**Here is the list of sample programs using C, This is list number 2. There are 21 more programs at this post.**

**http://www.nsnam.com/2014/10/simple-c-programs-for-beginners.html**

**C Program 1 – To find the reversal of a given number**

Assume the number 1234, after reversal the number becomes 4321.

Input: One number, Ex 1234, declared to

**num**

Output : One number, Ex 4321,

**rev**is the variable

Logic:

- separate each digit using modulo operator from the right and bring that number to front
- divide the given number by 10, so the right most digit will be discarded.
- run till the number becomes <=0

#include <stdio.h>

#include <conio.h>

int main()

{

int num,rev=0;

printf("enter the number to be reversed");

scanf("%d", &num);

while(num>0)//till the number is positive, perform the process{//separate each digit and bring it to the first

rev=rev*10 + (num%10);num=num/10;

}

printf("The reversal is %d ",rev);

getch();

return 0;

}

**C Program 2 – Electricity Bill calculation**

This program is to display the electricity bill calculation based on the number of units consumed every month

Input : the number of units – variable name – unit

Output – Amount of rupee – variable name – amount

Logic:

Units | Rupees |

1-50 units | 0.75/unit |

51-100 | 0.85/unit |

101-200 | 1.50/unit |

201-300 | 2.20/unit |

>300 | 3.00/unit |

#include <stdio.h>

#include <conio.h>

int main()

{

float amount=0,units;;

printf("Enter the number of units");

scanf("%f", &units);

if(units <=50)

{

amount = units * 0.75;

}

else if(units >50 && units <=100)

{

amount=0.75 * 50 + 0.85*(units-50);

}

else if(units >100 && units <200)

{

amount=(0.75*50) + (0.85*50 )+ (1.5 *(units-100));

}

else if(units >200 && units <300)

{

amount=(0.75*50) + (0.85*50 )+ (1.5 *100) + (2.20 *(units-200));

}

else

{

amount=(0.75*50) + (0.85*50 )+ (1.5 *100) + (2.20 * 100) +(3.0*(units-300));

}

amount=amount+(0.2*amount);

printf("The total electricity bill is %f", amount);

getch();

return 0;

}

**C Program 3 – To check whether a number is Armstrong Number**

The armstrong number is of the form 153= 1

^{3}+ 5

^{3}+ 3

^{3}

The input is : 153 or any other number

output: The number is armstrong or not.

Processing: take 153 as an example, remove 3, 5 and 1 in the reverse order (using % operator) and take the power of 3 and add to the sum variable.

if the total sum and the original number, both are same, then that is the arm strong number.

if else, the number is not an armstrong number

#include <stdio.h>

#include <conio.h>

int main()// Get the original number

{

int original_num, check, temp, sum=0;

printf("Enter the number to check for armstrong number");

scanf("%d", &original_num);temp=original_num;//run the loop till the number becomes 0

while(original_num>0){//remove the last digit using modulo operator

check=original_num%10;sum=sum+check*check*check;//the last digit is taken power to 3 and added to sumoriginal_num=original_num/10;//truncate the last digit and run the loop again}

if(sum==temp)

printf("This is an armstrong number\n");

else

printf("This is not an armstrong number \n");

getch();

return 0;

}

**C Program 4 – To check whether a given number is prime or not**

A prime number can be divided by 1 and itself, there are no other divisors,

Examples are : 2 3, 5, 7, 11, …..

To find out whether a given number is prime or not, here is the logic

1. Get the number

2. divide the given number from 2 to n-1 (Example if 6 is the number divided by 2,3,4,5 will get the remainder respectively 0,0,2,3)

3. increment a counter to 1 if the remainder is 0

4. if there counter variable is 0, then the given number is prime (because we didn’t get any remainder) else non prime

Here is the program

#include <stdio.h>

#include <conio.h>

int main()

{

int a,i,count=0;

printf("enter a");//Let the given number is a

scanf("%d",&a);//get the numberfor(i=2;i<a;i++)//divide the number a from 2 to a-1

{

if(a%i==0)

count++;//increment a counter if the divisibility is 0}

if(count !=0)//if the counter is not zero, then prime

printf("a is not a prime number");

else

printf("a is a prime number");

getch();

return 0;

