The document provides an overview of C programming concepts including data types, variables, operators, conditional statements, loops, functions, arrays and more. It begins with basic explanations and examples of comments, print statements, data types, input/output functions. Next, it covers conditional statements like if-else and examples to check even/odd and prime numbers. Loops like for and while are explained along with examples to iterate, count and print patterns. Other topics include functions, arrays, strings, structures and examples programs for calculations, pattern printing and sorting. Overall, the document serves as a useful reference for learning C programming fundamentals and common programming concepts.

1. C programs
Note:
// Comment – used to write comment, not executed.
n --- Newline, prints text goto next line. Eg: printf(“n”);
t --- Tab space , prints 5 spaces where t is used. printf(“t”);
getch() can be used at last line of program. Optional.
Data types are int , float, char, double.
Variables of various data type  %d for int , %f for float, %c for
char, %s for String(char array), %lf for double
int r=5;
float pi=3.14;
printf(“ r = %d , pi = %f n ”, r, pi);
-----------------------------------------------------------------------printf(“ Hello n world”);
- Prints output as
Hello
World
----------------------------------------------------------------------Condition statement is „ if ‟ . Execute any one set of statements based
on condition.
if( condition is true)
if (n>0)
{
Execute statements – true condition
}
else //condition is false
printf(“ positive number”);
else
printf(“ negative number”);
{
Execute statements –false condition
}
------------------------------------------------------------------------
Looping „for‟ loop format is
for (index = start_value; index < end_value; index +=
increment_value )
for(i=1; i<=10; i++) - counts i from 1 to 10. All statements inside ‘
for i ’ loop is executed 10 times. Equivalent to count x = 1 to 10
---------------------------------------------------------------------------for(j=1; j<=10; j=j+2)
- counts variable j from 1 to 10, only odd numbers, 1,3,5,7,9. All
statements in ‘for j’ loop executed only 5 times as j values 1,3,5,7,9.
----------------------------------------------------------------------------------
Simple program: // program gets integer „a‟ as input and prints it
#include <stdio.h>
Void main()
{
int a;
printf("Enter an integer n");
scanf("%d", &a);
printf("Integer entered is %d ", a);
}
Output:
Enter an Integer 85
Integer Entered is 85
c program to check odd or even:
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).
#include<stdio.h>
void main()
{
int n;
printf("Enter an integer ");
scanf("%d", &n);
if ( n%2 == 0 )
printf("n Even");
else
printf("n Odd");
}
Output:
Enter an integer 9
Odd
C program for to check prime number or not
Prime number logic: a number(n) is prime if it is divisible only by one and
itself(n). So, a number, say 6 is not a prime, if it is divisible by 2 or 3 or 4 or 5 ,
=> 2 to n-1. For checking whether number is divisible by 2 to n-1 using „for i
loop‟. (n%i) gives reminder, if n is not a prime, then remainder is zero. Say, for
number 6, 6%2 is 0, 6%3 is 0. Remember two is the only even and also the
smallest prime number. First few prime numbers are 2, 3, 5, 7, 11, 13, 17....etc.
To print prime, we need to check whether i is equal to n. as , if n is prime, „for‟
loop fully executes till i is n,say n=7.
We need 2 variables n, i declared as int (integer).
#include<stdio.h>
void main()
{
int n, i;
printf("Enter a number to check if it is prime ");
scanf("%d",&n);
for ( i = 2 ; i <= n - 1 ; i++ )
{
if ( n%i == 0 )
{
printf("%d is not prime.n", n);
break;
}
}
if ( i == n )
printf("n %d is prime n", n);
}
Output:
Enter a number to check if it is prime 7
7 is prime.
To print all primes between 2 to n. Prime num between 2 to 10 are 2,3,5,7.
1st for loop – to use numbers from 2 to n.
2nd „for‟ loop – to check given number is prime, by dividing it by 2 to n-1.
In 2nd for loop, condition can be i<=n-1 or i<n , both are same.
#include<stdio.h>
void main()
{
int n, i = 3, num;
printf("Enter the number n");
scanf("%d",&num);
printf(“Prime numbers till %d are”,num);
for ( n = 2 ; n <= num ;n++ )
{
for ( i = 2 ; i <= n - 1 ; i++ )
{
if ( n%i == 0 )
break;
}
if ( i == n )
printf("n %d is prime ",n);
}
}
Output:
Enter the number 10
Prime numbers till 10 are
2 is prime
3 is prime
5 is prime
7 is prime
Common mistakes:
1). In if condition, to check a is equal to 5 , we need to use ==.
int a=5;
if( a==5 )
{
printf(“ Value of a = %d”, a);
}
Note: here in „if‟ condition, only one statement inside if condition.
