.: CP-Lab Programs

1. 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: Read num as integer

Step 3: while (num > 0)

begin

dig ← num MOD 10;

sumOfDigits ← sumOf Digits + dig;

num ← num DIV 10;

end

Step 4: Print sumOfDigits

Step 5: Stop

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main ()

{

int number = 0, digit = 0, sumOfDigits = 0;

clrscr();

printf("Enter any number\n ");

scanf("%d", &number);

while (number != 0)

{

digit = number % 10;

sumOfDigits = sumOfDigits + digit;

number = number / 10;

}

printf ("Sum of individual digits of a given number is %d", sumOfDigits);

getch();

}

Input & Output:

Enter any number

1234

Sum of individual digits of a given number is 10

========================================================================

2 .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 fibinacci sequence

t1 = 0, t2 = 1

tn = tn - 2 + tn - 1 …… where n > 2

Algorithm:

Step 1: Start

Step 2: Read a, b, sum, lengthOfSeries values as integers

Step 3: Set a as 0 and b as 1

Step 4: for counter: 2 to no increment counter by 1

begin

sum ← a + b;

Print sum

a ← b;

b ← sum;

end

Step 5: Stop

Program:

#include<stdio.h>

#include<conio.h>

void main()

{

int a = 0, b = 1, lengthOfSeries = 0, counter, sum = 0;

clrscr();

printf("Enter the length of series \n ");

scanf("%d", &lengthOfSeries);

printf("Fibonacci series\n");

printf("%d %d", a, b);

for(counter = 2; counter < lengthOfSeries; counter++)

{

sum = a + b;

printf(" %d",sum);

a = b;

b = sum;

}

getch();

}

Input & Output:

Enter the length of series

Fibonacci series

0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

========================================================================

3. 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 goto step 5

If not goto step 10

Step 5: initialize j = 1

Step 6: if j <= 1 do as the follow. If no goto step 7

i)if i%j == 0 increment c

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

Program:

#include<stdio.h>

#include<conio.h>

void main()

{

int n, i, j, count;

clrscr();

printf("Prime no.series\n");

printf("Enter any number\n");

scanf("%d", &n);

printf("The prime numbers between 1 to %d\n",n);

for(i = 1; i <= n; i++)

{

count = 0;

for(j = 1; j <=i; j++)

if(i % j == 0)

{

count++;

}

if(count == 2)

{

printf("%d\t", i);

}

getch();

}

Input & Output:

Prime no. series

Enter any number

The prime numbers between 1 to 10

2 3 5 7

========================================================================

4. Write a C program to find the roots of a quadratic equation.

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 A, B, C as integer

Step 3: Declare disc, deno, x1, x2 as float

Step 4: Assign disc = (B * B) - (4 * A * C)

Step 5: Assign deno = 2 * A;

Step 6: if( disc > 0 )

begin

Print “THE ROOTS ARE REAL ROOTS”

Assign x1 ← (-B / deno) + (sqrt(disc) / deno)

Assign x2 ← (-B / deno) - (sqrt(disc) / deno)

Print x1, x2

end

else if(disc = 0)

begin

Print “ THE ROOTS ARE REPEATED ROOTS"

Assign x1 ← -B / deno

Print x1

end

else Print “THE ROOTS ARE IMAGINARY ROOTS”

Step7: Stop

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float a, b, c, d, root1, root2;

clrscr();

printf("Enter the values of a, b, c\n");

scanf("%f%f%f", &a, &b, &c);

if(a == 0 || b == 0 || c == 0)

{

printf("Error: Roots can't be determined");

}

else

{

d = (b * b) - (4.0 * a * c);

if(d > 0.00)

{

printf("Roots are real and distinct \n");

root1 = -b + sqrt(d) / (2.0 * a);

root2 = -b - sqrt(d) / (2.0 * a);

printf("Root1 = %f \nRoot2 = %f", root1, root2);

}

else if (d < 0.00)

{

printf("Roots are imaginary");

root1 = -b / (2.0 * a) ;

root2 = sqrt(abs(d)) / (2.0 * a);

printf("Root1 = %f +i %f\n", root1, root2);

printf("Root2 = %f -i %f\n", root1, root2);

}

else if (d == 0.00)

{

printf("Roots are real and equal\n");

root1 = -b / (2.0 * a);

root2 = root1;

printf("Root1 = %f\n", root1);

printf("Root2 = %f\n", root2);

}

getch();

}

Input & Output:

Enter the values of a, b, c

1 2 3

Roots are imaginary

Root1 = -1.000 + i

Root2 = -1.000 - i

======================================================================

5.Write a C program to calculate the following Sum: Sum=1-x2/2! +x4/4!-x6/6!+x8/8!-x10/10!

