C lab-programs
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C lab-programs C lab-programs Document Transcript

  • kala Experiment No. 1 Date: AIM: TO FIND SIMPLE INTEREST AND COMPOUND INTEREST ALGORITHM: Step 1: Start Step 2: Declare variables p,r,t,si,ci,amount as float. Step 3: Accept values for p,r,t,n. Step 4: Calculate si=p*t*r/100 Step 5: Calculate amount=p* pow((1+r/n*100),n*t) Step 6: Calculate ci=amount-p Step 7: Print Step 8: Stop C’ software lab - 1 - MCA’05
  • /*Simple interest and compound interest*/ #include<stdio.h> #include<conio.h> #include<math.h> void main() { float p,r,t,si,ci,amount; int n; clrscr(); printf("nEnter principle :"); scanf("%f",&p); printf("nEnter rate :"); scanf("%f",&r); printf("nEnter time :"); scanf("%f",&t); printf("nEnter compoundings per year :"); scanf("%d",&n); si=(p*t*r)/100; printf("nSimple Interest =%7.2f",si); amount=p*pow((double)(1+r/(n*100)),(double)(n*t)); ci=amount-p; printf("nCompond Interest=%7.2fn",ci); getch(); } C’ software lab - 2 - MCA’05
  • 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
  • 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
  • /*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
  • 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
  • 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
  • /*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
  • 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
  • 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
  • /*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
  • 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
  • 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
  • printf("n Enter Matrics :n"); accept(a,r1,c1); display(a,r1,c1); printf("n Transposed Matrics :n"); for(i=0;i<r1;i++) { for(j=0;j<c1;j++) { printf("%dt",a[j][i]); } printf("n"); } } void display(int x[][10],int m,int n) { for(i=0;i<m;i++) { for(j=0;j<n;j++) { printf("%dt",x[i][j]); } printf("n"); } } void accept(int x[][10],int m,int n) { for(i=0;i<m;i++) { for(j=0;j<n;j++) { C’ software lab - 14 - MCA’05
  • scanf("%d",&x[i][j]); } } } C’ software lab - 15 - MCA’05
  • 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
  • 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
  • /*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
  • OUTPUT : Enter the Number :1234 reverse of number is 4321 C’ software lab - 19 - MCA’05
  • 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 : tA[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
  • /*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
  • for(j=0;j<i;j++) { printf("%dt",a[j]); } getch(); } C’ software lab - 22 - MCA’05
  • 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
  • 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
  • /*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
  • { if(a[j]==a[j+1]) { for(k=j;k<i;k++) { a[k]=a[k+1]; } i--;j--; } } printf("n Sorted Elements are : n"); for(j=0;j<i;j++) { printf("%dt",a[j]); } getch(); } C’ software lab - 26 - MCA’05
  • 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
  • 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
  • /*Fibonacci Series*/ #include<stdio.h> #include<conio.h> void main() { int f=1,s=0,t,r,i; clrscr(); printf("n Enter the range :"); scanf("%d",&t); printf("n Fibonacci Series...in %d numbers is...",t); printf("n%d ",s); for(i=0;i<t-1;i++) { r=f+s; f=s; s=r; printf(" %d ",r); } getch(); } C’ software lab - 29 - MCA’05
  • 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
  • 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
  • /*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
  • 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
  • OUTPUT : CALCULATOR 1.Addition 2.Substraction 3.Multiplication 4.Division 5.Exit Enter U'r choice :1 Enter two values:1 18 The sum is :19 CALCULATOR 1.Addition 2.Substraction 3.Multiplication 4.Division 5.Exit Enter U'r choice :4 Enter two values:18 1 THe division is :18 C’ software lab - 34 - MCA’05
  • 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
  • /*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
  • if(avg>=80) st->grade='A'; else if(avg>=60 && avg<80) st->grade='B'; else if(avg>=50 && avg<60) st->grade='C'; else st->grade='F'; st++; avg=0; } st=s; printf("n Student details..."); printf("nReg_notNamettMarkst1t2t3tgrade"); for(i=0;i<num;i++) { printf("n%dt%sttt%dt%dt%dt%c",st->regno,st->name,st->mark[0],st- >mark[1],st->mark[2],st->grade); st++; } getch(); } C’ software lab - 37 - MCA’05
  • 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
  • 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
  • /*Sorting a 1-D Charater array*/ #include<stdio.h> #include<conio.h> void main() { char str[20],tmp; int len,i,j; clrscr(); printf("n Enter the Char. array..."); scanf("%s",str); for(len=0;str[len]!='0';len++); printf("n Entered Char.array is...%s",str); printf("n Length of entered string is...%dn",len); for(i=0;i<len;i++) for(j=i;j<len;j++) if(str[i]<str[j]) { tmp=str[i]; str[i]=str[j]; str[j]=tmp; } printf("n Sorted Char.array is...%s",str); getch(); } C’ software lab - 40 - MCA’05
