**This is a sample C# program to print even numbers. There are 2 ways to to print even numbers.**

**1.**Start at Zero. Increment the value by 2, print it. Do this until you reach your target number.

**2.**Start at Zero. Divide the number by 2. If the remainder is zero, then you know the number is even. So print it and then increment it by 1 and divide the number by 2 again. Repeat this until we reach our target number.

This program can also be used to print odd numbers by making a minor logic change. Initialize the targetNumber variable to 1 instead of 0. If you are using dividing logic, then print the number if the remainder is not zero.

**Note:**This program will crash and throws an exception, if you enter a very big number or a string. To handle exceptions, we can make use of try catch blocks. We will look at this in another program.

using System;

namespace SamplePrograms

{

class EvenNumbers

{

public static void Main()

{

// Prompt the user to enter a target number. Target number is the

// number untill which the user want to have even and odd numbers printed

Console.WriteLine("Please enter your target");

// Declare a variable to hold the target number

int targetNumber = 0;

// Retrieve, Convert and store the target number

targetNumber = Convert.ToInt32(Console.ReadLine());

// Use a FOR or WHILE loop to print the even numbers, until our target number

for (int i = 0; i <= targetNumber; i = i + 2)

{

Console.WriteLine(i);

}

// You can also check if a number is even, by dividing it by 2.

//for (int i = 0; i <= targetNumber; i++)

//{

// if ((i % 2) == 0)

// {

// Console.WriteLine(i);

// }

//}

// You can also use a while loop to do the same as shown below.

//int start = 0;

//while (start <= targetNumber)

//{

// Console.WriteLine(start);

// start = start + 2;

//}

// This line is to make the program wait for user input, instead of immediately closing

Console.ReadLine();}

}

}

thank you so much sir

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ReplyDeleteOne more and possibly the most elegant way to determine even/odd numbers is by using the C# bitwise & (and) operator instead of dividing by 2, like so:

ReplyDeleteif ((i & 1) == 1)

{

Console.WriteLine(i); // odd number

}

// or

if ((i & 1) == 0)

{

Console.WriteLine(i); // even number

}

This works because odd numbers always have 1 in the loweset bit of their binary representation, whereas even numbers always have 0.

1 is binary 0...00000001, so the bitwise & Operator will zero out all bits it reads from i (x & 0 == 0) except for the lowest one:

– For odd i, its lowest bit (1) will also lead to 1 in the result's lowest bit (1 & 1 == 1), so the result will be 0...00000001 == 1.

– For even i, its lowest bit (0) will also lead to 0 in the result's lowest bit (0 & 1 == 0), so the result will be 0...00000000 == 0.

Don P

superbbbbbbbbbb

ReplyDeletenice

ReplyDeleteuseful

ReplyDelete