I need to convert a two byte array to SFloat format according to IEEE-11073.
How can I do that?
I answer my question here.
public float ToSFloat(byte[] value)
{
if (value.Length != 2)
throw new ArgumentException();
byte b0 = value[0];
byte b1 = value[1];
var mantissa = unsignedToSigned(ToInt(b0) + ((ToInt(b1) & 0x0F) << 8), 12);
var exponent = unsignedToSigned(ToInt(b1) >> 4, 4);
return (float)(mantissa * Math.Pow(10, exponent));
}
public int ToInt(byte value)
{
return value & 0xFF;
}
private int unsignedToSigned(int unsigned, int size)
{
if ((unsigned & (1 << size-1)) != 0)
{
unsigned = -1 * ((1 << size-1) - (unsigned & ((1 << size-1) - 1)));
}
return unsigned;
}
Loosely based on the C implementation by Signove on GitHub I have created this function in C#:
Dictionary<Int32, Single> reservedValues = new Dictionary<Int32, Single> {
{ 0x07FE, Single.PositiveInfinity },
{ 0x07FF, Single.NaN },
{ 0x0800, Single.NaN },
{ 0x0801, Single.NaN },
{ 0x0802, Single.NegativeInfinity }
};
Single Ieee11073ToSingle(Byte[] bytes) {
var ieee11073 = (UInt16) (bytes[0] + 0x100*bytes[1]);
var mantissa = ieee11073 & 0x0FFF;
if (reservedValues.ContainsKey(mantissa))
return reservedValues[mantissa];
if (mantissa >= 0x0800)
mantissa = -(0x1000 - mantissa);
var exponent = ieee11073 >> 12;
if (exponent >= 0x08)
exponent = -(0x10 - exponent);
var magnitude = Math.Pow(10d, exponent);
return (Single) (mantissa*magnitude);
}
This function assumes that the bytes are in little endian format. If not you will have to swap bytes[0]
and bytes[1]
in the first line of the function. Or perhaps even better remove the first line from the function and change the function argument to accept a UInt16
(the IEEE 11073 value) and then let the caller decide how to extract this value from the input.
I highly advise you to test this code because I do not have any test values to verify the correctnes of the conversion.