How to convert from IEEE Python float to Microsoft Basic float

Question

I got a Python float value and I need to convert it to Microsoft Basic Float (MBF) format. Fortunately, there is code from the Internet that does the opposite.

def fmsbin2ieee(self,bytes):
    """Convert an array of 4 bytes containing Microsoft Binary floating point
    number to IEEE floating point format (which is used by Python)"""
    as_int = struct.unpack("i", bytes)
    if not as_int:
        return 0.0
    man = long(struct.unpack('H', bytes[2:])[0])
    exp = (man & 0xff00) - 0x0200
    if (exp & 0x8000 != man & 0x8000):
        return 1.0
        #raise ValueError('exponent overflow')
    man = man & 0x7f | (man << 8) & 0x8000
    man |= exp >> 1
    bytes2 = bytes[:2]
    bytes2 += chr(man & 255)
    bytes2 += chr((man >> 8) & 255)
    return struct.unpack("f", bytes2)[0]

Now I need to cancel this process, but so far it will not work. Any help please.

Answer 1

If you intend to perform these conversions while working under Windows, you can quickly download and install mbf2ieee.exe and call it CVS, which is suggested by the result Mbf2ieee.dll(for example, via [ctypes] [2]).

Python, ( , MBF ), ( Python C- ):

def mbf2ieee(mbf_4bytestring):
  msbin = struct.unpack('4B', mbf_4bytestring)
  if msbin[3] == 0: return 0.0

  ieee = [0] * 4
  sign = msbin[2] & 0x80
  ieee_exp = msbin[3] - 2
  ieee[3] = sign | (ieee_exp >> 1)
  ieee[2] = (ieee_exp << 7) | (msbin[2] & 0x7f)
  ieee[:2] = msbin[:2]

  return struct.unpack('f', ieee)[0]

, ?

: , :

def float2mbf4byte(f):
  ieee = struct.pack('f', f)
  msbin = [0] * 4
  sign = ieee[3] & 0x80

  msbin_exp = (ieee[3] << 1) | (ieee[2] >> 7)
  # how do you want to treat too-large exponents...?
  if msbin_exp == 0xfe: raise OverflowError
  msbin_exp += 2

  msbin[3] = msbin_exp
  msbin[2] = sign | (ieee[2] & 0x7f)
  msbin[:2] = ieee[:2]
  return msbin

Answer 2

, float2mbf4byte() 2 :

• ,
• Python 2, ieee int, string

:

def float2mbf4byte(f):
    ieee = [ord(s) for s in struct.pack('f', f)]
    msbin = [0] * 4

    sign = ieee[3] & 0x80
    ieee[3] &= 0x7f

    msbin_exp = (ieee[3] << 1) | (ieee[2] >> 7)
    # how do you want to treat too-large exponents...?
    if msbin_exp == 0xfe: raise OverflowError
    msbin_exp += 2

    msbin[3] = msbin_exp
    msbin[2] = sign | (ieee[2] & 0x7f)
    msbin[:2] = ieee[:2]
    return msbin

def ieee2fmsbin(f):
    return struct.pack('4B', *float2mbf4byte(f))