So, no need to use { }.
if ( a==5 )

2. printf(“ Value of a = %d”, a);
C program to sum first n numbers: sum = 1 + 2 +3 + … +n
Here we need 3 variables – sum, i, n.
Variable i to count i from 1 to n using „for‟ loop. Sum to calculate total. Since,
sum is used , it should be initialized to 0.
No need to use { } in „for i‟ loop as only one statement inside „for i „ loop.
#include<stdio.h>
void main()
{
Explanation
int n, i, sum=0;
n=3
printf (“ Enter number n”);
First i=1, sum=0
scanf (“%d”, &n);
. Sum=sum+i  sum=0+1=1
for ( i=1; i <=n; i++)
i incrmentes to 2, i=2,
sum = sum + i ;
. sum = sum +i  sum=1+2=3
printf(“ Sum = %d”, sum);
Next, i=3, (previous)sum =3
}
. sum=sum+i,  sum=3+3 = 6
Output:
Output: sum =6
Enter number n : 3
Sum = 6
Note:
1. Sum of series 12 + 22 + 32 + ….. +n2
Replace „ sum = sum + i ‟ with
sum = sum + ( i *i )
2.For product of first n numbers. 1 * 2 * 3 * ……* n
int product =1;
in „for‟ loop, product = product * i;
3.Sum of series 1 + 3 + 5 + 7 + …. +n
Replace for loop with
for ( i=1; i <=n; i=i+2)
4. sum of series
Declare sum as float,  float sum=0;
Changes in for loop,
sum = sum + (float) (1/i) ;
C program to add n numbers, input by user
Firstly user will enter a number indicating how many numbers user wishes to
add and then user will enter n numbers. In the first c program to add numbers
we are not using an array, and using array in the second code.
Here we need 3 variables – sum, i, num. Sum to calculate total. Variable i to
count i from 1 to n.
#include <stdio.h>
void main()
{
int n, sum = 0, i, value;
printf("Enter number of integers to add n");
scanf("%d", &n);
printf("Enter %d integers n",n);
for (i = 1; i <= n; i++)
{
scanf("%d",&value);
sum = sum + value;
}
printf("Sum of entered integers = %d n",sum);
}
Output:
Enter number of integers to add n”);
3
Enter 3 integers
1
5
10
Sum of entered integers = 16.
Using array – to add n numbers , input by user.
#include <stdio.h>
void main()
{
int n, sum = 0, i, array[100];
scanf("%d", &n);
for (i = 0; i < n; i++)
{
scanf("%d", &array[i]);
sum = sum + array[i];
}
printf("Sum = %dn",sum);
}
C program to Check Armstrong Number
A positive integer is called an Armstrong number if the sum of cubes of
individual digit is equal to that number itself. For example:
153 = 1*1*1 + 5*5*5 + 3*3*3
// 153 is an Armstrong number.
12 is not equal to 1*1*1+2*2*2
// 12 is not an Armstrong number.
Here, we need to get 3,5 and 1 separately. So, to get 3 we apply modulus
153%10 gives 3. So, remainder of 153 %10 as 3, remainder of 15 %10 as 5 ,
and reminder of 1 %10 as 1.
In order to get 15 from 153, we divide 153 by 10 gives 15.  153 /10  15.
Variables needed n, rem(remainder), sum (to sum 13 +53+33 ).
/* C program to check a number entered by user is Armstrong or not. */
#include <stdio.h>
int main()
{
int n, temp, rem, sum=0;
printf("Enter a positive integer: ");
scanf("%d", &n);
temp=n;
while ( temp!=0 )
{
rem = temp%10;
sum = sum + rem*rem*rem;
temp = temp /10;
}
if(sum == n)
printf("%d is an Armstrong number.",n);
else
printf("%d is not an Armstrong number.",n);
}
output:
Enter a positive integer: 153
153 is an Armstrong number
Consider the pattern. Program to print below pattern
*
**
***
****
*****
First for i loop to count each line.
