Converting an Int to a BCD byte array

Question

I want to convert an int to a byte[4] array using BCD.

The int in question will come from an device id and his needed to speak to an device via serialport.

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

example:

int id= 29068082

would output:

byte[4]{0x82, 0x80, 0x06, 0x29};

Answer 1

Same version as Peter O. but in VB.NET

Public Shared Function ToBcd(ByVal pValue As Integer) As Byte()
    If pValue < 0 OrElse pValue > 99999999 Then Throw New ArgumentOutOfRangeException("value")

    Dim ret As Byte() = New Byte(3) {} 'All bytes are init with 0's

    For i As Integer = 0 To 3
      ret(i) = CByte(pValue Mod 10)
      pValue = Math.Floor(pValue / 10.0)
      ret(i) = ret(i) Or CByte((pValue Mod 10) << 4)
      pValue = Math.Floor(pValue / 10.0)
      If pValue = 0 Then Exit For
    Next

    Return ret
End Function

The trick here is to be aware that simply using pValue /= 10 will round the value so if for instance the argument is "16", the first part of the byte will be correct, but the result of the division will be 2 (as 1.6 will be rounded up). Therefore I use the Math.Floor method.

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.)

Answer 3

maybe a simple parse function containing this loop

i=0;
while (id>0)
{
    twodigits=id%100; //need 2 digits per byte
    arr[i]=twodigits%10 + twodigits/10*16;  //first digit on first 4 bits second digit shifted with 4 bits
    id/=100;

    i++;
}