Convert Delphi Real48 to C# double

Question

Hey,

I need to be able to convert from a Delphi Real48 to C# double.

I've got the bytes I need to convert but am looking for an elegant solution. to the problem.

Anybody out there had to do this before?

I'm needing to do the conversion in C#

Thanks in advance

Answer 1

static double GetDoubleFromBytes(byte[] bytes)
{
    var real48 = new long[6];
    real48[0] = bytes[0];
    real48[1] = bytes[1];
    real48[2] = bytes[2];
    real48[3] = bytes[3];
    real48[4] = bytes[4];
    real48[5] = bytes[5];

    long sign = (real48[0] & 0x80) >> 7;

    long significand = 
        ((real48[0] % 0x80) << 32) + 
         (real48[1] << 24) + 
         (real48[2] << 16) + 
         (real48[3] << 8) + 
         (real48[4]);

    long exponent = bytes[5];

    if (exponent == 0)
    {
        return 0.0;
    }

    exponent += 894;
    long bits = (sign << 63) + (exponent << 52) + (significand << 13);
    return BitConverter.Int64BitsToDouble(bits);
}

Answer 2

Ok

I've done some hunting around and I found some C++ code to do the job, converted it and it seems to be giving the right answer... damned if I understand it all though :S

    private static double Real48ToDouble(byte[] real48)
    {

        if (real48[0] == 0)
            return 0.0; // Null exponent = 0

        double exponent = real48[0] - 129.0;
        double mantissa = 0.0;

        for (int i = 1; i < 4; i++)
        {
            mantissa += real48[i];
            mantissa *= 0.00390625; // mantissa /= 256
        }


        mantissa += (real48[5] & 0x7F);
        mantissa *= 0.0078125; // mantissa /= 128
        mantissa += 1.0;

        if ((real48[5] & 0x80) == 0x80) // Sign bit check
            mantissa = -mantissa;

        return mantissa * Math.Pow(2.0, exponent);
    }

If somebody can explain it that would be great :D

Answer 3

I've changed the code you've posted into a more readable format so you can see how it works:

        Double exponentbase = 129d;
        Double exponent = real48[0] - exponentbase; // The exponent is offest so deduct the base.

        // Now Calculate the mantissa
        Double mantissa = 0.0;
        Double value = 1.0;
        // For Each Byte.
        for (int i = 5; i >= 1; i--)
        {
            int startbit = 7;
            if (i == 5)
            { startbit = 6; } //skip the sign bit.

            //For Each Bit
            for (int j = startbit; j >= 0; j--)
            {
                value = value / 2;// Each bit is worth half the next bit but we're going backwards.
                if (((real48[i] >> j) & 1) == 1) //if this bit is set.
                {
                    mantissa += value; // add the value.
                }

            }
        }

        if (mantissa == 1.0 && real48[0] == 0) // Test for null value
            return 0.0;

        if ((real48[5] & 0x80) == 1) // Sign bit check
            mantissa = -mantissa;

        return (1 + mantissa) * Math.Pow(2.0, exponent);