3. OUTPUT :
Enter principle :10000
Enter rate :8
Enter time :4
Enter compoundings per year :2
Simple Interest =3200.00
Compond Interest=3685.69
C’ software lab - 3 - MCA’05
4. Experiment No. 2
Date:
AIM : TO FIND THE AMSTRONG NUMBER AMONG THE GIVEN NUMBERS
ALGORITHM :
Step1: Start
Step2: Accept a Number
Step3: Assign the number to another variable
Step4: Check whether the no is greater than Zero go to step 8
Step5: Extract the last digit and add the cube to a sum variable
Step6: Divde the number by 10
Step7: Goto step 4
Step8: Check whether the sum = temporary variable and print result
Step9: Stop
C’ software lab - 4 - MCA’05
5. /*Armstrong number*/
#include<stdio.h>
#include<conio.h>
void main()
{
int num,arms=0,tmp,mod;
clrscr();
printf("n Enter the number :");
scanf("%d",&num);
tmp=num;
while(num>0)
{
mod=num%10;
arms+=mod*mod*mod;
num=num/10;
}
if(tmp==arms)
printf("n Number %d is armstrong",arms);
else
printf("n Number %d is not armstrong",tmp);
getch();
}
C’ software lab - 5 - MCA’05
6. OUTPUT :
1. Enter the number :153
Number 153 is armstrong
2. Enter the number :111
Number 111 is not armstrong
C’ software lab - 6 - MCA’05
7. Experiment No. 3
Date:
AIM : TO FIND THE ROOTS OF A QUADRATIC EQUATION
ALGORITHM :
Step1: Start
Step2: Accept values for the variables x2 x and constant
Step3: Apply the formula b2-4ac (b=cof x2,a=cof x,c=constant) as save to a variable
Step4: Check whether the value is less than 0 then print roots are imaginary
Step5: If value=0 then print the roots are equal –b/2a
Step6: If value greater than 0, 1st root is (-b+sqrt(value))/(2*a),
2nd root is (-b-sqrt(value))/(2*a).
Step7: Print the result
Step8: Stop
C’ software lab - 7 - MCA’05
8. /*Quadratic Equation*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float a,b,c,d,r1,r2;
clrscr();
printf("n Enter the Coeff. of the Eq. :");
scanf("%f%f%f",&a,&b,&c);
d=(b*b)-(4*a*c);
if(d==0)
{
printf("n Roots are REAL and EQUAL");
r1=-b/(2*a);
printf("n Root Is : %.2f",r1);
}
else if (d>0)
{
printf("n Roots are Real and Distinct");
r1=(-b+sqrt(d))/(2*a);
r2=(-b-sqrt(d))/(2*a);
printf("n Root 1 : %.2f ..... Root 2 : %.2f",r1,r2);
}
else
{
printf("n Roots are imaginary");
}
getch();
}
C’ software lab - 8 - MCA’05
9. OUTPUT :
1.Enter the Coeff. of the Eq. :1 -5 6
Roots are Real and Distinct
Root 1 : 3.00 ..... Root 2 : 2.00
2.Enter the Coeff. of the Eq. :4 4 1
Roots are REAL and EQUAL
Root Is : -0.50
C’ software lab - 9 - MCA’05
10. Experiment No. 4
Date:
AIM : TO PERFORM MATRICS ADDITION, MULTIPLICATION. TRANSPOSE.
ALGORITHM :
Step 1: Start
Step 2: Declare three integer 2 dimensional array of order 3*3 .
Step 3: Accept a choice from user
Step 4: If choice = 1 : goto Step 9
Step 5: If choice = 2 : goto Step 10
Step 6: If choice = 3 : goto step 11
Step 7: else if choice = 4 : goto Step 12
Step 8: Accept the elements of the two 2 D arrays
Step 9: If choice = 1
a. Add the elements that belong corresponding cells in the arrays and store in
third array’s respective cell.
b. Display the result.
Step 10 : If choice = 2
a. Multiply the elements of the first two matrices and store the result in third
array.
b. Display the result.
Step 11 : If choice = 3
a. Accept the elements of the matrics one.
b. Display the matrics in the order column*row format.
