C programming codes

• Hello world
• Print Integer
• Addition
• Odd or Even
• Add, subtract, multiply and divide
• Check vowel
• Leap year
• Add digits
• Factorial
• HCF and LCM
• Decimal to binary conversion
• ncR and nPr
• Add n numbers
• Swapping
• Reverse number
• Palindrome number
• Print Pattern
• Diamond
• Prime numbers
• Find armstrong number
• Generate armstrong number
• Fibonacci series
• Print floyd’s triangle
• Print pascal triangle
• Addition using pointers
• Maximum element in array
• Minimum element in array
• Linear search
• Binary search
• Reverse array
• Insert element in array
• Delete element from array
• Merge arrays
• Bubble sort
• Insertion sort
• Selection sort
• Add matrices
• Subtract matrices
• Transpose matrix
• Multiply two matrices
• Print string
• String length
• Compare strings
• Copy string
• Concatenate strings
• Reverse string
• Find palindrome
• Delete vowels
• C substring
• Sort a string
• Remove spaces
• Change case
• Swap strings
• Character’s frequency
• Anagrams
• Read file
• Copy files
• Merge two files
• List files in a directory
• Delete file
• Random numbers
• Add complex numbers
• Print date
• Get IP address
• Shutdown computer

C hello world example

#include 

#include 

 
int void main()
{
  printf("Hello world\n");
  getch();
}

Hello world program in C using character Array

#include 

#include 

int main()
{
  char string[] = "Hello World";

  printf("%s\n", string);

getch();
}

C program print integer

#include 

int main()
{
  int a;

  printf("Enter an integer\n");
  scanf("%d", &a);

  printf("Integer that you have entered is %d\n", a);

  return 0;
}

c program print integer

This  program takes an input as integer type and prints it. Input is done using scanf function and number is printed on screen using printf.

C programming code

#include 

# include 

 void main()

{

  int a;

  printf("Enter an integer\n");

  scanf("%d", &a);

  printf("Integer that you have entered is %d\n", a);

getch();

}

C program to add two numbers

C program to add two numbers: This c language program perform the basic arithmetic operation of addition on two numbers and then prints the sum on the screen. For example if the user entered two numbers as 5, 6 then 11 ( 5 + 6 ) will be printed on the screen.

C programming code

#include

#include

Void main()

{

   int a, b, c;

   printf("Enter two numbers to add\n");

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

   c = a + b;

   printf("Sum of entered numbers = %d\n",c);

getch();

}

Add two nos without using third variable

#include

#include

Void main()

{

   int a = 1, b = 2;

    /* Storing result of addition in variable a */

    a = a + b;

    /* Not recommended because original value of a is lost

    * and you may be using it some where in code considering it

    * as it was entered by the user.

    */

    printf("Sum of a and b = %d\n", a);

 getch();

}

C program to add two numbers repeatedly

#include

#include

main()

{

   int a, b, c;

   char ch;

   while(1)

   {

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

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

      c = a + b;

      printf("a + b = %d\n", c);

      printf("Do you wish to add more numbers(y/n)\n");

      scanf(" %c",&ch);

      if ( ch == 'y' || ch == 'Y' )

         continue;

      else

          break;

   }

getch();

}

Adding numbers in c using function

We have used long data type as it can handle large numbers.

#include

#include

long addition(long, long);

main()

{

   long first, second, sum;

   scanf("%ld%ld", &first, &second);

   sum = addition(first, second);

   printf("%ld\n", sum);

getch();

}

long addition(long a, long b)

{

   long result;

   result = a + b;

   return result;

}

C program to check odd or even

c program to check odd or even: We will determine whether a number is odd or even by using different methods all are provided with a code in c language. As you have study in mathematics that in decimal number system even numbers are divisible by 2 while odd are not so we may use modulus operator(%) which returns remainder, For example 4%3 gives 1 ( remainder when four is divided by three). Even numbers are of the form 2*p and odd are of the form (2*p+1) where p is is an integer.

