Get The "bits" Of A Float In Python?

I am looking for the Python equivalent of Java's Float.floatToBits. I found this Python: obtain & manipulate (as integers) bit patterns of floats but does anyone know of a les

Solution 1:

The answer that Alex Martelli gives in that question is really pretty simple -- you can reduce it to:

>>>import struct>>>>>>>>>deffloatToBits(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:

>>>defbitsToFloat(b):...    s = struct.pack('>l', b)...return struct.unpack('>f', s)[0]>>>bitsToFloat(0x432d5000)
173.3125

Solution 2:

Here is the 64-bit, little endian representation of a python float 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

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

Solution 3:

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