How to add even parity bit on 7-bit binary number

Question

I am continuing from my previous question. I am making a c# program where the user enters a 7-bit binary number and the computer prints out the number with an even parity bit to

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.