The above equation looks like a COSINE Equation of Taylor Sries i.e.,

Algorithm:

Step 1: Start

Step 2: Read x, n values as integers

Step 3: Set i = 2, s = 1, pwr = 1, nr = 1

Step 4: Convert x1 into degrees

Step 5: Assign x1 as sum

Step 6: while (i <= n)

begin

pwr ← pwr + 2;

dr ← dr * pwr * (pwr - 1);

sum ← sum + (nr DIV dr) * s;

s ← s * (-1);

nr ← nr * x1 * x1;

i ← i + 2;

end

Step 7: Print sum

Step 8: Stop

Program:

# include<stdio.h>

# include<conio.h>

# include<math.h>

void main()

{

int i, n ;

float x, val, sum = 1, t = 1 ;

clrscr() ;

printf("Enter the value for x : ") ;

scanf("%f", &x) ;

printf("\nEnter the value for n : ") ;

scanf("%d", &n) ;

val = x ;

x = x * 3.14159 / 180 ;

for(i = 1 ; i < n + 1 ; i++)

{

t = t * pow((double) (-1), (double) (2 * i - 1)) *

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

sum = sum + t ;

}

printf("\nCosine value of sin %f is : %8.4f", val, sum) ;

getch() ;

}

Input & Output:

Enter the Value of x: 2

Enter the limit of n: 4

The sum of sin 2.000000 series is 0.9994

========================================================================

6.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

Step 2: Read interval as integer

Step 3: for counter: 1 to interval increment counter by 1

begin

Read time, velocity, acceleration

Distance += (velocity * time + (accelerations * pow(time, 2)) / 2);

end

Step 4: Print Distance

Step 5: Stop

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main() e

{

int i, n, sec;

float d, u, a;

clrscr();

printf("Enter the no. of intervals\n");

scanf("%d", &n);

for(i = 1; i <= n; i++)

{

printf("interval: %d \n", i);

printf("Enter the time in seconds \n");

scanf("%d",&sec);

printf("Enter the velocity \n");

scanf("%f", &u);

printf("Enter the acceleration \n");

scanf("%f", &a);

d= d + (u * sec + (a * (pow(sec, 2))) / 2);

}

printf("Total distance travelled is %.2f", d);

getch();

}

Input & Output:

Enter the number of intervals: 2

Interval: 1

Enter the time in seconds

Enter the velocity

Enter the acceleration

Interval: 2

Enter the time in seconds

Enter the velocity

Enter the acceleration

Total distance travelled is 35850.00

========================================================================

7.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 x and y values

Step 3: Read option + or – or * or / or %

Step 4: If option is ‘+’ res = x + y

Step 5: If option is ‘-’ res = x - y

Step 6: If option is ‘*’ res = x * y

Step 7: If option is ‘/’ res = x / y

Step 8: If option is ‘%’ res = x % y

Step 9: If option does not match with + or – or * or / or %

Print select option +, -, *, /, /, % only

Step 10: Print x, option, y, res values

Step 11: Stop

Program:

#include <stdio.h>

#include <conio.h>

void main()

{

int a, b, c;

char ch;

clrscr() ;

printf("Enter your operator(+, -, /, *, %)\n");

scanf("%c", &ch);

printf("Enter the values of a and b\n");

scanf("%d%d", &a, &b);

switch(ch)

{

case '+': c = a + b;

printf("addition of two numbers is %d", c);

break;

case '-': c = a - b;

printf("substraction of two numbers is %d", c);

break;

case '*': c = a * b;

printf("multiplication of two numbers is %d", c);

break;

case '/': c = a / b;

printf("remainder of two numbers is %d", c);

break;

case '%': c = a % b;

printf("quotient of two numbers is %d", c);

break;

default: printf("Invalid operator");

break;

}

getch();

}

Input & Output:

Enter you operator(+, -, /, *, %)

Enter the values of a and b

1 3

addition of two numbers is 4

8. 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

i) To find the factorial of a given integer.