Here we need another for j loop to print stars in a
line. Number of stars in each line is equal to line
number. So, j loop will count till i (line number)
#include <stdio.h>
void main()
{
int n, i, j;
printf("Enter number of rowsn");
scanf("%d",&n);
for ( i = 1 ; i <= n ; i++ )
{
for( j = 1 ; j <= i ; j++ )
printf("*");
printf("n");
}
}
note: for loop and its equivalent while loop
To print numbers 1 to 5
for( i=1 ; i<=5 ; i++)
printf(“n %d”, i);
output:
1
2
3
To print numbers 1 to 5
i=1;
while(i<=5)
{
printf("n %d", i);
i++; //or i=i+1
}

3. 4
5
Program to print below pattern
*
**
***
****
*****
3 loops needed. 1 loop for each line. 2nd loop for
spaces. 3rd loop for printing stars. For spaces last
line has zero or no spaces. First line has 4 spaces.
Number of lines be n. Number of spaces = n - line
number  5 – 1 =1 (for 1st line has one space).
#include<stdio.h>
void main()
{
int n, line, j, k;
printf("Enter number of rowsn");
scanf("%d",&n);
for ( line = 1 ; line <= n ; line++ )
{
for(k = 1 ; k <= n-line ; k++ )
printf(" ");
for( j = 1 ; j <= line ; j++ )
printf("*");
}
printf("n");
}
}
Output:
Enter the number of rows for pascal triangle : 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
// program for merge sort
#include<stdio.h>
#include<conio.h>
void mergesort(int *,int,int);
void merge(int *,int,int,int);
void main()
{
int i,n;
int a[20];
clrscr();
printf("n Merge Sort");
printf("n Enter number of elements");
scanf("%d",&n);
printf("n Enter the elements ");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("n");
}
}
Pascal Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
This triangle can be constructed by adding 1 to first and last element of each
row. Other elements can be calculated by adding the number above and to the
left with the number above and to the right to find the new value.
To calculate a[6][3]=a[5][2] + a[5][3]  10, where a[5][2] =4, a[5][3]=6
So,
a[i][j]=a[i-1][j-1]+a[i-1][j]
#include<stdio.h>
void main()
{
int a[10][10];
int no , i, j, k;
printf("Enter the number of rows for pascal triangle : ");
scanf("%d", &n);
//taking input
for(i=1; i<=n; i++)
{
for(j=1; j<=i; j++)
{
if( j==1 || j==i )
{
a[i][j]=1; //first and last value in triangle is 1
}
else
// taking arrays value , i is row/line. J is val ue position
{
// sum of previous row 2 num
a[i][j]=a[i-1][j-1]+a[i-1][j];
}
}
}
for(i=1; i<=n; i++)
{
for(k=1; k<=n-i; k++)
{
printf(" ");
}
for(j=1; j<=i ; j++)
{
printf("%d ",a[i][j]);
//printing pascal triangle
mergesort(a,0,n-1);
printf("n Sorted list");
for(i=0;i<n;i++)
printf("%d t",a[i]);
}
void mergesort(int a[], int low, int high)
{
if (low < high)
{
int m = (high + low)/2;
mergesort(a, low, m);
mergesort(a, m + 1, high);
merge(a, low, m, high);
}
}
/* We need 2 arrays 'a' and 'b', of equal size
*Here 'a' is the primary array (i.e which contains initial
input, and final output)
and 'b' is the temporary array,used for merging 2 sorted half's in 'a'
*/
void merge(int a[], int low, int mid, int high)
{
int b[10000];
int i = low, j = mid + 1, k = low;
while (i <= mid && j <= high)
{
if (a[i] <= a[j])
b[k++] = a[i++];
else
b[k++] = a[j++];
}
while (i <= mid)
b[k++] = a[i++];
while (j <= high)
b[k++] = a[j++];
for(k=low;k<=high;k++)
a[k]=b[k];
}

4. First 4 levels is mersgesort() function operation
Last 4 levels is merge operation
C[i][j] = A[i][j] + B[i][j]
Matrix multiplication
Fibonacci recursive function executes as
3 for loops
For I to 1 to row1
For j = 1 to col2
For k 1 to row2
C[i][j] = c[i][j] + a[i][k] + b[k][j]
For loop vs While loop.
While loop -> initialize done outside the loop.
For loop  initialize done inside the loop at first time.
Matrix transpose
B[i[][j] = A[j][i]
Matrix addition
Eg: to print numbers 1 to 5.
For(int i=1;i<=5;i++)
{
Printf(“n i= %d”, i);
}
i=1;
while(i<=5)
{
printf(“n i = %d”,i);
i++;
}
Prepared by Dr.B.Surendiran
Anna university
FOCP – Computer programming subject notes. Subject Code : GE6151