We can use bitwise AND (&) operator to check odd or even, as an example consider binary of 7 (0111) when we perform 7 & 1 the result will be one and you may observe that the least significant bit of every odd number is 1, so ( odd_number & 1 ) will be one always and also ( even_number & 1 ) is zero.

In c programming language when we divide two integers we get an integer result, For example the result of 7/3 will be 2.So we can take advantage of this and may use it to find whether the number is odd or even. Consider an integer n we can first divide by 2 and then multiply it by 2 if the result is the original number then the number is even otherwise the number is odd. For example 11/2 = 5, 5*2 = 10 ( which is not equal to eleven), now consider 12/2 = 6 and 6 *2 = 12 ( same as original number). These are some logic which may help you in finding if a number is odd or not.

C program to check odd or even using modulus operator

#include

#include

Void main()

{

   int n;

   printf("Enter an integer\n");

   scanf("%d",&n);

   if ( n%2 == 0 )

      printf("Even\n");

   else

      printf("Odd\n");

 getch();

   }

C program to check odd or even using bitwise operator

#include

#include

main()

{

   int n;

   printf("Enter an integer\n");

   scanf("%d",&n);

   if ( n & 1 == 1 )

      printf("Odd\n");

   else

      printf("Even\n");

 getch();

}

C program to check odd or even without using bitwise or modulus operator

#include

#include

main()

{

   int n;

   printf("Enter an integer\n");

   scanf("%d",&n);

   if ( (n/2)*2 == n )

      printf("Even\n");

   else

      printf("Odd\n");

 getch();

}

Find odd or even using conditional operator

#include

#include

main()

{

   int n;

   printf("Enter an integer\n");

   scanf("%d",&n);

   n%2 == 0 ? printf("Even number\n") : printf("Odd number\n");

    getch();

}

C program to perform addition, subtraction, multiplication and division

C program to perform basic arithmetic operations i.e. addition , subtraction, multiplication and division of two numbers. Numbers are assumed to be integers and will be entered by the user.

C programming code

#include 

#include

int main()

{

   int first, second, add, subtract, multiply;

   float divide;

   printf("Enter two integers\n");

   scanf("%d%d", &first, &second);

   add      = first + second;

   subtract = first - second;

   multiply = first * second;

   divide   = first / (float)second;   //typecasting

   printf("Sum = %d\n",add);

   printf("Difference = %d\n",subtract);

   printf("Multiplication = %d\n",multiply);

   printf("Division = %.2f\n",divide);

   getch();

}

In c language when we divide two integers we get integer result for example 5/2 evaluates to 2. As a general rule integer/integer = integer and float/integer = float or integer/float = float. So we convert denominator to float in our program, you may also write float in numerator. This explicit conversion is known as typecasting.

c program to check whether input alphabet is a vowel or not

This code checks whether an input alphabet is a vowel or not. Both lower-case and upper-case are checked.

C programming code

#include 

#include

 main()

{

  char ch;

  printf("Enter a character\n");

  scanf("%c", &ch);

  if (ch == 'a' || ch == 'A' || ch == 'e' || ch == 'E' || ch == 'i' || ch == 'I' || ch =='o' || ch=='O' || ch == 'u' || ch == 'U')

    printf("%c is a vowel.\n", ch);

  else

    printf("%c is not a vowel.\n", ch);

  getch();

}

Check vowel using switch statement

#include 

Void main()

{

  char ch;

  printf("Enter a character\n");

  scanf("%c", &ch);

  switch(ch)

  {

    case 'a':

    case 'A':

    case 'e':

    case 'E':

    case 'i':

    case 'I':

    case 'o':

    case 'O':

    case 'u':

    case 'U':

      printf("%c is a vowel.\n", ch);

      break;

    default:

      printf("%c is not a vowel.\n", ch);

  }

 getch();

  }

Function to check vowel

int check_vowel(char a)

{

    if (a >= 'A' && a 

       a = a + 'a' - 'A';   /* Converting to lower case or use a = a + 32 */

    if (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')

       return 1;

    return 0;

}

c program to check leap year

c program to check leap year: c code to check leap year, year will be entered by the user. Please read theleap year article before reading the code, it will help you to understand the program.

