Convert Bytes to Floating Point Numbers WITHOUT using STRUCT

Question

I'm trying to write my "personal"(without using any modules or functions : struct,float....,int....,...) python version of STL binary file reader, according to WIKIPEDIA : A binary STL file contains :

a 80-character (byte) headern which is generally ignored.
a 4-byte unsigned integer indicating the number of triangular facets in the file.
Each triangle is described by twelve 32-bit floating-point numbers: three for the normal and then three for the X/Y/Z coordinate of each vertex – just as with the ASCII version of STL. After these follows a 2-byte ("short") unsigned integer that is the "attribute byte count" – in the standard format, this should be zero because most software does not understand anything else. (((3+3+3)+3)*4+2=50 bytes for each point)

--Floating-point numbers are represented as IEEE floating-point numbers and are assumed to be little-endian--

With the help of two saviors I discovered how unsigned integers are stored, I can figure out the number of triangular facets in the file with 3 methods (computed by hand ) :

def int_from_bytes(inbytes): # ShadowRanger's
     res = 0
     for i, b in enumerate(inbytes):
         res |= b << (i * 8)  
     return res

or

def int_from_bytes(inbytes): # ShadowRanger's
    res = 0
    for b in inbytes:
        res <<= 8  # Adjust bytes seen so far to make room for new byte
        res |= b   # Mask in new byte
    return res

or

def unsigned_int(s): # Robᵩ's
    result = 0
    for ch in s[::-1]:
        result *= 256
        result += ch
    return result

Now I have to convert the rest of the file (3rd item in the list):Floating-point numbers

for the first point the 50-bytes are :

b'\x9a'b'\xa3' b'\x14' b'\xbe' b'\x05' b'$' b'\x85' b'\xbe' b'N' b'b'  
b't' b'?' b'\xcd' b'\xa6' b'\x04' b'\xc4' b'\xfb' b';' b'\xd4' b'\xc1' 
b'\x84' b'w' b'\x81' b'A' b'\xcd' b'\xa6' b'\x04' b'\xc4' b'\xa5' b'\x15' 
b'\xd3' b'\xc1' b'\xb2' b'\xc7' b'\x81' b'A' b'\xef' b'\xa6' b'\x04' 
b'\xc4' b'\x81' b'\x14' b'\xd3' b'\xc1' b'Y' b'\xc7' b'\x81' b'A' b'\x00' 
b'\x00'

How can I convert this by hand, What is the principle of the representation, what rules should I know to do the conversion by hand (some bytes don't start with \x ). ?

Thank you for your time.

Is there an actual purpose to this or are you just trying to see how far you can get? — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Dec 9, 2016 at 18:45
Also why are you trying to avoid using the struct module? — martineau
– martineau, Commented Dec 9, 2016 at 18:45
Some of the bytes don't start with \x because their value happens to be the character code of an ascii character, so they're simply displayed as a (b-prefixed) quoted version character itself. — martineau
– martineau, Commented Dec 9, 2016 at 18:55
struct is part of the set of modules that comes with Python so avoiding it is silly. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Dec 9, 2016 at 19:07

Matt Timmermans · Accepted Answer · 2016-12-09 19:27:40Z

2

Like this:

def int_from_bytes(inbytes):
    res = 0
    shft = 0
    for b in inbytes:
        res |= ord(b) << shft
        shft += 8
    return res

def float_from_bytes(inbytes):
    bits = int_from_bytes(inbytes)
    mantissa = ((bits&8388607)/8388608.0)
    exponent = (bits>>23)&255
    sign = 1.0 if bits>>31 ==0 else -1.0
    if exponent != 0:
        mantissa+=1.0
    elif mantissa==0.0:
        return sign*0.0
    return sign*pow(2.0,exponent-127)*mantissa


print float_from_bytes('\x9a\xa3\x14\xbe')
print float_from_bytes('\x00\x00\x00\x40')
print float_from_bytes('\x00\x00\xC0\xbf')

output:

-0.145155340433
2.0
-1.5

The format is IEEE-754 floating point. Try this out to see what each bit means: https://www.h-schmidt.net/FloatConverter/IEEE754.html

edited Dec 9, 2016 at 19:27

answered Dec 9, 2016 at 19:17

Matt Timmermans

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3 Comments

Dietrich Epp Over a year ago

One should note that this doesn't handle denormals or non-finite numbers, but these are all somewhat unusual cases so you might be fine without such support.

Matt Timmermans Over a year ago

@DietrichEpp oh yeah -- fixed to handle denormals.

Mark Ransom Over a year ago

I would use hex for the magic constants instead of decimal. But congratulations for giving a good answer to a bad question.

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