C# 0 The Complete Reference

if

 statements, the value of 

num

 is ANDed with 1. If bit zero in 

num

 is set, the result 

of

num & 1

 is 1; otherwise, the result is zero. Therefore, the 

if

 statement can succeed only 

when the number is odd.

You can use the bit-testing capability of the bitwise 

&

 to create a program that uses the 

bitwise

&

 to show the bits of a 

byte

 value in binary format. Here is one approach:

// Display the bits within a byte.

using System;

class ShowBits {

  static void Main() {

    int t;

    byte val;

    val = 123;

    for(t=128; t > 0; t = t/2) {

      if((val & t) != 0) Console.Write("1 ");

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      if((val & t) == 0) Console.Write("0 ");

    }

  }

}

The output is shown here:

0 1 1 1 1 0 1 1

The

for

 loop successively tests each bit in 

val

, using the bitwise AND, to determine if it is on 

or off. If the bit is on, the digit 

1

 is displayed; otherwise, 

0

 is displayed.

The bitwise OR can be used to turn bits on. Any bit that is set to 1 in either operand will 

cause the corresponding bit in the variable to be set to 1. For example

      1 1 0 1  0 0 1 1

|    1 0 1 0  1 0 1 0

      --------------------

      1 1 1 1  1 0 1 1

You can make use of the OR to change the make-even program shown earlier into a 

make-odd program, as shown here:

//  Use bitwise OR to make a number odd.

using System;

class MakeOdd {

  static void Main() {

    ushort num;

    ushort i;

    for(i = 1; i <= 10; i++) {

      num = i;

      Console.WriteLine("num: " + num);

      num = (ushort) (num | 1);

      Console.WriteLine("num after turning on bit zero: "

                        +  num + "\n");

    }

  }

}

The output from this program is shown here:

num: 1

num after turning on bit zero: 1

num: 2

num after turning on bit zero: 3

num: 3

num after turning on bit zero: 3

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