Go Program to Check Even or Odd Number

In Go, 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, divide it by 2 and look at the remainder. If the remainder is 0, the number is even because it divides evenly by 2. If the remainder is 1, the number is odd because it leaves a leftover when divided by 2.

Algorithm

1

Get the input number.

2

Calculate the remainder when the number is divided by 2 using the modulus operator.

3

If the remainder is 0, the number is even.

4

Otherwise, the number is odd.

5

Print the result.

Code

go

package main

import "fmt"

func main() {
    var num int
    fmt.Print("Enter a number: ")
    fmt.Scan(&num)

    if num%2 == 0 {
        fmt.Printf("%d is even\n", num)
    } else {
        fmt.Printf("%d is odd\n", num)
    }
}

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 = 1 (since 7 divided by 2 leaves remainder 1)

3

Check remainder

Since remainder is 1, condition num%2 == 0 is false

4

Print result

Print '7 is odd'

num	num % 2	Condition (num % 2 == 0)	Output
7	1	false	"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

go

package main

import "fmt"

func main() {
    var num int
    fmt.Print("Enter a number: ")
    fmt.Scan(&num)

    if num&1 == 0 {
        fmt.Printf("%d is even\n", num)
    } else {
        fmt.Printf("%d is odd\n", num)
    }
}

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

Using a function to check even or odd

go

package main

import "fmt"

func isEven(num int) bool {
    return num%2 == 0
}

func main() {
    var num int
    fmt.Print("Enter a number: ")
    fmt.Scan(&num)

    if isEven(num) {
        fmt.Printf("%d is even\n", num)
    } else {
        fmt.Printf("%d is odd\n", num)
    }
}

This approach separates logic into a function, improving code reuse and clarity.

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

The program uses a fixed amount of memory for variables, so space complexity is O(1).

Which Approach is Fastest?

Using bitwise AND is slightly faster than modulus but both are constant time and efficient for this task.

Approach	Time	Space	Best For
Modulus operator (%)	O(1)	O(1)	Simplicity and clarity
Bitwise AND (&)	O(1)	O(1)	Performance and low-level operations
Function abstraction	O(1)	O(1)	Code reuse and readability