C programming code

#include 

#include 

int main()

{

  int year;

  printf("Enter a year to check if it is a leap year\n");

  scanf("%d", &year);

  if ( year%400 == 0)

    printf("%d is a leap year.\n", year);

  else if ( year%100 == 0)

    printf("%d is not a leap year.\n", year);

  else if ( year%4 == 0 )

    printf("%d is a leap year.\n", year);

  else

    printf("%d is not a leap year.\n", year);

  getch();

}

add digits of number in c

C program to add digits of a number: Here we are using modulus operator(%) to extract individual digits of number and adding them.

C programming code

#include 

main()

{

   int n, sum = 0, remainder;

   printf("Enter an integer\n");

   scanf("%d",&n);

   while(n != 0)

   {

      remainder = n % 10;

      sum = sum + remainder;

      n = n / 10;

   }

   printf("Sum of digits of entered number = %d\n",sum);

   return 0;

}

For example if the input is 98, sum(variable) is 0 initially
98%10 = 8 (% is modulus operator which gives us remainder when 98 is divided by 10).
sum = sum + remainder
so sum = 8 now.
98/10 = 9 because in c whenever we divide integer by another integer we get an integer.
9%10 = 9
sum = 8(previous value) + 9
sum = 17
9/10 = 0.
So finally n = 0, loop ends we get the required sum.

Output of program:

Add digits using recursion

#include 

#include

int add_digits(int);

int main() {

  int n, result;

   scanf("%d", &n);

   result = add_digits(n);

   printf("%d\n", result);

 getch();

}

int add_digits(int n) {

  static int sum = 0;

  if (n == 0) {

    return 0;

  }

   sum = n%10 + add_digits(n/10);

   return sum;

}

Factorial program in c

Factorial program in c: c code to find and print factorial of a number, three methods are given, first one uses a for loop, second uses a function to find factorial and third using recursion. Factorial is represented using !, so five factorial will be written as 5!, n factorial as n!. Also
n! = n*(n-1)*(n-2)*(n-3)…3.2.1 and zero factorial is defined as one i.e. 0!=1.

Factorial program in c using for loop

: Here we find factorial using for loop.

#include 

#include

int main()

{

  int c, n, fact = 1;

   printf("Enter a number to calculate it's factorial\n");

  scanf("%d", &n);

   for (c = 1; c 

    fact = fact * c;

   printf("Factorial of %d = %d\n", n, fact);

   getch();

}

Factorial
Output of code:

Factorial program in c using function

#include 

#include 

long factorial(int);

int main()

{

  int number;

  long fact = 1;

   printf("Enter a number to calculate it's factorial\n");

  scanf("%d", &number);

   printf("%d! = %ld\n", number, factorial(number));

   return 0;

}

long factorial(int n)

{

  int c;

  long result = 1;

   for (c = 1; c 

    result = result * c;

   return result;

}

Factorial program in c using recursion

#include

#include

long factorial(int);

 int main()

{

  int num;

  long f;

   printf("Enter a number to find factorial\n");

  scanf("%d", &num);

   if (num 

    printf("Negative numbers are not allowed.\n");

  else

  {

    f = factorial(num);

    printf("%d! = %ld\n", num, f);

  }

getch();

}

long factorial(int n)

{

  if (n == 0)

    return 1;

  else

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

}

c program to find hcf and lcm

C program to find hcf and lcm: The code below finds highest common factor and least common multiple of two integers. HCF is also known as greatest common divisor(GCD) or greatest common factor(gcf).

C programming code

#include 

#include

 int main() {

  int a, b, x, y, t, gcd, lcm;

   printf("Enter two integers\n");

  scanf("%d%d", &x, &y);

   a = x;

  b = y;

   while (b != 0) {

    t = b;

    b = a % b;

    a = t;

  }

   gcd = a;

  lcm = (x*y)/gcd;

  printf("Greatest common divisor of %d and %d = %d\n", x, y, gcd);

  printf("Least common multiple of %d and %d = %d\n", x, y, lcm);

 getch();