}

**C Program 5 – Arrays with Functions (passing Array elements to function)**

/*

Program to demonstrate arrays with function in which the array elements are passed as function parameters

*/

#include <stdio.h>

#include <conio.h>

void display(int);//function prototype

int main()

{

int a[10],i;

for(i=0;i<10;i++)//getting the array elements

{

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

}

for(i=0;i<10;i++)

display(a[i]);//call the function inside the loop, this function called for 10 timesgetch();

return 0;

}

void display(int a)//function implementation to print the elements, here the parameter is an integer only(means we are passing the array elements)

{

printf("%d",a);

}

**C Program 6 – Passing an entire array to a function**

/*In the function call ie

This program is to pass the entire array to a function

*/

#include <stdio.h>

#include <conio.h>

int maximum(int[],int);//function prototype, 1st parameter is array and the second is integer

int main()

{

int a[100],i,n,max=0;

printf("Enter the numbers");

scanf("%d",&n);//get the index number

for(i=0;i<n;i++)//get the array elements

{

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

}

max=maximum(a,n);//function call

printf("Maximum number is %d",max);

getch();

return 0;

}

int maximum(int b[],int a)//function implementation

{

int i,max=b[0];

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

if(b[i]>max)

max=b[i];

return max;

}

the above line can be written asmax=maximum(a,n)where a is the array and n is the number of array elements

since the array is just a memory address, it is enough to mention the base address (address of the first element) of the arraymax = maximum(&a[0],n)

so the array name is just equivalent to the base address

in the above example

a means &a[0]

**C Program 7 – To sort a Given set of numbers in ascending order (BubbleSort)**

/* Program to sort the given set of numbers in

ascending order, this sorting is called as bubble sort algorithm

*/

#include <stdio.h>

#include <conio.h>

int main()

{

int a[10],i,j,temp=0;

printf("Enter all the 10 numbers");

for(i=0;i<10;i++)

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

for(i=0;i<10;i++) //This loop is for total array elements (n)

{

for(j=0;j<9;j++) //this loop is for total combinations (n-1)

{

if(a[j]>a[j+1]) //if the first number is bigger then swap the two numbers

{

temp=a[j];

a[j]=a[j+1];

a[j+1]=temp;

}

}

}

printf("The ordered array is");

for(j=0;j<10;j++) //Finally print the ordered array

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

getch();

return 0;

}

**C Program 8 – To search a number in a given array**

//This program is to search a given number in an array

#include <stdio.h>

#include <conio.h>

int main()

{

int a[10],i,num;

printf("enter the array elements");

for(i=0;i<10;i++) //get all the numbers

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

printf("Enter the number to search");

scanf("%d",&num);

for(i=0;i<10;i++)

{

if(a[i]==num) //given num is matched in the array

{

printf("The number is found in the %d position",i+1);

getch();

exit(0); //to go the end of the program

}

}

printf("The number is not found"); //if num not found, this will be displayed

getch();

return 0;

}

**C Program 9 – Adding two matrices**

/* Program to add two matrices */

#include <stdio.h>

#include <conio.h>

int main()

{

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

printf("Enter a");

for(i=0;i<2;i++) //get the matrix A

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

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

printf("Enter b");

for(i=0;i<2;i++) //get the matrix B

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

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

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

{

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

{

c[i][j] = a[i][j] +b[i][j]; //adding two matrices

}

}

printf("Added Matrix is \n");

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

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

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

getch();

return 0;

}

**C Program–Multiplying two matrices**

/* Program to multiply two matrices */

#include <stdio.h>

#include <conio.h>

int main()

{

int a[2][3],b[3][2],c[2][2],k,j,i;

printf("enter a");

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

**//Get array A**

{

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

{

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

}

}

printf("enter b");

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

**//Get array B**

{

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

{

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

}

}

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

{

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

{

c[i][j]=0;

**//to Hold a temporary multiplication**

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

{

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

**//Multiplication algorithm**

}

}

}

printf("C is ");

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

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

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

getch();

return 0;

}

T S Pradeep Kumar

## Comments

## Post a Comment