# Check Whether a Number is Even or Odd in C++

## Check Whether a Number is Even or Odd in C++

Given an integer input num, the objective is to write a code to Check Whether a Number is Even or Odd in C++. To do so we check if the number is divisible by 2 or not, it’s Even if it’s divisible otherwise Odd.

Example
Input : num = 12
Output : Even

## Check Whether a Number is Even or Odd in C++

Given an integer input the objective is to write a C++ code to Check Whether a Number is Even or Odd. To do so the main idea is to divide the number by 2 and check if it’s divisible or not. It’s an Even number is it’s perfectly divisible by 2 or an Odd number otherwise.

Here are the Methods to solve the above mentioned problem,

• Method 1 : Using Brute Force
• Method 2 : Using Ternary Operator
• Method 3 : Using Bitwise Operators
We’ll discuss the above mentioned methods in detail in the next section.

## Method 1 : Using Brute Force

This method simply checks if the given input integer is divisible by 2 or not. If it’s divisible then print Even or Odd otherwise.

### C++ Code

Run
#include <iostream>
using namespace std;
int main ()
{
int number;
cout << "Enter a number:"; cin >> number;

//checking whether the number is even or odd
if (number % 2 == 0)
cout << number << " : Even";
else
cout << number << " : Odd";

return 0;
}

### Output

Enter a number: 4
24 : Even

### Working

The working of the above code is mentioned below

1. Input an integer input “number
2. Check whether the number is divisible by 2
3. This means using modulo/remainder operator leaves 0 as a remainder
4. Do : if (number % 2 == 0)
1. if yes, print “Even number”
2. if not, print “Odd number”

## Method 2 : Using Ternary Operator

This Method uses the ternary operator to check if the integer input is divisible by 2, If true print Even or Odd otherwise.

### C++ Code

Run
#include <iostream>
namespace std;

int main ()
{
int number;
cout << "Enter a number:"; cin >> number;

//Checking if the number is divisible by 2
number % 2 == 0 ? cout << "Even":cout << "Odd";

return 0;
}

### Output

Enter a number: 17
Odd

### Working

The working of the above code is as follows,

1. Input an integer input “number
2. Check whether the number is divisible by 2 using the ternary operator
3. (number % 2) ? (cout <<“Even”) : (cout << “Odd”)

## Method 3 : Using Bitwise Operator

This Method uses bitwise operators to check if a given number is Even or Odd.

### C++ Code

Run
#include <iostream>
using namespace std;

// Returns true if n is even, else odd
bool isEven(int number)
{

// n & 1 is 1, then odd, else even
return (!(number & 1));
}

// Driver code
int main()
{
int number;

cout << "Enter the number: "; cin >> number;

if(isEven(number))
cout << "Even";
else
cout << "Odd";

//below can also be used instead of if else conditions
//isEven(number)? cout << "Even" : cout << "Odd";

return 0;
}

### Output

Enter a number: 13
Odd

### Working

The working of the above code is as follows,

1. If we have any number ‘n‘ doing bitwise ‘&‘ operation will give resultant as
• 1: If n is odd
• 0: if n is even

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