How to convert a float into hex

Question

In Python I need to convert a bunch of floats into hexadecimal. It needs to be zero padded (for instance, 0x00000010 instead of 0x10). Just like http://gregstoll.dyndns.org/~gregstoll/floattohex/ does. (sadly i can't use external libs on my platform so i can't use the one provided on that website)

What is the most efficient way of doing this?

Answer 1

In Python float is always double-precision.

If you require your answer to be output in the form of a hexadecimal integer, the question was already answered:

import struct

# define double_to_hex as in the other answer

double_to_hex(17.5)   # Output: '0x4031800000000000'
double_to_hex(-17.5)  # Output: '0xc031800000000000'

However you might instead consider using the builtin function:

(17.5).hex()    # Output: '0x1.1800000000000p+4'
(-17.5).hex()   # Output: '-0x1.1800000000000p+4'

# 0x1.18p+4 == (1 + 1./0x10 + 8./0x100) * 2**4 == 1.09375 * 16 == 17.5

This is the same answer as before, just in a more structured and human-readable format.

The lower 52 bits are the mantissa. The upper 12 bits consists of a sign bit and an 11-bit exponent; the exponent bias is 1023 == 0x3FF, so 0x403 means '4'. See Wikipedia article on IEEE floating point.

Answer 2

if you are on micropython (which is not said in the question, but I had trouble finding) you can use this

import binascii
def float_to_hex(f):
    binascii.hexlify(struct.pack('<f', f))

float_to_hex(17.5) #Answer: 0x418c0000

Answer 3

This is a bit tricky in python, because aren't looking to convert the floating-point value to a (hex) integer. Instead, you're trying to interpret the IEEE 754 binary representation of the floating-point value as hex.

We'll use the pack and unpack functions from the built-in struct library.

A float is 32-bits. We'll first pack it into a binary¹ string, and then unpack it as an int.

def float_to_hex(f):
    return hex(struct.unpack('<I', struct.pack('<f', f))[0])

float_to_hex(17.5)    # Output: '0x418c0000'

We can do the same for double, knowing that it is 64 bits:

def double_to_hex(f):
    return hex(struct.unpack('<Q', struct.pack('<d', f))[0])

double_to_hex(17.5)   # Output: '0x4031800000000000L'

---

^{1 - Meaning a string of raw bytes; not a string of ones and zeroes.}

Answer 4

Further to Jonathon Reinhart's very helpful answer. I needed this to send a floating point number as bytes over UDP

import struct

# define double_to_hex (or float_to_hex)
def double_to_hex(f):
    return hex(struct.unpack('<Q', struct.pack('<d', f))[0])

# On the UDP transmission side
doubleAsHex = double_to_hex(17.5)
doubleAsBytes = bytearray.fromhex(doubleAsHex.lstrip('0x').rstrip('L'))

# On the UDP receiving side
doubleFromBytes = struct.unpack('>d', doubleAsBytes)[0] # or '>f' for float_to_hex