Algorithm:

Step 1: Start

Step 2: Read n value as integer

Step 3: Call function factorial (int n)

Step 4: End

Call function factorial(int n)

begin

if (n = 0)

return 1;

else

return (n * factorial(n - 1));

end

Program:

#include <stdio.h>

#include <conio.h>

void main()

{

int n, a, b;

clrscr();

printf("Enter any number\n");

scanf("%d", &n);

a = recfactorial(n);

printf("The factorial of a given number using recursion is %d \n", a);

b = nonrecfactorial(n);

printf("The factorial of a given number using nonrecursion is %d ", b);

getch();

}

int recfactorial(int x)

{

int f;

if(x == 0)

{

return(1);

}

else

{

f = x * recfactorial(x - 1);

return(f);

}

int nonrecfactorial(int x)

{

int i, f = 1;

for(i = 1;i <= x; i++)

{

f = f * i;

}

return(f);

}

Input & Output:

Enter any number

The factorial of a given number using recursion is 120

The factorial of a given number using nonrecursion is 120

========================================================================

9.ii) To find the GCD (greatest common divisor) of two given integers

Algorithm:

Step 1: Start

Step 2: Read a, b values as integers

Step 3: Call function gcd (a, b) and assign it to res

Step 4: Print res

Step 5: Stop

Called function gcd (a, b)

begin

while(v != 0)

begin

temp ← u MOD v;

u ← v;

v ← temp;

end

return(u);

end

Program:

#include <stdio.h>

#include <conio.h>

void main()

{

int a, b, c, d;

clrscr();

printf("Enter two numbers a, b\n");

scanf("%d%d", &a, &b);

c = recgcd(a, b);

printf("The gcd of two numbers using recursion is %d\n", c);

d = nonrecgcd(a, b);

printf("The gcd of two numbers using nonrecursion is %d", d);

getch();

}

int recgcd(int x, int y)

{

if(y == 0)

{

return(x);

}

else

{

return(recgcd(y, x % y));

}

int nonrecgcd(int x, int y)

{

int z;

while(x % y != 0)

{

z = x % y;

x = y;

y = z;

}

return(y);

}

Input & Output:

Enter two numbers a, b

3 6

The gcd of two numbers using recursion is 3

The gcd of two numbers using no recursion is 3
10 .Binary search program

#include <stdio.h>

int main()

{

int c, first, last, middle, n, search, array[100];

printf("Enter number of elements\n");

scanf("%d",&n);

printf("Enter %d integers\n", n);

for (c = 0; c < n; c++)

scanf("%d",&array[c]);

printf("Enter value to find\n");

scanf("%d", &search);

first = 0;

last = n - 1;

middle = (first+last)/2;

while (first <= last) {

if (array[middle] < search)

first = middle + 1;

else if (array[middle] == search) {

printf("%d found at location %d.\n", search, middle+1);

break;

}

else

last = middle - 1;

middle = (first + last)/2;

}

if (first > last)

printf("Not found! %d is not present in the list.\n", search);

return 0;

}

Sample Input:

Sample out put:

11. Bubble sort program

#include<stdio.h>

#include<conio.h>

int main( )

{

int a[100];

int i, j, temp, n ;

printf("how many numbers you want to sort : \n");

scanf("%d",&n);

printf("Enter %d number values you want to sort\n", n);

for(j=0; j<n; j++)

scanf("%d",&a[j]);

for(j=1;j<n;j++)

{

for(i=0; i<n; i++)

{

if(a[i]>a[i+1])

{

temp=a[i];

a[i]=a[i+1];

a[i+1]=temp;

}

printf ( "\n\nArray after sorting:\n") ;

for ( i = 0 ; i <n ; i++ )

printf ( "%d\t", a[i] ) ;

getch();

}

12.Addition of two matrixes

#include<stdio.h>

int main(){

int a[3][3],b[3][3],c[3][3],i,j;

printf("Enter the First matrix->");

for(i=0;i<3;i++)

for(j=0;j<3;j++)

scanf("%d",&a[i][j]);

printf("\nEnter the Second matrix->");

for(i=0;i<3;i++)

for(j=0;j<3;j++)

scanf("%d",&b[i][j]);

printf("\nThe First matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",a[i][j]);