Step 13 : Repeat step 3 to 11.
Step 14 : Stop.
C’ software lab - 10 - MCA’05
11. /*Matrics Addition, Multiplication, Transpose*/
#include<stdio.h>
#include<conio.h>
int a[10][10],b[10][10],c[10][10],r1,r2,c1,c2,i,j,k,ch;
void add();
void mul();
void tran();
void accept(int [][10],int,int);
void display(int [][10],int,int);
void main()
{
clrscr();
do
{
clrscr();
printf("n Matrix Manipulator");
printf("n 1.Addition");
printf("n 2.Multiplication");
printf("n 3.Transpose");
printf("n press any other key to exit...");
printf("n Enter U'r choice :");
scanf("%d",&ch);
if (ch>=1&&ch<=3)
{
printf("n Enter order of the matrices :");
printf("n Mat 1 :");
scanf("%d%d",&r1,&c1);
printf("n Mat 2 :");
scanf("%d%d",&r2,&c2);
}
switch(ch)
{
C’ software lab - 11 - MCA’05
12. case 1:
if((r1==r2)&&(c1==c2))
add();
else
printf("n addition not possible...");
getch();
break;
case 2:
if(c1==r2)
mul();
else
printf("n multiplication not possible...");
getch();
break;
case 3: tran();getch();break;
default: printf("n Wrong choice Exit...");
}
}while(ch>0&&ch<4);
getch();
}
void add()
{
printf("n Matrics addition");
printf("n Enter Matrics 1:");
accept(a,r1,c1);
printf("n Enter Matrics 2:");
accept(b,r2,c2);
for(i=0;i<r1;i++)
for(j=0;j<c1;j++)
c[i][j]= a[i][j]+b[i][j];
printf("n Matrics A :n");
C’ software lab - 12 - MCA’05
13. display(a,r1,c1);
printf("n Matrics B :n");
display(b,r2,c2);
printf("n Matrics C :n");
display(c,r1,c2);
}
void mul()
{
printf("n Matrics multiplication");
printf("n Enter Matrics 1:");
accept(a,r1,c1);
printf("n Enter Matrics 2:");
accept(b,r2,c2);
for(i=0;i<r1;i++)
for(j=0;j<c2;j++)
{
c[i][j]=0;
for(k=0;k<c1;k++)
c[i][j] += a[i][k]*b[k][j];
}
printf("n Matrics A :n");
display(a,r1,c1);
printf("n Matrics B :n");
display(b,r2,c2);
printf("n Matrics C :n");
display(c,r1,c2);
}
void tran()
{
printf("n Matrics Transpose");
C’ software lab - 13 - MCA’05
16. OUTPUT :
Matrix Manipulator
1.Addition
2.Multiplication
3.Transpose
press any other key to exit...
Enter U'r choice :3
Enter order of the matrices :
Mat 1 :3 3
Mat 2 :2 2
Matrics Transpose
Enter Matrics :
1 2 3
2 3 4
3 4 5
1 2 3
2 3 4
3 4 5
Transposed Matrics :
1 2 3
2 3 4
3 4 5
C’ software lab - 16 - MCA’05
17. Experiment No. 5
Date:
AIM : TO FIND OUT REVERSE OF A GIVEN NUMBER
ALGORITHM :
Step 1: Start
Step 2: Declare variables n, dig, rev=0 as integer.
Step 3: Input the number, n.
Step 4: If n>0 then goto step 5 else goto step 8
Step 5: Modular division is done on the number with 10 .The value obtained
is assigned to variable
Step 6: The variable, rev is multiplied by 10 and then added with the variable dig,
thisvalue is assigned to rev.
