Programming egs

PROGRAMMING EXAMPLES
Check Whether a Number is Even or Odd
/*C program to check whether a number entered by user is even or odd. */
#include <stdio.h>
int main(){
int num;
printf("Enter an integer you want to check: ");
scanf("%d",&num);
if((num%2)==0) /* Checking whether remainder is 0 or not. */
printf("%d is even.",num);
else
printf("%d is odd.",num);
return 0;
}
Output 1
Enter an integer you want to check: 25
25 is odd.
Output 2
Enter an integer you want to check: 12
12 is even.
In this program, user is asked to enter an integer which is stored in variable num. Then,
the remainder is found when that number is divided by 2 and checked whether remainder
is 0 or not. If remainder is 0 then, that number is even otherwise that number is odd.
This task is performed using if...else statement in C programming and the result is
displayed accordingly.
This program also can be solved using conditional operator[ ?: ] which is the shorthand
notation for if...else statement.
/* C program to check whether an integer is odd or even using conditional operator */

#include <stdio.h>
int main(){
int num;
printf("Enter an integer you want to check: ");
scanf("%d",&num);
((num%2)==0) ? printf("%d is even.",num) : printf("%d is odd.",num);
return 0;
}
C Program to Print a Integer Entered by a User
#include <stdio.h>
int main()
{
int num;
printf("Enter a integer: ");
scanf("%d",&num); /* Storing a integer entered by user in variable num */
printf("You entered: %d",num);
return 0;
}
Output
Enter a integer: 25
You entered: 25
In this program, a variable num is declared of type integer using keyword int.
The printf() function prints the content inside quotation mark which is "Enter a integer: ".
Then, the scanf() takes integer value from user and stores it in variable num. Finally, the
value entered by user is displayed in the screen using printf().
C Program to Count Number of Digits of an Integer
#include <stdio.h>
int main()
{
int n,count=0;
printf("Enter an integer: ");
scanf("%d", &n);
while(n!=0)
{
n/=10; /* n=n/10 */
++count;
}
printf("Number of digits: %d",count);
}

Output
Enter an integer: 34523
Number of digits: 5
This program takes an integer from user and stores that number in variable n. Suppose,
user entered 34523. Then, while loop is executed because n!=0 will be true in first
iteration. The codes inside while loop will be executed. After first iteration, value of n
will be 3452 and count will be 1. Similarly, in second iteration n will be equal to 345
and count will be equal to 2. This process goes on and after fourth iteration, n will be
equal to 3 and count will be equal to 4. Then, in next iteration n will be equal to 0
and count will be equal to 5 and program will be terminated as n!=0 becomes false.
Check Character is an alphabet or not
/* C programming code to check whether a character is alphabet or not.*/
#include <stdio.h>
int main()
{
char c;
printf("Enter a character: ");
scanf("%c",&c);
if( (c>='a'&& c<='z') || (c>='A' && c<='Z'))
printf("%c is an alphabet.",c);
else
printf("%c is not an alphabet.",c);
return 0;
}
Output 1
Enter a character: *
* is not an alphabet
Output 2
Enter a character: K
K is an alphabet

When a character is stored in a variable, ASCII value of that character is stored instead
of that character itself. For example: If 'a' is stored in a variable, ASCII value of 'a'
which is 97 is stored. If you see the ASCII table, the lowercase alphabets are from 97 to
122 and uppercase letter are from 65 to 90. If the ASCII value of number stored is
between any of these two intervals then, that character will be an alphabet. In this
program, instead of number 97, 122, 65 and 90; we have used 'a', 'z', 'A' and 'Z'
respectively which is basically the same thing.
This program also can be performed using standard library function isalpha().
Find HCF or GCD
/* C Program to Find Highest Common Factor. */
#include <stdio.h>
int main()
{
int num1, num2, i, hcf;
printf("Enter two integers: ");
scanf("%d %d", &num1, &num2);
for(i=1; i<=num1 || i<=num2; ++i)
{
if(num1%i==0 && num2%i==0) /* Checking whether i is a factor of both number */
hcf=i;
}
printf("H.C.F of %d and %d is %d", num1, num2, hcf);
return 0;
}
In this program, two integers are taken from user and stored in variable num1 and num2.
Then i is initialized to 1 and for loop is executed until i becomes equal to smallest of
two numbers. In each looping iteration, it is checked whether i is factor of both numbers
or not. If i is factor of both numbers, it is stored to hcf. When for loop is completed, the
H.C.F of those two numbers will be stored in variable hcf.
Find Factorial of a Number
/* C program to display factorial of an integer if user enters non-negative integer. */
#include <stdio.h>
int main()
{
int n, count;
unsigned long long int factorial=1;
scanf("%d",&n);
if ( n< 0)

