 0 ; j < q ; j++)
{
c[i][j]=0;
for(k = 0 ; k < n ; k++)
c[i][j] += a[i][k] * b[k][j];
}
}
printf("\n MATRIX A \n");
for(i = 0 ; i < m ; i++)
{
for(j = 0 ; j < n ; j++)
{
printf(" %d \t", a[i][j]);
}
printf("\n");
}
printf("\n MATRIX B \n");
for(i = 0 ; i < p ; i++)
{
for(j = 0 ; j < q ; j++)
{
printf(" %d \t", b[i][j]);
}
printf("\n");
}
printf("\n MATRIX C \n");
for(i = 0 ; i < m ; i++)
{
for(j = 0 ; j < q ; j++)
{
printf(" %d \t", c[i][j]);
}
printf("\n");
}}
else
printf("Matrix A & B is not multiplicable");
}
Sample Output
Enter the order of the matrix A :2 2
Enter the order of the matrix B :2 2
Enter the elements of matrix A
1 2
3 4
Enter the elements of matrix B
2 3
4 5
MATRIX A
1 2
3 4
MATRIX B
2 3
4 5
MATRIX C
10 13
22 29
Enter the order of the matrix A :2 3
Enter the order of the matrix B :2 3
Matrix A & B is not multiplicable

11) Design, develop and execute a parallel program in C to add, element-wise, two
one-dimensional arrays A and B of N integer elements and to store the result in
another one-dimensional array C of N integer elements.
Algorithm
1. Start
2. Read n
3. For i ← 0 to n in steps of 1
Read elements of Array A
END For
4. For j ← 0 to n in steps of 1
Read elements of Array B
5. For i←0, to n in steps of 1
c[i]=a[i]+b[i]
Print C
6. Stop.

program
#include<stdio.h>
#include<omp.h>
void main()
{
int i,j,k,n,a[10],b[10],c[10];
printf("enter the size of an arrays\n");
scanf("%d",&n);
printf("enter array elements of A\n");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("enter array elements of B\n");
for(j=0;j<n;j++)
scanf("%d",&b[j]);
#pragma omp parallel for
for(i=0;i<n;i++)
{
c[i]= a[i]+b[i];
printf ("c[%d]=%d,threadno=%d\n",i,c[i],omp_get_thread_num);
}
}
Sample output
enter the size of an arrays
5E
nter array elements of A
24681
0
Enter array elements of B
12345T
he array elements of C are
C [0] =3
C [1] =6
C [2] =9
C [3] =12
C [4] =15

12) Design and develop a function rightrot(x,n) in C that returns the value of the
integer x rotated to the right by n bit positions as an unsigned integer. Invoke the
function from the main with different values for x and n and print the results with
suitable headings.
Algorithm
1. Start
2. Read x and n
3. Call function RR(x, n)
6. Print the result
7. Stop
Algorithm to rotate value by n bits
1. IF n==0
Return x
ELSE
Return ((x >> n) | (x << (32- n)))
2. Return


Program
#include<stdio.h>
void main ( )
{
unsigned int ans,x,n;
printf("enter x and n values\n");
scanf("%d%d",&x,&n);
ans=rightrot(x,n);
printf("\nThe value after rotating %d bit is : ",n);
printf("%d",ans);
}
int rightrot(unsigned int x,unsigned int n)
{
if (n == 0)
return x;
else
return ((x >> n) | (x << ( 32- n)));
}
Sample Output
Enter x and n values
8 3
The value after rotating 3 bit is: 1

13) Design and develop a function isprime(x) that accepts an integer argument
and returns 1 if the argument is prime and 0 otherwise. The function is to use
plain division checking approach to determine if a given number is prime. Invoke
this function from the main with different values obtained from the user and print
appropriate messages.
Algorithm
1. Start
2. Read the number
3. Call function prime(n)
P= prime(n)
4. IF p ← 1
Print “ Number is prime”
ELSE
Print “Number not a prime”
End IF
5. Stop.
Algorithm to check the prime number
1. FOR i ←2 to n/2 in steps of 1
IF (n mod i) is equal to zero
Return 0
Return 1
End For
2. Return.

