float to binary <-> binary to float conversion

Question

I want to convert a float number to a binary string and back.

I tried this:

import struct
from ast import literal_eval


float_to_binary = bin(struct.unpack('!i',struct.pack('!f', 3.14))[0])
print (float_to_binary)

binary_to_float = float(int(float_to_binary, 0))
print (binary_to_float)


result = float(literal_eval(float_to_binary))
print (result) #wrong, prints 1078523331.0, should be 3.14

user1129665 · Accepted Answer · 2014-02-06 16:30:25Z

2

This doesn't do what you think:

binary_to_float = float(int(float_to_binary, 0))

it doesn't reinterpret the data, rather it converts integer to float, e.g. 1234 to 1234.0.

you can use:

>>> from struct import *
>>>
>>> # float -> binary
>>> bin( unpack('I', pack('f', 3.14))[0] )
'0b1000000010010001111010111000011'
>>>
>>>
>>> # binary -> float
>>> unpack('f', pack('I', 0b1000000010010001111010111000011) )[0]
(3.140000104904175,)
>>>

answered Feb 6, 2014 at 16:30

user1129665

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3 Comments

andandandand Over a year ago

cool, though when I use variables, like this: original_float = 3.14 binary_string = int(bin( unpack('I', pack('f', original_float))[0] )) float_value = unpack('f', pack('I', binary_string) )[0] doesn't work. ValueError: invalid literal for int() with base 10: '0b1000000010010001111010111000011'

user1129665 Over a year ago

@omgzor You need to tell int that the integer in the string is integer of base 2 using int(..., 2), so it should be binary_string = int(bin( unpack('I', pack('f', original_float))[0] ), 2). You can remove int(bin(...), 2) if you are just interested in the base 10 value, binary_string = unpack('I', pack('f', original_float))[0].

Scott Stafford Over a year ago

I don't think this works, not exactly. int is 32 bit and float is 64 bit.

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