printf("Error!!! Factorial of negative number doesn't exist.");
else
{
for(count=1;count<=n;++count) /* for loop terminates if count>n */
{
factorial*=count; /* factorial=factorial*count */
}
printf("Factorial = %lu",factorial);
}
return 0;
}
Output 1
Enter an integer: -5
Error!!! Factorial of negative number doesn't exist.
Output 2
Factorial = 3628800
Here the type of factorial variable is declared as: unsigned long long. It is because, the
factorial is always positive, so unsigned keyword is used and the factorial of a number
can be pretty large. For example: the factorial of 10 is 3628800 thus, long long keyword is
used.
Find Number of Vowels, Consonants, Digits and
White Space Character
#include<stdio.h>
int main(){
char line[150];
int i,v,c,ch,d,s,o;
o=v=c=ch=d=s=0;
printf("Enter a line of string:n");
gets(line);
for(i=0;line[i]!='0';++i)
{
if(line[i]=='a' || line[i]=='e' || line[i]=='i' || line[i]=='o' || line[i]=='u' || line[i]=='A' ||
line[i]=='E' || line[i]=='I' || line[i]=='O' || line[i]=='U')
++v;
else if((line[i]>='a'&& line[i]<='z') || (line[i]>='A'&& line[i]<='Z'))
++c;
else if(line[i]>='0'&&c<='9')
++d;
else if (line[i]==' ')

++s;
}
printf("Vowels: %d",v);
printf("nConsonants: %d",c);
printf("nDigits: %d",d);
printf("nWhite spaces: %d",s);
return 0;
}
Output
Enter a line of string:
This program is easy 2 understand
Vowels: 9
Consonants: 18
Digits: 1
White spaces: 5
Calculate Length without Using strlen() Function
#include <stdio.h>
int main()
{
char s[1000],i;
printf("Enter a string: ");
scanf("%s",s);
for(i=0; s[i]!='0'; ++i);
printf("Length of string: %d",i);
return 0;
}
Output
Enter a string: Programiz
Length of string: 9
Copy String Manually
#include <stdio.h>
int main()
{
char s1[100], s2[100], i;
printf("Enter string s1: ");
scanf("%s",s1);
for(i=0; s1[i]!='0'; ++i)
{

s2[i]=s1[i];
}
s2[i]='0';
printf("String s2: %s",s2);
return 0;
}
Output
Enter String s1: programiz
String s2: programiz
This above program copies the content of string s1 to string s2 manually.
Calculate Sum of Natural Numbers
/* This program is solved using while loop. */
#include <stdio.h>
int main()
{
int n, count, sum=0;
scanf("%d",&n);
count=1;
while(count<=n) /* while loop terminates if count>n */
{
sum+=count; /* sum=sum+count */
++count;
}
printf("Sum = %d",sum);
return 0;
}
Calculate Sum Using for Loop
/* This program is solved using for loop. */
#include <stdio.h>
int main()
{
scanf("%d",&n);
{
}
return 0;
}

Output
Sum = 5050
Both program above does exactly the same thing. Initially, the value of count is set to 1.
Both program have test condition to perform looping iteration until
condition count<=n becomes false and in each iteration ++count is executed.
This program works perfectly for positive number but, if user enters negative number or
0, Sum = 0is displayed but, it is better is display the error message in this case. The
above program can be made user-friendly by adding if else statement as:
/* This program displays error message when user enters negative number or 0 and displays
the sum of natural numbers if user enters positive number. */
#include <stdio.h>
int main()
{
scanf("%d",&n);
if ( n<= 0)
printf("Error!!!!");
else
{
{
}
}
return 0;
}
Display Fibonacci series up to n terms
/* Displaying Fibonacci sequence up to nth term where n is entered by user. */
#include <stdio.h>
int main()
{
int count, n, t1=0, t2=1, display=0;
printf("Enter number of terms: ");
scanf("%d",&n);
printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
count=2; /* count=2 because first two terms are already displayed. */
while (count<n)
{

display=t1+t2;
t1=t2;
t2=display;
++count;
printf("%d+",display);
}
return 0;
}
Output
Enter number of terms: 10
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+
Suppose, instead of number of terms, you want to display the Fibonacci series util the
term is less than certain number entered by user. Then, this can be done using source
code below:
/* Displaying Fibonacci series up to certain number entered by user. */
#include <stdio.h>
int main()
{
int t1=0, t2=1, display=0, num;
scanf("%d",&num);
printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
display=t1+t2;
while(display<num)
{
printf("%d+",display);
t1=t2;
t2=display;
display=t1+t2;
}
return 0;
}
Output
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+
Generate Multiplication Table
/* C program to find multiplication table up to 10. */

