Get the “bits” of a float in Python?

Question

I am looking for the Python equivalent of Java\'s Float.floatToBits.

I found this Python: obtain & manipulate (as integers) bit patterns of floats

Answer 1

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 specifiers
• binascii.hexlify

---

_{[1] for example I needed this specifically for interoperability with .NET's BitConverter on intel (ie little endian)}

Answer 2

>>> import ctypes
>>> f = ctypes.c_float(173.3125)
>>> ctypes.c_int.from_address(ctypes.addressof(f)).value
1127043072

Answer 3

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