如何在7位二进制数上添加偶校验位

Question

I am continuing from my previous question. I am making ac# program where the user enters a 7-bit binary number and the computer prints out the number with an even parity bit to the right of the number. I am struggling. I have a code, but it says BitArray is a namespace but is used as a type. Also, is there a way I could improve the code and make it simpler?

namespace BitArray
{
    class Program
    {    
        static void Main(string[] args)    
        {
            Console.WriteLine("Please enter a 7-bit binary number:");
            int a = Convert.ToInt32(Console.ReadLine());
            byte[] numberAsByte = new byte[] { (byte)a };
            BitArray bits = new BitArray(numberAsByte);
            int count = 0;

            for (int i = 0; i < 8; i++)
            {
                if (bits[i])
                {
                    count++;
                }
            }

            if (count % 2 == 1)
            {
                bits[7] = true;
            }

            bits.CopyTo(numberAsByte, 0);
            a = numberAsByte[0];
            Console.WriteLine("The binary number with a parity bit is:");
            Console.WriteLine(a);

Answer 1

Might be more fun to duplicate the circuit they use to do this..

bool odd = false;

for(int i=6;i>=0;i--)
  odd ^= (number & (1 << i)) > 0;

Then if you want even parity set bit 7 to odd, odd parity to not odd.

or

bool even = true;

for(int i=6;i>=0;i--)
  even ^= (number & (1 << i)) > 0;

The circuit is dual function returns 0 and 1 or 1 and 0, does more than 1 bit at a time as well, but this is a bit light for TPL....

PS you might want to check the input for < 128 otherwise things are going to go well wrong.

ooh didn't notice the homework tag, don't use this unless you can explain it.

Answer 2

Almost the same process, only much faster on a larger number of bits. Using only the arithmetic operators (SHR && XOR), without loops:

public static bool is_parity(int data)
{
    //data ^= data >> 32; // if arg >= 64-bit (notice argument length)
    //data ^= data >> 16; // if arg >= 32-bit 
    //data ^= data >> 8;  // if arg >= 16-bit
    data ^= data >> 4;
    data ^= data >> 2;
    data ^= data >> 1;
    return (data & 1) !=0;
}

public static byte fix_parity(byte data)
{
    if (is_parity(data)) return data;
    return (byte)(data ^ 128);
}

Answer 3

Using a BitArray does not buy you much here, if anything it makes your code harder to understand. Your problem can be solved with basic bit manipulation with the & and | and << operators.

For example to find out if a certain bit is set in a number you can & the number with the corresponding power of 2. That leads to:

int bitsSet = 0;
for(int i=0;i<7;i++)
    if ((number & (1 << i)) > 0)
        bitsSet++;

Now the only thing remain is determining if bitsSet is even or odd and then setting the remaining bit if necessary.