Step 7: The variable, n is divided by 10 and the result is stored in n. Goto step 4
Step 8: The result is displayed on the screen
Step 9: Stop
C’ software lab - 17 - MCA’05
18. /*Reverse of a Number*/
#include<stdio.h>
#include<conio.h>
void main()
{
int num,rnum=0,mod;
clrscr();
printf("n Enter the Number :");
scanf("%d",&num);
while(num>0)
{
mod=num%10;
rnum=rnum*10+mod;
num=num/10;
}
printf("n reverse of number is %d",rnum);
getch();
}
C’ software lab - 18 - MCA’05
19. OUTPUT :
Enter the Number :1234
reverse of number is 4321
C’ software lab - 19 - MCA’05
20. Experiment No. 6
Date:
AIM : PERFORM SORTING A ONE DIMENSIONAL
ALGORITHM :
Step 1 : Start
Step 2 : Read the limit, N
Step 3 : Read elements of A[N] using for loop
Step 4 : Using for loop check whether A[I]>A[J] if yes then proceed
Step 5 : tA[I]
Step 6 : A[I]A[J]
Step 7 : A[J]t
Step 8 : End If
Step 9 : End for
Step 10 : Print Sorted array A[N]
Step 11 : Stop
C’ software lab - 20 - MCA’05
21. /*Sorting a 1-D array*/
#include<stdio.h>
#include<conio.h>
void main()
{
int a[15],i,tmp,j,k;
clrscr();
printf("n Enter the Size :");
scanf("%i",&i);
printf("n Enter the elements in the array : n");
for(j=0;j<i;j++)
{
scanf("%d",&a[j]);
}
printf("n Entered Elements are : n");
for(j=0;j<i;j++)
{
printf("%dt",a[j]);
}
for(j=0;j<i;j++)
{
for(k=j;k<i;k++)
{
if(a[j]>a[k])
{
tmp = a[j];
a[j]= a[k];
a[k]=tmp;
}
}
}
printf("n Sorted Elements are : n");
C’ software lab - 21 - MCA’05
23. OUTPUT :
Enter the Size :8
Enter the elements in the array :
3 7 5 9 7 4 3
8
Entered Elements are :
3 7 5 9 7 4 3 8
Sorted Elements are :
3 3 4 5 7 7 8 9
C’ software lab - 23 - MCA’05
24. Experiment No. 7
Date:
AIM : TO PROGRAM FOR SORT THE ENTERED STRINGS IN ALPHABETIC
ORDER.
ALGORITHM :
Step 1: Start
Step 2: Accept a single dimensional array, also declare another temporary array
Step 3: Sort the elements of the vector in ascending order
Step 4: Extract numbers one by one from main vector and check whether
the number is equal to the previous number if no’s are different insert it into
the
temporary array.
step 5: Print the temporary Array to see the result
step 6: Stop
C’ software lab - 24 - MCA’05
25. /*Deleting Duplicates in a 1-D array*/
#include<stdio.h>
#include<conio.h>
void main()
{
int a[15],i,tmp,j,k;
clrscr();
printf("n Enter the Size :");
scanf("%i",&i);
printf("n Enter the elements in the array : n");
for(j=0;j<i;j++)
{
scanf("%d",&a[j]);
}
printf("n Entered Elements are : n");
for(j=0;j<i;j++)
{
printf("%dt",a[j]);
}
for(j=0;j<i;j++)
{
for(k=j;k<i;k++)
{
if(a[j]>a[k])
{
tmp = a[j];
a[j]= a[k];
a[k]=tmp;
}
}
}
for(j=0;j<i;j++)
C’ software lab - 25 - MCA’05
27. OUTPUT :
Enter the Size :8
Enter the elements in the array :
6
5
8
4
2
6
8
5
Entered Elements are :
6 5 8 4 2 6 8 5
Sorted Elements are :
2 4 5 6 8
C’ software lab - 27 - MCA’05
28. Experiment No. 8
Date:
AIM : WRITE A PROGRAM TO PRINT FIBONACCI SERIES.
ALGORITHM :
Step 1: Start
Step 2: Declare variables n,a=0,b=1,c,i=1 as integer.
Step 3: Input value of n.
Step 4: Value of n is decremented by 1.
Step 5: The sum of a and b is assigned to c.
Step 6: The value c is displayed.
Step 7: The value of b is assigned to a and value of c is assigned to b.
Step 8: value of i is incremented by 1
Step 9: If i<n then goto step 6 Else goto step 10.
Step 10: Stop.