}

C program to find hcf and lcm using recursion

#include 

#include

long gcd(long, long);

int main() {

  long x, y, hcf, lcm;

   printf("Enter two integers\n");

  scanf("%ld%ld", &x, &y);

   hcf = gcd(x, y);

  lcm = (x*y)/hcf;

   printf("Greatest common divisor of %ld and %ld = %ld\n", x, y, hcf);

  printf("Least common multiple of %ld and %ld = %ld\n", x, y, lcm);

   getch();

}

long gcd(long a, long b) {

  if (b == 0) {

    return a;

  }

  else {

    return gcd(b, a % b);

  }

}

C program to find hcf and lcm using function

#include 

#include

long gcd(long, long);

int main() {

  long x, y, hcf, lcm;

   printf("Enter two integers\n");

  scanf("%ld%ld", &x, &y);

   hcf = gcd(x, y);

  lcm = (x*y)/hcf;

   printf("Greatest common divisor of %ld and %ld = %ld\n", x, y, hcf);

  printf("Least common multiple of %ld and %ld = %ld\n", x, y, lcm);

   getch();

}

 long gcd(long x, long y) {

  if (x == 0) {

    return y;

  }

   while (y != 0) {

    if (x > y) {

      x = x - y;

    }

    else {

      y = y - x;

    }

  }

  return x;

}

Decimal to binary conversion

C program to convert decimal to binary: c language code to convert an integer from decimal number system(base-10) to binary number system(base-2). Size of integer is assumed to be 32 bits. We use bitwise operators to perform the desired task. We right shift the original number by 31, 30, 29, …, 1, 0 bits using a loop and bitwise AND the number obtained with 1(one), if the result is 1 then that bit is 1 otherwise it is 0(zero).

C programming code

#include 

#include 

 int main()

{

  int n, c, k;

   printf("Enter an integer in decimal number system\n");

  scanf("%d", &n);

   printf("%d in binary number system is:\n", n);

   for (c = 31; c >= 0; c--)

  {

    k = n >> c;

     if (k & 1)

      printf("1");

    else

      printf("0");

  }

   printf("\n");

   return 0;

}

Above code only prints binary of integer, but we may wish to perform operations on binary so in the code below we are storing the binary in a string. We create a function which returns a pointer to string which is the binary of the number passed as argument to the function.

C programm to store decimal to binary conversion in a string

#include 

#include 

#include

char *decimal_to_binary(int);

 main()

{

   int n, c, k;

   char *pointer;

    printf("Enter an integer in decimal number system\n");

   scanf("%d",&n);

    pointer = decimal_to_binary(n);

   printf("Binary string of %d is: %s\n", n, t);

    free(pointer);

 getch();

}

char *decimal_to_binary(int n)

{

   int c, d, count;

   char *pointer;

    count = 0;

   pointer = (char*)malloc(32+1);

    if ( pointer == NULL )

      exit(EXIT_FAILURE);

    for ( c = 31 ; c >= 0 ; c-- )

   {

      d = n >> c;

       if ( d & 1 )

         *(pointer+count) = 1 + '0';

      else

         *(pointer+count) = 0 + '0';

      count++;

   }

   *(pointer+count) = '\0';

    return  pointer;

}

c program to find ncr and npr

c program to find nCr and nPr: This code calculate nCr which is n!/(r!*(n-r)!) and nPr = n!/(n-r)!

C program to find nCr using function

#include

#include

long factorial(int);

long find_ncr(int, int);

long find_npr(int, int);

main()

{

   int n, r;

   long ncr, npr;

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

   scanf("%d%d",&n,&r);

    ncr = find_ncr(n, r);

   npr = find_npr(n, r);

    printf("%dC%d = %ld\n", n, r, ncr);

   printf("%dP%d = %ld\n", n, r, npr);

    getch();

}

long find_ncr(int n, int r)

{

   long result;

    result = factorial(n)/(factorial(r)*factorial(n-r));

    return result;

}

long find_npr(int n, int r)

{

   long result;

    result = factorial(n)/factorial(n-r);

    return result;

}

long factorial(int n)

{

   int c;

   long result = 1;

    for( c = 1 ; c 

      result = result*c;

    return ( result );

}