  • 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
  • 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
  • /*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
  • 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
  • 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
  • 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
  • /*palindrome*/ #include<stdio.h> #include<conio.h> #include<string.h> void main() { char str[10],str1[10]; int i,n,j; clrscr(); printf("Enter a string:"); scanf("%s",str); n=strlen(str); for(i=0,j=n-1;i<n/2;i++,j--) str[i]=tolower(str[i]); str[j]=tolower(str[j]); if(str[i]!=str[j]) { printf("n String is Not a palindrome") ; getch(); exit(0); } printf("n String is a Palindrome"); getch(); } C’ software lab - 47 - MCA’05
  • 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
  • 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, conc0,dig0 Step 4: Input the line Step 5: Print the given line Step 6: lenlength 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 vowcvowc+1 else go to step 8 Step 8: check whether the a[I]is between A and Z. if yes concconc+1 Step 9: check whether the a[I]is between 0 and 9 If yes digdig+1 Step 10: print the vowc, conc, dig. Step 11: Stop C’ software lab - 49 - MCA’05
  • /*count vowels*/ #include<stdio.h> #include<conio.h> #include<string.h> void main() { int i=0,j; char a[100],c; int chr=0,vow=0,dig=0,word=0,whites=0,others=0,cons=0; clrscr(); printf("Enter a text: "); scanf("%[^n]",a); while((c=tolower(a[i++]))!='0') { chr++; if(c== 'a' || c== 'e' || c== 'i'|| c=='o'|| c=='u') vow++; else if(c >='a' && c<='z') ++cons; else if(c>='0' && c<='9') ++dig; else if(c==' ') { ++word; ++whites; while((a[i]==' ') || (a[i]=='t')) { i++; whites++; } } } word++; printf("n Entered String Consists of :n"); printf("nVowels =%d",vow); C’ software lab - 50 - MCA’05
  • printf("nconsonants =%d",cons); printf("ndigits =%d",dig); printf("nwords =%d",word); printf("nwhite spaces =%d",whites); printf("nothers =%d",others); printf("nit takes %d bytes for Storage",chr); getch(); } C’ software lab - 51 - MCA’05
  • 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
  • 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
  • /*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
  • 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
  • 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
  • /*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
  • OUTPUT : Enter the number :6 Factorial of 6 is 720 C’ software lab - 58 - MCA’05
  • 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
  • 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
  • /*convert figure into word*/ #include<stdio.h> #include<string.h> #include<conio.h> void main() { char a[10][10]={"one","two","three","four", "five","six","seven","eight","nine","ten"}; char b[10][10]={"eleven","twelve","thirteen","fourteen", "fifiteen","sixteen","seventeen","eighteen", "nineteen"}; char c[10][10]={"ten","twenty","thirty","forty","fifty", "sixty","seventy","eighty","ninty","hundred"}; int r,s,t; long n; clrscr(); printf("Enter no."); scanf("%ld",&n); if(n>9999) { r=n/1000; if(r>10 && r<20) { r=r%10; printf("%s thousand ",b[r-1]); } else { s=r/10; t=r%10; printf("%s",c[s-1]); printf("%s thousand ",a[t-1]); C’ software lab - 61 - MCA’05
  • } n=n%1000; } if(n>1000) { r=n/1000; printf("%s thousand ",a[r-1]); n=n%1000; } if(n>100) { r=n/100; printf("%s hundred ",a[r-1]); n=n%100; } if(n>10 && n<20) { r=n%10; printf("%s ",b[r-1]); } if(n>19 && n<=100) { r=n/10; n=n%10; printf("%s",c[r-1]); } if(n>0 && n<=10) printf("%s ",a[n-1]); getch(); } C’ software lab - 62 - MCA’05
  • OUTPUT : Enter no.15018 fifiteen thousand eighteen Enter no.44444 fortyfour thousand four hundred fortyfour C’ software lab - 63 - MCA’05
  • 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
  • /*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
  • 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
  • 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
  • /*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
  • 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
  • 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
  • /*even and odd file*/ #include<stdio.h> #include<conio.h> void main() { FILE *fp1,*fp2,*fp3; int c,num,n; clrscr(); fp1=fopen("number","w"); printf("Input numbers:"); while((scanf("%d",&c))!=EOF) putw(c,fp1); fclose(fp1); fp1=fopen("number","r"); fp2=fopen("even","w"); fp3=fopen("odd","w"); printf("nnContents of file 'number':n "); while((c=getw(fp1))!=EOF) { printf("%dt",c); num=c%2; if(num==0) putw(c,fp2); else putw(c,fp3); } fclose(fp1); fclose(fp2); fclose(fp3); fp2=fopen("even","r"); fp3=fopen("odd","r"); printf("nnContents of file 'even':n "); C’ software lab - 71 - MCA’05
  • while((c=getw(fp2))!=EOF) printf("%dt",c); printf("nnContents of file 'odd':n "); while((c=getw(fp3))!=EOF) printf("%dt",c); getch(); } C’ software lab - 72 - MCA’05
  • OUTPUT : Input numbers:3 2 4 5 6 7 8 2 1 ^Z Contents of file 'number': 3 2 4 5 6 7 8 2 1 Contents of file 'even': 2 4 6 8 2 Contents of file 'odd': 3 5 7 1 C’ software lab - 73 - MCA’05
  • C’ software lab - 74 - MCA’05