Python binary float to integer conversion using ctypes

Question

Please help me to understand this code snippet:

def binary_float_to_int(float_number: float) -> int:

    return ctypes.c_uint.from_buffer(ctypes.c_float(float_number)).value

The results from these inputs:

print(binary_float_to_int(7.1746481e-43)) 
print(binary_float_to_int(5.3809861e-43))

Are: 512 & 384

Why does the simple Python conversion int(7.1746481e-43) not work? Are there any other ways to do this type of conversion?

int(7.1746481e-43) works fine so far as I can see. int() rounds towards zero, and 7.1746481e-43 is less than 1. I'm not sure what binary_float_to_int() is doing, but 7.1746481e-43 isn't anywhere close to 512. — brunns
– brunns, Commented Jul 30, 2020 at 12:40
@brunns the conversion is the 32-bit integer value of the 32-bit IEEE 754 floating point value representing that floating point number. — Mark Tolonen
– Mark Tolonen, Commented Jul 31, 2020 at 5:32

Mark Tolonen · Accepted Answer · 2020-07-31 05:32:50Z

2

The ctypes code is:

Put the floating point number in a 32-bit C (IEEE 754-format) ctypes.c_float(float_number)
Treat that same 4-byte value, as an C unsigned int. ctypes.c_uint.from_buffer()
Extract that unsigned integer value .value

Your numbers are correct, if you want the raw 32-bit value of those floating point numbers expressed as integers. Here's another way to do it:

>>> import struct
>>> struct.unpack('i',struct.pack('f',7.1746481e-43))[0]
512
>>> struct.unpack('i',struct.pack('f',5.3809861e-43))[0]
384

These generate the 4-byte float32 value, then unpack it as an integer.

7.1746481e-43 is a very small value close to zero. int() returns the integer portion...in this case, zero, so that's as expected as well.

answered Jul 31, 2020 at 5:14

Mark Tolonen

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