C’ software lab - 28 - MCA’05
30. OUTPUT :
Enter the range :8
Fibonacci Series...in 8 numbers is...
0 1 1 2 3 5 8 13
C’ software lab - 30 - MCA’05
31. Experiment No. 9
Date :
AIM : TO PERFORM THE FUNCTIONS OF A CALCULATOR USING SWITCH
CASE AND FUNCTIONS.
ALGORITHM :
Step 1 : Start
Step 2 : Read choice 1.Add 2.Sub 3.Mul 4.Div.
Step 3 : If choice =1 then add the numbers
Step 4 : If choice =2 then subtract the numbers.
Step 5 : If choice = 3 then multiply the numbers.
Step 6 : If choice = 4 then divide the numbers
Step 7 : Display respective output.
Step 8 : Stop
C’ software lab - 31 - MCA’05
32. /*Calculator*/
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int a,b;
int ch;
do
{
clrscr();
printf("nnnnnnn");
printf("ttttCALCULATORnnnn");
printf("ttt1.Additionn");
printf("ttt2.Substractionn");
printf("ttt3.Multiplicationn");
printf("ttt4.Divisionn");
printf("ttt5.Exit");
printf("nnntttEnter U'r choice :");
scanf("%d",&ch);
if(ch<5)
{ printf("ntttEnter two values:");
scanf("%d %d",&a,&b);
}
else exit(0);
switch(ch)
{
case 1:
printf("tttThe sum is :%d",a+b); getch();break;
case 2:
printf("tttThe diference is :%d",a-b);getch();break;
case 3:
C’ software lab - 32 - MCA’05
33. printf("tttThe multiplication is :%d",a*b);getch();break;
case 4:
if(b==0) printf("tttDivide by zero error");
else
printf("tttTHe division is :%d",a/b);getch();
break;
case 5: exit(0);
}
}while(ch<6);
getch();
}
C’ software lab - 33 - MCA’05
35. Experiment No.10
Date:
AIM : TO WRITE A PROGRAM TO CREATE AND MANIPULATE STUDENTS
RECORD USING STRUCTURES AND POINTERS.
ALGORITHM :
Step 1 : Start
Step 2 : Declare structure Stud
Roll_no as Integer
Name as Character array
Grade as Character
Mark, as an integer array
Step 3 : Create student object as an array as st[10], and also a pointer object *sd
Step 4 : Declare tot_mark, n
Step 5 : Accept number of entries as n
Step 6 : Accept student details
Step 7 : Calculate the average of students as sum of marks / 3
Step 8 : Calculate the grades and store to structure as on the conditions
if average > = 80 : grade = A
if average > = 60 and average < 80 : grade = B
if average > = 50 and average < 60 : grade = c
else if average < 50 : grade = Failed
Step 9 : Print details of students as per fields in the structure
Step 10 : Stop
C’ software lab - 35 - MCA’05
36. /*Student’s Record Using Poinetrs*/
#include<stdio.h>
#include<conio.h>
struct stud
{
int regno;
char name[20];
int mark[3];
char grade;
}s[10],*st;
void main()
{
int i,avg=0,j,num;
clrscr();
printf("n Enter the number of students...");
scanf("%d",&num);
st = s;
for(i=0;i<num;i++)
{
printf("n Enter student's Details :");
printf("n Roll no :");
scanf("%d",&st->regno);
printf(" Name :");
scanf("%s",st->name);
for(j=0;j<3;j++)
{
printf(" mark %d :",j+1);
scanf("%d",&st->mark[j]);
avg+=st->mark[j];
}
printf("Average of %s is %d",st->name,avg=avg/3);
C’ software lab - 36 - MCA’05
38. OUTPUT :
Enter the number of students...2
Enter student's Details :
Roll no :1
Name :Sachin
mark 1 :75
mark 2 :85
mark 3 :80
Average of Sachin is 80
Enter student's Details :
Roll no :2
Name :Rohit
mark 1 :80
mark 2 :65
mark 3 :82
Average of Rohit is 75
Student details...
