Basic C Programming Practice - Practice 2

=> Write a program to check the user inputted number is even or odd number?

Solution :

#include<stdio.h>

#include<conio.h>

main()

{

      int numb;

      printf("Let's Know The Number Is Even Or Odd.  \nPlease Enter Your Number : ");

      scanf("%d",&numb);

      if(numb%2==0)

      printf("The Number Is Even.");

      else

      printf("The Number Is Odd.");

      getch();

}

Program Analysis :

We know if any number get divided by 2 and the reminder (%) become 0 then the number is EVEN so here I use this logic in this line :

   if(numb%2==0)

Here numb is a variable inputted by user and the expression is if the number get divided by 2 and the remainder become 0 than the number is EVEN:

We can also do it like the following :

#include<stdio.h>

#include<conio.h>

main()

{

      int numb;

      printf("Let's Know The Number Is Even Or Odd.  \nPlease Enter Your Number : ");

      scanf("%d",&numb);

      if(numb%2==1)

      printf("The Number Is Odd.");

      else

      printf("The Number Is Even.");

      getch();

}

Program Analysis :

if(numb%2==1)

This expression says that if the given number get divided by 2 then the reminder will be 1 and it will be an odd number otherwise it will be an Even number.

Example Program :                                              

#include<stdio.h>

#include<conio.h>

main()

{

      int numb;

      printf("Let's Know The Number Is Even Or Odd.  \nPlease Enter Your Number : ");

      scanf("%d",&numb);

      if(numb%2==1)

      printf("The Number Is Odd.");

      else

      printf("The Number Is Even.");

      if(numb>5)

      printf("\nThe Number Is Grater Than 5");

      else

      if(numb<5)

      printf("\nThe Number Is Less Than 5");

      else

      if(numb=5)

      printf("\nThe Number Is 5");

      getch();

      }

Hope you get a clear concept on this decision control statement " if "