Odd numbers are not even. With modulo division, we can see if the number is evenly divisible by 2. If it is not, it must be odd.
We demonstrate the IsOdd
method and then the IsEven
method. It is important to test methods like these—one can be implemented with the negation of the other.
The IsOdd
static
method performs a modulo division on the parameter, which returns the remainder of a division operation.
using System; class Program { static void Main() { for (int i = 0; i <= 100; i++) { if (IsOdd(i)) { Console.WriteLine(i); } } } public static bool IsOdd(int value) { return value % 2 != 0; } }1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99
All even numbers are divisible by two. This means we can use the modulo division operator to see if there is any remainder when the number is divided by two.
using System; class Program { static void Main() { for (int i = 0; i <= 100; i++) { if (IsEven(i)) { Console.WriteLine(i); } } } public static bool IsEven(int value) { return value % 2 == 0; } }0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100
IsOdd
You could implement IsOdd
by using the IsEven
method and returning its boolean opposite. In other words, you could simply return "!IsEven."
A surprising number of programs require that you test for even and odd numbers. For example, if data must be entered in pairs, its count will always be even.