Reg_no Name Marks 1 2 3 grade
1 Sachin 75 85 80 A
2 Rohit 80 65 82 B
C’ software lab - 38 - MCA’05
39. Experiment No. 11
Date:
AIM : TO PERFORM, SORT A CHARACTER ARRAY IN ACSENDING ORDER
ALGORITHM :
Step 1: Start
Step 2: Accept a character array
Step 3: use a for loop to extract characters from the array and check it with each
characters of the array if value is greater interchange the value in the positon
Step 4: Display the result
Step 5: Stop
C’ software lab - 39 - MCA’05
41. OUTPUT :
Enter the Char. array...chinagate
Entered Char.array is...chinagate
Length of entered string is...9
Sorted Char.array is...tnihgecaa
C’ software lab - 41 - MCA’05
42. Experiment No. 12
Date:
AIM : TO PROGRAM TO CHECK THE OCCURRENCE OF A PARTICULAR
NUMBER IN AN ARRAY.
ALGORITHM :
Step 1: Start
Step 2: Declare array a[25], variables n,i,num,count=0 as integer.
Step 3: Store the limit of the array in the variable n.
Step 4: Store the values of the array.
Step 5: Enter the value to be checked and store it in the variable, num.
Step 6: Start a for loop with loop variable, i.Compare each element of the array with
variable num. For each a[i]=num increment the counter count by 1.
Step 7: Display count on the output screen as the number of occurrence.
Step 8: Stop.
C’ software lab - 42 - MCA’05
43. /*occurance of particular number in an array*/
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,a[25],n,num,count=0,temp,flag=0,ch;
clrscr();
do
{
clrscr();
printf("nnEnter number of elements:");
scanf("%d",&n);
printf("nEnter %d array elements:n",n);
for(j=0;j<n;j++)
scanf("%d",&a[j]);
printf("nEnter the number to be searched:");
scanf("%d",&num);
printf("nThe positions of the given number are: ");
for(i=0;i<n;i++)
{
if(a[i]==num)
{
printf("%5d",i+1);
count++; flag=1;
}
}
if(flag==1)
{
printf("nn%d is repeated %d times in the array",num,count);
count = 0;
flag = 0;
}
C’ software lab - 43 - MCA’05
44. else
printf("0 nnNo element found");
printf("nnnnnttDo you want to continue?(Press 1 to continue):");
scanf("%d",&ch);
}while(ch==1);
getch();
}
C’ software lab - 44 - MCA’05
45. OUTPUT :
Enter number of elements:5
Enter 5 array elements:
3
5
4
3
6
Enter the number to be searched:3
The positions of the given number are: 1 4
3 is repeated 2 times in the array
Do you want to continue?(Press 1 to continue):1
Enter number of elements:3
Enter 3 array elements:
6
5
23
Enter the number to be searched:8
The positions of the given number are: 0
No element found
Do you want to continue?(Press 1 to continue):1
C’ software lab - 45 - MCA’05
46. Experiment No. 13
Date:
AIM : TO CHECK WHETHER A GIVEN STRING IS PALINDROME OR NOT
ALGORITHM :
Step 1: Start
Step 2: Initialize a string array a[100],flag=0
Step 3: Calculate length of the string using strlen() function
Step 4: Using for loop from I=0,J=len-1 to j<len/2 perform step 5
Step 5: Check whether A[I] != A[J] if false Set flag=1
Step 6: End for
Step 7: If flag=0 print Palindrome else Not Palindrome
Step 8: Stop
C’ software lab - 46 - MCA’05
48. OUTPUT :
1. Enter a string:INDIAN
String is Not a palindrome
2. Enter a string:Malayalam
String is a Palindrome
C’ software lab - 48 - MCA’05
49. Experiment No.14
Date:
AIM : TO COUNT THE VOWELS, CONSONANTS AND DIGITS.
ALGORITHM :
Step 1: Start
Step 2: Read str, vowc,conc,I,dig
Step 3: vowc o, conc0,dig0
Step 4: Input the line
Step 5: Print the given line
Step 6: lenlength of the string.
Step 7: Using for loop perform the step 7 to step 10 until len.