}

printf("\nThe Second matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",b[i][j]);

}

for(i=0;i<3;i++)

for(j=0;j<3;j++)

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

printf("\nThe Addition of two matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",c[i][j]);

}

return 0;

}

Sample out put

13.Multiplication of two matrixes

#include<stdio.h>

int main() {

int a[10][10], b[10][10], c[10][10], i, j, k;

int sum = 0;

printf("\nEnter First Matrix : n");

for (i = 0; i < 3; i++) {

for (j = 0; j < 3; j++) {

scanf("%d", &a[i][j]);

}

printf("\nEnter Second Matrix:n");

for (i = 0; i < 3; i++) {

for (j = 0; j < 3; j++) {

scanf("%d", &b[i][j]);

}

printf("The First Matrix is: \n");

for (i = 0; i < 3; i++) {

for (j = 0; j < 3; j++) {

printf(" %d ", a[i][j]);

}

printf("\n");

}

printf("The Second Matrix is : \n");

for (i = 0; i < 3; i++) {

for (j = 0; j < 3; j++) {

printf(" %d ", b[i][j]);

}

printf("\n");

}

//Multiplication Logic

for (i = 0; i <= 2; i++) {

for (j = 0; j <= 2; j++) {

sum = 0;

for (k = 0; k <= 2; k++) {

sum = sum + a[i][k] * b[k][j];

}

c[i][j] = sum;

}

printf("\nMultiplication Of Two Matrices : \n");

for (i = 0; i < 3; i++) {

for (j = 0; j < 3; j++) {

printf(" %d ", c[i][j]);

}

printf("\n");

}

return (0);

}

14.Insert a substring in main string

#include <stdio.h>

#include <conio.h>

#include <string.h>

void main()

{

char a[10];

char b[10];

char c[10];

int p=0,r=0,i=0;

int t=0;

int x,g,s,n,o;

clrscr();

puts("Enter First String:");

gets(a);

puts("Enter Second String:");

gets(b);

printf("Enter the position where the item has to be inserted: ");

scanf("%d",&p);

r = strlen(a);

n = strlen(b);

i=0;

// Copying the input string into another array

while(i <= r)

{

c[i]=a[i];

i++;

}

s = n+r;

o = p+n;

// Adding the sub-string

for(i=p;i<s;i++)

{

x = c[i];

if(t<n)

{

a[i] = b[t];

t=t+1;

}

a[o]=x;

o=o+1;

}

printf("%s", a);

getch();

}

Sample input : Enter the first string: welcome

Enter ther second string:eee

Enter the position where the item has to be inserted:3

Sample input : “Welcomeeee”

15.To delete n characters from a given position in a given string

#include <stdio.h>

#include <conio.h>

#include <string.h>

void delchar(char *x,int a, int b);

void main()

{

char string[10];

int n,pos,p;

clrscr();

puts("Enter the string");

gets(string);

printf("Enter the position from where to delete");

scanf("%d",&pos);

printf("Enter the number of characters to be deleted");

scanf("%d",&n);

delchar(string, n,pos);

getch();

}

void delchar(char *x,int a, int b)

{

if ((a+b-1) <= strlen(x))

{

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

puts(x);

}

Sample input: enter the string:amar yengala

Enter the position from where to delete 6

Enter the number of characters to be deleted 3

Sample Output:amaryngala

16.Using Non recursive given number is palindrome or not

#include<stdio.h>

int main(){

int num,reverse_number;

//User would input the number

printf("\nEnter any number:");

scanf("%d",&num);

//Calling user defined function to perform reverse

reverse_number=reverse_function(num);

printf("\nAfter reverse the no is :%d",reverse_number);

return 0;

}

int sum=0,rem;

reverse_function(int num){

if(num){

rem=num%10;

sum=sum*10+rem;

reverse_function(num/10);

}

else

return sum;

}

  Sample input: Enter any number: 23456

Sample out put: After reverse the no is :65432

17 .Write a C program to replace a substring with another in a given line of text.

