Converting a int to a BCD byte array

Question

I want to convert an int to a byte[2] array using BCD.

The int in question will come from DateTime representing the Year and must be converted to two bytes.

Is there any pre-made function that does this or can you give me a simple way of doing this?

example:

int year = 2010

would output:

byte[2]{0x20, 0x10};

Answer 1

    static byte[] Year2Bcd(int year) {
        if (year < 0 || year > 9999) throw new ArgumentException();
        int bcd = 0;
        for (int digit = 0; digit < 4; ++digit) {
            int nibble = year % 10;
            bcd |= nibble << (digit * 4);
            year /= 10;
        }
        return new byte[] { (byte)((bcd >> 8) & 0xff), (byte)(bcd & 0xff) };
    }

Beware that you asked for a big-endian result, that's a bit unusual.

Answer 2

Use this method.

    public static byte[] ToBcd(int value){
        if(value<0 || value>99999999)
            throw new ArgumentOutOfRangeException("value");
        byte[] ret=new byte[4];
        for(int i=0;i<4;i++){
            ret[i]=(byte)(value%10);
            value/=10;
            ret[i]|=(byte)((value%10)<<4);
            value/=10;
        }
        return ret;
    }

This is essentially how it works.

• If the value is less than 0 or greater than 99999999, the value won't fit in four bytes. More formally, if the value is less than 0 or is 10^(n*2) or greater, where n is the number of bytes, the value won't fit in n bytes.
• For each byte:
  • Set that byte to the remainder of the value-divided-by-10 to the byte. (This will place the last digit in the low nibble [half-byte] of the current byte.)
  • Divide the value by 10.
  • Add 16 times the remainder of the value-divided-by-10 to the byte. (This will place the now-last digit in the high nibble of the current byte.)
  • Divide the value by 10.

(One optimization is to set every byte to 0 beforehand -- which is implicitly done by .NET when it allocates a new array -- and to stop iterating when the value reaches 0. This latter optimization is not done in the code above, for simplicity. Also, if available, some compilers or assemblers offer a divide/remainder routine that allows retrieving the quotient and remainder in one division step, an optimization which is not usually necessary though.)