Check whether the a[I] is any of a,e,i,o,u if yes
vowcvowc+1 else go to step 8
Step 8: check whether the a[I]is between A and Z.
if yes concconc+1
Step 9: check whether the a[I]is between 0 and 9
If yes digdig+1
Step 10: print the vowc, conc, dig.
Step 11: Stop
C’ software lab - 49 - MCA’05
52. OUTPUT :
Enter a text: Believe in LOVE
Entered String Consists of :
Vowels =7
consonants =6
digits =0
words =3
white spaces =2
others =0
it takes 15 bytes for Storage
C’ software lab - 52 - MCA’05
53. Experiment No. 15
Date:
AIM : TO SWAP THE VALUE OF TWO VARIABLES USING POINTERS.
ALGORITHM :
Step1: Start
Step2: Accept two numbers
Step 3: Declare two pointer variables
Step4: Intialise the pointer variable with address of variable
Step5: Use a temporary variable to interchange the address of variable
Step6: Print the Result
Step7: Stop
C’ software lab - 53 - MCA’05
54. /*Swapping Values Using Pointers*/
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b;
void swap(int *,int *);
clrscr();
printf("n Enter the values of A & B :");
scanf("%d%d",&a,&b);
printf("n the values entered are A : %d & B : %d",a,b);
swap(&a,&b);
printf("n the swapped values are A : %d & B : %d",a,b);
getch();
}
void swap(int *x,int *y)
{
int *temp;
*temp=*x;
*x=*y;
*y=*temp;
}
C’ software lab - 54 - MCA’05
55. OUTPUT :
Enter the values of A & B :1 18
the values entered are A : 1 & B : 18
the swapped values are A : 18 & B : 1
C’ software lab - 55 - MCA’05
56. Experiment No. 16
Date:
AIM : TO FIND FACTORIAL OF A NUMBER USING RECURSION.
ALGORITHM :
Step1 : Start
Step2 : Declare variables and the function Fact();
Step3 : Accept the values, as num
Step4 : Print outputs as fact(num)
Step5 : In fact check for num to be 1
If true return(1)
If false return(num*fact(num))
Step6 : Stop
C’ software lab - 56 - MCA’05
57. /*Factorial Using Recursion*/
#include<stdio.h>
#include<conio.h>
void main()
{
int fact(int);
int num;
clrscr();
printf("n Enter the number :");
scanf("%d",&num);
printf("n Factorial of %d is %d",num,fact(num));
getch();
}
int fact(int x)
{
if(x==1)
return(1);
else
return(x*fact(x-1));
}
C’ software lab - 57 - MCA’05
58. OUTPUT :
Enter the number :6
Factorial of 6 is 720
C’ software lab - 58 - MCA’05
59. Experiment No.17
Date:
AIM : PROGRAM TO CONVERT NUMBERS INTO DIGITS
ALGORITHM :
Step 1 : Start
Step 2 : Declare separate 2D arrays for representing
characters from : one, two, three, …….,nine.
eleven, twelve, …………, nineteen
ten, twenty, ……………., hundred
Step 3 : Declare variables required
Step 4 : Accept the number less than 10000
Step 5 : Extract the left most digits by modular division of the number by 1000 to a
variable num
a. if num less than 10 give corresponding output
b. if num between eleven and nineteen both inclusive give corresponding
output
c. if number is a multiple of 10 give the result.
d. else split num as one’s and tens’s digit
repeat the step 5.a and 5.c
Step 7 : Add to the output a thousand
Step 8 : Extract the digits in hundreds the number by 1000 to a variable nnum
Step 9 : Extract the left most digits by modular division of the number by 100 to a
variable nunum
a. the number nunum is less than 10, hence give corresponding output
Step 10 : Add to the output a ‘hundred’.
Step 11 : Extract the right most digits by division of the number by 100 to a variable
nunum
a. the number nunum is less than 10, hence give corresponding output
C’ software lab - 59 - MCA’05
60. b. if num between eleven and nineteen both inclusive give corresponding
output
c. if number is a multiple of 10 give the result.
d. else split num as one’s and tens’s digit
repeat the step 11.a and 11.c
Step 12 : Display the output
Step 13 : Stop
C’ software lab - 60 - MCA’05
63. OUTPUT :
Enter no.15018
fifiteen thousand eighteen
Enter no.44444
fortyfour thousand four hundred fortyfour
C’ software lab - 63 - MCA’05
64. Experiment No. 18
Date:
AIM : CONCATENATING TWO STRINGS USING A CHARACTER POINTER.
