Parsing a hex formated DEC 32 bit single precision floating point value in python

2
votes

I'm having problems parsing a hex formatted DEC 32bit single precision floating point value in python, the value I'm parsing is represented as D44393DB in hex. The original floating point value is ~108, read from a display of the sending unit.

The format is specified as: 1bit sign + 8bit exponent + 23bit mantissa. Byte 2 contains the sign bit + the 7 most significant bits of the exponent Byte 1 contains the least significant bit of the exponent + the starting most significant bits of the mantissa.

The only thing I have found that differs in the two formats is the bias of the exponent which is 128 in DEC32 and 127 in IEEE-754 (http://www.irig106.org/docs/106-07/appendixO.pdf)

Using http://babbage.cs.qc.edu/IEEE-754/32bit.html does not give the expected result.

/Kristofer

4
pythonfloating-point
I believe the mantissa is also represented differently (the IEEE format has an implicit 0.5 added). I have an LSI-11/02 handbook at home - I will post an answer if I find anything tonight. – MikeJ-UK
The sign bit must be in an end byte, byte 2 can't be on the end no matter whether you count 1023 or 1234. Which byte is transmitted first, D4 or DB? Better still, tell us the order of transmission of each byte. Do you really have only one known value? – John Machin
The received order is B1B2B3B4 or in actual hex: D44393DB The order to be parsed is B2B1B4B3, which in this case is: 43D4DB93, the sign bit is the MSB of B2. – Kristofer

4 Answers

4
votes

Is it possible that the bytes got shuffled somehow? The arrangements of bits that you describe (sign bit in byte 2, LSB of exponent in byte 1) is different from Appendix O that you link to. It looks like byte 1 and 2 were exchanged.

I'll assume that byte 3 and 4 were also exchanged, so that the real hex value is 43D4DB93. This translates as 0100 0011 1101 0100 1101 1011 1001 0011 in binary, so the sign bit is 0, indicating a positive number. The exponent is 10000111 (binary) = 135 (decimal), indicating a factor of 2^(135-128) = 128. Finally, the mantissa is 0.1101 0100 1101 1011 1001 0011 (binary), using that Appendix O says that you have to add 0.1 in front, which is approximately 0.8314 in decimal. So your number is 0.8314 * 128 = 106.4 under my assumptions.

Added: Some Python 2 code might clarify:

input = 0xD44393DB;
reshuffled = ((input & 0xFF00FF00) >> 8) | ((input & 0x00FF00FF) << 8);
signbit = (reshuffled & 0x80000000) >> 31;
exponent = ((reshuffled & 0x7F800000) >> 23) - 128;
mantissa = float((reshuffled & 0x007FFFFF) | 0x00800000) / 2**24;
result = (-1)**signbit * mantissa * 2**exponent;

This yields result = 106.42885589599609.

Here is an explanation for the line computing the mantissa. Firstly, reshuffled & 0x007FFFFF yield the 23 bits encoding the mantissa: 101 0100 1101 1011 1001 0011. Then ... | 0x00800000 sets the hidden bit, yielding 1101 0100 1101 1011 1001 0011. We now have to compute the fraction 0.1101 0100 1101 1011 1001 0011. By definition, this equals 1*2^(-1) + 1*2^(-2) + 0*2^(-3) + ... + 1*2^(-23) + 1*2^(-24). This can also be written as (1*2^23 + 1*2^22 + 0*2^21 + ... + 1*2^1 + 1*2^0) / 2^24. The expression in brackets is the value of 1101 0100 1101 1011 1001 0011 (binary), so we can find the mantissa by dividing (reshuffled & 0x007FFFFF) | 0x00800000 by 2^24.

1
votes

From my copy of "Microcomputers and Memories" (DEC, 1981), you are correct about the difference between the two formats. The DEC mantissa is normalized to 0.5<=f<1 and the IEEE format mantissa is normalized to 1<=f<2, both with the MSB implicit and not stored. Thus the mantissa bit-layouts are the same. Jitse Niesens assumptions look like a plausible explanation since the value of D44393DB would be -0.7639748 X 2^40 (which is -8.3999923E11).

1
votes

Found under "Related" on the RHS: these answers from last month

One of the references helps understanding the "wired" (weird?) byte2 byte1 notation.

0
votes

Is it definitely a DEC32 value? The sign bit seems to be 1, which indicates negative by this format. However, you do get a result very close to your 108 value if you ignore this and assume that the exponent bias is 15, retaining the 0.1 factor on the mantissa:

def decode(x):
    exp = (x>>30) & 0xff
    mantissa = x&((2**24)-1)
    return 0.1 * mantissa * (2**(exp-15))

>>> decode(0xD44393DB)
108.12409668