#include<stdio.h>

#include<conio.h>

void strreplace(char *,char,char);

int main()

{

char s[10],chr,repl_chr;

printf("\nEnter a string: ");

scanf("%s", &s);

printf("\nEnter character to be replaced: ");

scanf("%s", &chr);

printf("\nEnter replacement character: ");

scanf("%s", &repl_chr);

printf("\nModified string after replacement is: ");

strreplace(s,chr,repl_chr);

getch();

return 0;

}

void strreplace(char s[], char chr, char repl_chr)

{

int i=0;

while(s[i]!='\0')

{

if(s[i]==chr)

{

s[i]=repl_chr;

}

i++;

}

printf("%s",s);

}

Sample output:

Enter line of text

Snehal IS A GOOD GIRL

Enter the word to replace

GIRL

Enter the word to replace with

BOY

Text

IS A GOOD BOY

18.Write a C program that reads 15 names each of up to 30 characters, stores them in an array, and uses an array of pointers to display them in ascending (ie. alphabetical) order

#include <stdio.h>

#include <string.h>

void main()

{

char name[10][8], tname[10][8], temp[8];

int i, j, n;

printf("Enter the value of n \n");

scanf("%d", &n);

printf("Enter %d names n", \n);

for (i = 0; i < n; i++)

{

scanf("%s", name[i]);

strcpy(tname[i], name[i]);

}

for (i = 0; i < n - 1 ; i++)

{

for (j = i + 1; j < n; j++)

{

if (strcmp(name[i], name[j]) > 0)

{

strcpy(temp, name[i]);

strcpy(name[i], name[j]);

strcpy(name[j], temp);

}

printf("\n----------------------------------------\n");

printf("Input NamestSorted names\n");

printf("------------------------------------------\n");

for (i = 0; i < n; i++)

{

printf("%s\t\t%s\n", tname[i], name[i]);

}

printf("------------------------------------------\n");

}

Sample out put:

$ cc pgm32.c

$ a.out

Enter the value of n

Enter 7 names

heap

stack

queue

object

class

program

project

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

Input Names Sorted names

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

heap class

stack heap

queue object

object program

class project

program queue

project stack

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

19.Write a C program to convert a positive integer to a roman numeral. Ex. 11 is converted to XI.

#include <stdio.h>

void predigit(char num1, char num2);

void postdigit(char c, int n);

char romanval[1000];

int i = 0;

int main()

{

int j;

long number;

printf("Enter the number: ");

scanf("%d", &number);

if (number <= 0)

{

printf("Invalid number");

return 0;

}

while (number != 0)

{

if (number >= 1000)

{

postdigit('M', number / 1000);

number = number - (number / 1000) * 1000;

}

else if (number >= 500)

{

if (number < (500 + 4 * 100))

{

postdigit('D', number / 500);

number = number - (number / 500) * 500;

}

else

{

predigit('C','M');

number = number - (1000-100);

}

else if (number >= 100)

{

if (number < (100 + 3 * 100))

{

postdigit('C', number / 100);

number = number - (number / 100) * 100;

}

else

{

predigit('L', 'D');

number = number - (500 - 100);

}

else if (number >= 50 )

{

if (number < (50 + 4 * 10))

{

postdigit('L', number / 50);

number = number - (number / 50) * 50;

}

else

{

predigit('X','C');

number = number - (100-10);

}

else if (number >= 10)

{

if (number < (10 + 3 * 10))

{

postdigit('X', number / 10);

number = number - (number / 10) * 10;

}

else

{

predigit('X','L');

number = number - (50 - 10);

}

else if (number >= 5)

{

if (number < (5 + 4 * 1))

{

postdigit('V', number / 5);

number = number - (number / 5) * 5;

}

else

{

predigit('I', 'X');

number = number - (10 - 1);

}

else if (number >= 1)

{

if (number < 4)

{

postdigit('I', number / 1);

number = number - (number / 1) * 1;

}

else

{

predigit('I', 'V');

number = number - (5 - 1);

}

printf("Roman number is: ");

for(j = 0; j < i; j++)

printf("%c", romanval[j]);

return 0;

}

void predigit(char num1, char num2)

{

romanval[i++] = num1;

romanval[i++] = num2;

}

void postdigit(char c, int n)

{

int j;

for (j = 0; j < n; j++)

romanval[i++] = c;

}

Sample Output:

$ cc pgm8.c

$ a.out

Enter the number: 500

Roman number is be: D

20.Write a C program to display the contents of a file to standard output device.