Program
#include<stdio.h>
int prime(int n)
{
int n,i;
for(i=2;i<=n/2;i++)
if(n%i ==0)
return 0;
return 1;
}
void main()
{
int n,p;
printf("enter the no\n");
scanf("%d",&n);
p=prime(n);
if(p==1)
printf("The given no is prime\n");
else
printf("The given no is not prime\n");
}
Sample output
Enter the number
17
The given number is prime
Enter the number
33
The given number is not prime

14) Design, develop and execute a parallel program in C to determine and print
the prime numbers which are less than 100 making use of algorithm of the sieve
of Eratosthenes.
Algorithm
1. Start
2. Initialize n ← 100
3. For i←0 to n in steps of 1
num[i] ← i
END For
4. For i←2 to in steps of 1
IF num[i] != 0
For j← i*i to n in steps of j← j+i
num[j] ← 0
END For
END IF
END For
5. For i←0 to n in steps of 1
iF num[i] != 0
Print non zero numbers
END IF
END For
6. Stop

Program
#include<stdio.h>
#include<math.h>
#include<omp.h>
int main()
{
int num[120], i, j,n;
printf("enter the value of n\n”);
scanf(“%d”,&n);
#pragma omp parallel for
for(i=0;i<=n;i++)
{
num[i]=i;
}
#pragma omp parallel for
for(i=2;i<=sqrt(n);i++)
{
if(num[i]!=0)
{
#pragma omp parallel for
for(j=(i*i);j<=n;j=j+i)
{
num[j]=0;
}}
}
#pragma omp parallel for
printf(“The prime numbers that are less 100\n”);
for(i=0;i<=n;i++)
{
if(num[i]!=0)
printf("\nprimeno=%d threadID=%d\n",num[i],omp_get_thread_num());
}
}

Sample output
The prime numbers that are less than 100
Prime no=2 threadID=0
Prime no=3 threadID=0
Prime no=5 threadID=0
Prime no=7 threadID=0
Prime no=11 threadID=0
Prime no=13 threadID=0
Prime no=17 threadID=0
Prime no=19 threadID=0
Prime no=23 threadID=0
Prime no=29 threadID=0
Prime no=31 threadID=0
Prime no=37 threadID=0
Prime no=41 threadID=0
Prime no=43 threadID=0
Prime no=47 threadID=0
Prime no=53 threadID=0
Prime no=59 threadID=0
Prime no=61 threadID=0
Prime no=67 threadID=0
Prime no=71 threadID=0
Prime no=73 threadID=0
Prime no=79 threadID=0
Prime no=83 threadID=0
Prime no=89 threadID=0
Prime no=97 threadID=0

15) Design and develop a function reverses(s) in C to reverse the string s in
place. Invoke this function from the main for different strings and print the
original and reversed strings.
Algorithm
1. Start
2. Read the string
3. Print the string
4. Call function reverse(s)
5. Stop
Algorithm to reverse a string
1. Start
2. Calculate the length
3. For i← 0, i<string length/2 in step of 1
temp← str[i]
str[i]←str[len]
str[len]←temp
decrement length
END For
4. Print the reverse string
5. Stop
Flowchart


Program
#include <stdio.h>
#include <string.h>
void reverse(char s[])
{
char s[10],temp;
int i,len;
len = strlen(s)-1;
for(i=0;i<strlen(s)/2;i++)
{
temp=s[i];
s[i]=s[len];
s[len]=temp;
len--;
}
printf("the reversed string is = %s",s);
}
void main()
{
char s[10];
int i;
printf("Enter String : ");
scanf("%s",s);
printf("the original string is=%s\n",s);
reverse(s);
}
Sample Output
Enter String : computer
the original string is=computer
the reversed string is=retupmoc
Enter String : magic
the original string is=magic
the reversed string is=cigam

16) Design and develop a function matchany(s1,s2) which returns the first
location in the string s1 where any character from the string s2 occurs, or -1 if s1
contains no character from s2. Do not use the standard library function which
does a similar job! Invoke the function matchany(s1,s2) from the main for
different strings and print both the strings and the return value from the function
matchany(s1,s2).
Algorithm
1. Start
2. Read string in str1
3. Read string in str2
4. Call function x = matchany(str1,str2)
5. IF x is equal -1
Print no character matching
ELSE
Print position and matched character
End if
6. Stop.
Algorithm of matchany function
1.For i ← 0 to s1[i] != null in steps of 1
For j ← 0 to s2[j] != null in steps of 1
IF s1[i] = s2[j]
Return i
End IF
End For
End For
2. Return -1
3. Return

Program
#include <stdio.h>
#include <string.h>
int matchany(char s1[ ], char s2[ ]);
void main()
{
char str1[20],str2[20];
int x;
printf("\nEnter first String:");
scanf("%s",str1);
printf("\nEnter second String:");
scanf("%s",str2);
x=matchany(str1,str2);
if(x==-1)printf("No charecter matching\n");
else
printf("The position is %d and charecter is %c\n",x+1,str1[x]);
}
int matchany(char s1[10], char s2[10])
{
int i,j;
for(i=0;s1[i]!='\0';i++)
{
for(j=0;s2[j]!='\0';j++)
{
if(s1[i] == s2[j])
return i;
}
}
return -1;
}
Sample Output
Enter first String: india
Enter second String: mandia
The position is 1 and character is i
Enter first String: hkbk
Enter second String: class
No character matching
END