#include <stdio.h>
int main()
{
int n, i;
printf("Enter an integer to find multiplication table: ");
scanf("%d",&n);
for(i=1;i<=10;++i)
{
printf("%d * %d = %dn", n, i, n*i);
}
return 0;
}
Output
Enter an integer to find multiplication table: 9
9 * 1 = 9
9 * 2 = 18
9 * 3 = 27
9 * 4 = 36
9 * 5 = 45
9 * 6 = 54
9 * 7 = 63
9 * 8 = 72
9 * 9 = 81
9 * 10 = 90
This program generates multiplication table of an integer up 10. But, this program can be
made more user friendly by giving option to user to enter number up to which, he/she
want to generate multiplication table. The source code below will perform this task.
#include <stdio.h>
int main()
{
int n, range, i;
printf("Enter an integer to find multiplication table: ");

scanf("%d",&n);
printf("Enter range of multiplication table: ");
scanf("%d",&range);
for(i=1;i<=range;++i)
{
printf("%d * %d = %dn", n, i, n*i);
}
return 0;
}
Output
Enter an integer to find multiplication table: 6
Enter range of multiplication table: 4
6 * 1 = 6
6 * 2 = 12
6 * 3 = 18
6 * 4 = 24
Reverse an Integer
#include <stdio.h>
int main()
{
int n, reverse=0, rem;
scanf("%d", &n);
while(n!=0)
{
rem=n%10;
reverse=reverse*10+rem;
n/=10;
}
printf("Reversed Number = %d",reverse);
return 0;
}
Output
Reversed Number = 5432

Check Palindrome Number
/* C program to check whether a number is palindrome or not */
#include <stdio.h>
int main()
{
int n, reverse=0, rem,temp;
scanf("%d", &n);
temp=n;
while(temp!=0)
{
rem=temp%10;
reverse=reverse*10+rem;
temp/=10;
}
/* Checking if number entered by user and it's reverse number is equal. */
if(reverse==n)
printf("%d is a palindrome.",n);
else
printf("%d is not a palindrome.",n);
return 0;
}
Output
12321 is a palindrome.
Check Armstrong Number
/* C program to check whether a number entered by user is Armstrong or not. */
#include <stdio.h>
int main()
{
int n, n1, rem, num=0;
printf("Enter a positive integer: ");
scanf("%d", &n);
n1=n;
while(n1!=0)
{
rem=n1%10;
num+=rem*rem*rem;
n1/=10;
}
if(num==n)
printf("%d is an Armstrong number.",n);
else
printf("%d is not an Armstrong number.",n);
}

Output
Enter a positive integer: 371
371 is an Armstrong number.
Make Simple Calculator in C
/* Source code to create a simple calculator for addition, subtraction, multiplication and
division using switch...case statement in C programming. */
# include <stdio.h>
int main()
{
char o;
float num1,num2;
printf("Enter operator either + or - or * or divide : ");
scanf("%c",&o);
printf("Enter two operands: ");
scanf("%f%f",&num1,&num2);
switch(o) {
case '+':
printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);
break;
case '-':
printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);
break;
case '*':
printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);
break;
case '/':
printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);
break;
default:
/* If operator is other than +, -, * or /, error message is shown */
printf("Error! operator is not correct");
break;
}
return 0;
}
Output
Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.4
3.4 - 8.4 = -5.0

This program takes an operator and two operands from user. The operator is stored in
variable operator and two operands are stored in num1 and num2 respectively. Then,
switch...case statement is used for checking the operator entered by user. If user enters +
then, statements for case: '+' is executed and program is terminated. If user enters - then,
statements for case: '-' is executed and program is terminated. This program works
similarly for * and / operator. But, if the operator doesn't matches any of the four
character [ +, -, * and / ], default statement is executed which displays error message.

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