Get the "bits" of a float in Python?
Here is the 64-bit, little endian representation of a python float1 just to add to the discussion:
>>> import struct
>>> import binascii
>>> print('0x' + binascii.hexlify(struct.pack('<d', 123.456789)))
0x0b0bee073cdd5e40
References:
struct.pack
endianness and byte size format specifiersbinascii.hexlify
[1] for example I needed this specifically for interoperability with .NET's BitConverter
on intel (ie little endian)
The answer that Alex Martelli gives in that question is really pretty simple -- you can reduce it to:
>>> import struct
>>>
>>>
>>> def floatToBits(f):
... s = struct.pack('>f', f)
... return struct.unpack('>l', s)[0]
...
...
>>> floatToBits(173.3125)
1127043072
>>> hex(_)
'0x432d5000'
Once you have it as an integer, you can perform any other manipulations you need to.
You can reverse the order of operations to round-trip:
>>> def bitsToFloat(b):
... s = struct.pack('>l', b)
... return struct.unpack('>f', s)[0]
>>> bitsToFloat(0x432d5000)
173.3125
>>> import ctypes
>>> f = ctypes.c_float(173.3125)
>>> ctypes.c_int.from_address(ctypes.addressof(f)).value
1127043072