C++ Program to Check Even or Odd Number

In C++, you can check if a number is even or odd by using the modulus operator % like this: if (num % 2 == 0) { /* even */ } else { /* odd */ }.

Examples

Input4

Output4 is even

Input7

Output7 is odd

Input0

Output0 is even

How to Think About It

To check if a number is even or odd, think about dividing it by 2. If the remainder is zero, the number is even because it divides evenly. If the remainder is not zero, the number is odd because it leaves a leftover.

Algorithm

1

Get the input number from the user

2

Calculate the remainder when the number is divided by 2

3

If the remainder is 0, the number is even

4

Otherwise, the number is odd

5

Print the result

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    if (num % 2 == 0) {
        cout << num << " is even" << endl;
    } else {
        cout << num << " is odd" << endl;
    }
    return 0;
}

Output

Enter a number: 7
7 is odd

Dry Run

Let's trace the input 7 through the code

1

Input number

User enters 7, so num = 7

2

Calculate remainder

7 % 2 equals 1 (since 7 divided by 2 leaves remainder 1)

3

Check remainder

Since remainder is 1 (not 0), number is odd

4

Print result

Output: "7 is odd"

num	num % 2	Condition	Output
7	1	1 != 0 (odd)	7 is odd

Why This Works

Step 1: Using modulus operator

The % operator gives the remainder of division, which helps us find if a number divides evenly by 2.

Step 2: Even number check

If the remainder is 0, the number is even because it divides exactly by 2.

Step 3: Odd number check

If the remainder is not 0, the number is odd because it leaves a remainder when divided by 2.

Alternative Approaches

Using bitwise AND operator

cpp

#include <iostream>
using namespace std;

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    if ((num & 1) == 0) {
        cout << num << " is even" << endl;
    } else {
        cout << num << " is odd" << endl;
    }
    return 0;
}

This method uses bitwise AND to check the last bit; it's faster but less intuitive for beginners.

Using ternary operator

cpp

#include <iostream>
using namespace std;

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    cout << num << " is " << ((num % 2 == 0) ? "even" : "odd") << endl;
    return 0;
}

This approach is concise and uses a single line to print the result.

Complexity: O(1) time, O(1) space

Time Complexity

The check uses a single modulus operation which takes constant time, so the time complexity is O(1).

Space Complexity

Only a few variables are used, so the space complexity is O(1).

Which Approach is Fastest?

The bitwise AND method is slightly faster than modulus but both are effectively constant time and suitable for this task.

Approach	Time	Space	Best For
Modulus operator (%)	O(1)	O(1)	Clarity and simplicity
Bitwise AND (&)	O(1)	O(1)	Performance and low-level operations
Ternary operator	O(1)	O(1)	Concise code