ALGORITHM :
Step 1: Start
Step 2: Declare 3 character array and 2 character pointer
Step 3: Accept 2 Strings
Step 4: Save the address of the strings to the pointers
Step 5: Save character by character of the first array to new array till the length by
incrementing address
Step 6: Save character by character of the second array by incrementing address of
2nd variable to the resultant array
Step 7: Add a null character at the end of resultant array
Step 8: Print the new array
Step 9: Stop
C’ software lab - 64 - MCA’05
65. /*String Concatenation using Pointers*/
#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define LEN 10
void main()
{
char *s1,*s2,*s3;
int i=0,j=0;
char c;
clrscr();
s1=(char *)malloc(LEN * sizeof(char));
s2=(char *)malloc(LEN * sizeof(char));
s3=(char *)malloc(LEN * sizeof(char));
printf("n Enter the First String :");
scanf("%s",s1);
printf("n Enter the Second String :");
scanf("%s",s2);
while((c=s1[i])!='0')
{
s3[i]=c;
i++;
}
while((c=s2[j])!='0')
{
s3[i+j]=c;
j++;
} s3[i+j]='0';
printf("n Entered Strings '%s' and '%s' are concatenated to ...%s",s1,s2,s3);
getch();
}
C’ software lab - 65 - MCA’05
66. OUTPUT :
Enter the First String :Lovingly
Enter the Second String :Your's
Entered Strings 'Lovingly' and 'Your's' are concatenated to ...LovinglyYour's
C’ software lab - 66 - MCA’05
67. Experiment No. 19
Date:
AIM : TO PROGRAM TO CREATE AND COUNT NUMBER OF CHARACTERS
IN A FILE.
ALGORITHM :
Step 1: start
Step 2: Declare fp as file pointer
Step 3: Declare count=0 as integer and c as character variable.
Step 4: Open the file charcount.txt in write mode (w) as fp.
Step 5: Read characters to be written into the file from the user
Step 6: Write the entered characters into the file using putc( ).
Step 7: close the file charcount.txt and reopen in read mode (r).
Step 8: Read the characters one by one from the file using getc( ) upto EOF and
Increment the count by 1.
Step 9: print the count as number of characters in the file.
Step 10: stop.
C’ software lab - 67 - MCA’05
68. /*create and count number of charachers in the file*/
#include<stdio.h>
void main()
{
FILE *fp;
char c;
int count =0;
clrscr();
printf("Enter characters:");
fp=fopen("chcount","w");
while((c=getchar())!=EOF)
putc(c,fp);
fclose(fp);
fp=fopen("chcount","r");
printf("String in the file is:");
while((c=getc(fp))!=EOF)
{
printf("%c",c);
++count;
}
fclose(fp);
printf("nNumber of characters in the file is %dn",count);
getch();
}
C’ software lab - 68 - MCA’05
69. OUTPUT :
Enter characters:Arun is a
goob boy.
^Z
String in the file is:Arun is a
goob boy.
Number of characters in the file is 20
C’ software lab - 69 - MCA’05
70. Experiment No. 20
Date:
AIM : TO READ A FILE AND WRITE THE ODD VALUE TO A ODDFILE AND
EVEN VALUES TO A EVEN FILE.
ALGORITHM :
Step1: Start
Step2: Declare 3 File pointers (mainfile,oddfile,evenfile)
Step 3: Open the first file in w mode,oddfile and evenfile in write mode
Step4: Write the integer values in the file
Step5: Close the file
Step6: Reopen the file in read mode
Step7: Retrieve the value one by one from the file
Step8: Check whether there is any remainder when divided by two
Step9: If the remainder =0 Write the value to the even file
Step10: Otherwise write the value to the odd file
Step11: Stop
C’ software lab - 70 - MCA’05
