# C++ Programs to Check Entered Number is EVEN or ODD

In this part of the short tutorial series on C++, we will show you how to check whether a number entered by the user is Even or Odd.

• The number which can be divided by 2 is even
• The number that can’t be divided by 2 is odd

### First Program – Using % operator in if else

The % operator is used to return the remainder after division.

As such, the remainder of a number that can be divided by two is always zero – we can use it to achieve our task i.e. a number input by the user is even or odd.

See the C++ program below that uses the % operator with if..else statements. For taking user input, we used C++ cin as shown in the program below:

```#include <iostream>
using namespace std;

int main() {
int number;

//Asking user input to check for even or odd

cout << "Please enter a number? ";
cin >> number;

//Using if with % operator to check if number is even or odd

if ( number % 2 == 0)
cout <<"The given number " << number << " is EVEN!" <<"\n\n";
else
cout <<"The given number " << number << " is ODD!" <<"\n\n";

return 0;

}```

Sample Output 1:

Sample Output 2

### Second Approach – Using Bitwise operator AND

This C++ program uses the AND bitwise operator to check if the number is Even or Odd:

```#include <iostream>

using namespace std;

int main() {

int number;

//Asking user input to check for even or odd

cout << "Please enter a number? ";

cin >> number;

//Using bitwise operator AND

if((number & 1) == 0)

cout <<"The given number " << number << " is EVEN!" <<"\n\n";

else

cout <<"The given number " << number << " is ODD!" <<"\n\n";

return 0;

}```

Sample output 1

Sample output 1

### Third Approach – Using ternary operator example

This program also tells if the entered number is even or odd by using the ternary operator.

Have a look at the code and sample outputs:

```#include <iostream>
using namespace std;

int main() {

int number;

//Asking user input to check for even or odd
cout << "Please enter a number? ";

cin >> number;

//Ternary operator to check a number is even or odd

(number % 2 == 0) ? cout << number << " is an EVEN number!" :  cout << number << " is and ODD number!";

return 0;

}```

Output:

### Fourth Solution – Using Switch Case Statement

As such, the switch case is also a decision-making statement in C++. We include this example just to add another option if you get this assignment.

This example also uses the modulus operator (%), just like the if statement and remainder options are used as cases.

See the code and sample output below:

```#include <iostream>

using namespace std;

int main(){

int number;

//Asking user input to check for even or odd

cout << "Please enter a number? ";

cin >> number;

switch(number % 2)

{
case 0: cout <<"The given number " << number << " is EVEN!" <<"\n\n";
break;
case 1: cout <<"The given number " << number << " is ODD!" <<"\n\n";
break;
}
return 0;

}```

Output:

##### Author - Atiq Zia

Atiq is the writer at jquery-az.com, an online tutorial website started in 2014. With a passion for coding and solutions, I navigate through various languages and frameworks. Follow along as we solve the mysteries of coding together!