如何更改二进制表示中的第三位？

Question

I'm trying to write code in which code the third bit is always one.

I've done the following to detect the third bit, but i don't know how to change it to show one , it should be done without using loops, because this exercise is from first part of book for beginners and in this chapter there is information about variables and operators in c#.

int decimalNumber = int.Parse(Console.ReadLine());
string binaryConv = Convert.ToString(decimalNumber, 2);
char thirdBite = binaryConv[2];

Answer 1

int decimalNumber = int.Parse(Console.ReadLine()) | 4;

The 4 is the 3rd binary digit (100 base 2 = 4 base 10).

The | is a binary OR

Answer 2

What if you just OR the number with 4 (100 in binary)? This will ensure that 3rd bit is always set:

using System;

public class Program
{
    public static void Main()
    {
        Console.Write("Enter integer value: ");
        int intVal = Convert.ToInt32(Console.ReadLine());
        Console.WriteLine("You entered: {0} (In binary this number is: {1})", intVal, Convert.ToString(intVal, 2));
        Console.WriteLine("{0} | 4 (100 in binary) = {1} (In binary this number is: {2})", intVal, intVal | 4, Convert.ToString(intVal | 4, 2));
    }
}

Example Usage 1:

Enter integer value: 2
You entered: 2 (In binary this number is: 10)
2 | 4 (100 in binary) = 6 (In binary this number is: 110)

Example Usage 2:

Enter integer value: 49
You entered: 49 (In binary this number is: 110001)
49 | 4 (100 in binary) = 53 (In binary this number is: 110101)

Example Usage 3:

Enter integer value: -6
You entered: -6 (In binary this number is: 11111111111111111111111111111010)
-6 | 4 (100 in binary) = -2 (In binary this number is: 11111111111111111111111111111110)

Answer 3

Welcome. Your solution will give you the third bit. But it does it by converting to a string, which is way more expensive then it needs to be.

Get the value of the third bit like this:

 bool thirdBit = (decimalNumber & (1 << 2)) != 0;

Make sure the 3rd bit is set ON:

 decimalNumber = decimalNumber | (1 << 2);

Make sure the 3rd bit is set OFF:

 decimalNumber = decimalNumber & ~(1 << 2);