#include <stdio.h>

#include <errno.h>

long count_characters(FILE *);

void main(int argc, char * argv[])

{

int i;

long cnt;

char ch, ch1;

FILE *fp1, *fp2;

if (fp1 = fopen(argv[1], "r"))

{

printf("The FILE has been opened...\n");

fp2 = fopen(argv[2], "w");

cnt = count_characters(fp1); // to count the total number of characters inside the source file

fseek(fp1, -1L, 2); // makes the pointer fp1 to point at the last character of the file

printf("Number of characters to be copied %d\n", ftell(fp1));

while (cnt)

{

ch = fgetc(fp1);

fputc(ch, fp2);

fseek(fp1, -2L, 1); // shifts the pointer to the previous character

cnt--;

}

printf("\n**File copied successfully in reverse order**\n");

}

else

{

perror("Error occured\n");

}

fclose(fp1);

fclose(fp2);

}

// count the total number of characters in the file that *f points to

long count_characters(FILE *f)

{

fseek(f, -1L, 2);

long last_pos = ftell(f); // returns the position of the last element of the file

last_pos++;

return last_pos;

}

Sample out put:

$ gcc file12.c

$ cat test2

The function STRERROR returns a pointer to an ERROR MSG STRING whose contents are implementation defined.

THE STRING is not MODIFIABLE and maybe overwritten by a SUBSEQUENT Call to the STRERROR function.

$ a.out test2 test_new

The FILE has been opened..

Number of characters to be copied 203

**File copied successfully in reverse order**

$ cat test_new

.noitcnuf RORRERTS eht ot llaC TNEUQESBUS a yb nettirwrevo ebyam dna ELBAIFIDOM ton si GNIRTS EHT

.denifed noitatnemelpmi era stnetnoc esohw GNIRTS GSM RORRE na ot retniop a snruter RORRERTS noitcnuf ehT

$ ./a.out test_new test_new_2

The FILE has been opened..

Number of characters to be copied 203

**File copied successfully in reverse order**

$ cat test_new_2

The function STRERROR returns a pointer to an ERROR MSG STRING whose contents are implementation defined.

THE STRING is not MODIFIABLE and maybe overwritten by a SUBSEQUENT Call to the STRERROR function.

$ cmp test test_new_2

21.Write a C program which copies one file to another, replacing all lowercase characters with their uppercase equivalents.

#include<stdio.h>

#include<conio.h>

void main()

{

FILE *fp1,*fp2;

char *fn1,*fn2,ch;

clrscr();

printf("\n enter the source file");

gets(fn1);

printf("\n enter the destination file");

gets(fn2);

fp1=fopen(fn1,"r");

fp2=fopen(fn2,"w");

if(fp1==NULL||fp2==NULL)

{

printf("\n unable to open file");

exit(0);

}

while(!feof(fp1))

{

ch=fgetc(fp1);

if(ch>='a'&& ch<='z')

ch=ch-32;

fputc(ch,fp2);

}

printf("\n file successfully copied");

fcloseall();

getch();

}

22.Write a C program to merge two files into a third file

#include <stdio.h>

#include <stdlib.h>

int main()

{

// Open two files to be merged

FILE *fp1 = fopen("file1.txt", "r");

FILE *fp2 = fopen("file2.txt", "r");

// Open file to store the result

FILE *fp3 = fopen("file3.txt", "w");

char c;

if (fp1 == NULL || fp2 == NULL || fp3 == NULL)

{

puts("Could not open files");

exit(0);

}

// Copy contents of first file to file3.txt

while ((c = fgetc(fp1)) != EOF)

fputc(c, fp3);

// Copy contents of second file to file3.txt

while ((c = fgetc(fp2)) != EOF)

fputc(c, fp3);

printf("Merged file1.txt and file2.txt into file3.txt");

fclose(fp1);

fclose(fp2);

fclose(fp3);

return 0;

}

Sample Output:

Merged file1.txt and file2.txt into file3.txt

.

JSF sample 1

CP-Lab Programs

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