有没有办法填充偶数个数字？

Question

I'm trying to create a hex representation of some data that needs to be transmitted (specifically, in ASN.1 notation). At some points, I need to convert data to its hex representation. Since the data is transmitted as a byte sequence, the hex representation has to be padded with a 0 if the length is odd.

Example:

>>> hex2(3)
'03'
>>> hex2(45)
'2d'
>>> hex2(678)
'02a6'

The goal is to find a simple, elegant implementation for hex2 .

Currently I'm using hex , stripping out the first two characters, then padding the string with a 0 if its length is odd. However, I'd like to find a better solution for future reference. I've looked in str.format without finding anything that pads to a multiple.

Answer 1

def hex2(n):
  x = '%x' % (n,)
  return ('0' * (len(x) % 2)) + x

Answer 2

To be totally honest, I am not sure what the issue is. A straightforward implementation of what you describe goes like this:

def hex2(v):
  s = hex(v)[2:]
  return s if len(s) % 2 == 0 else '0' + s

I would not necessarily call this "elegant" but I would certainly call it "simple."

Answer 3

Python's binascii module's b2a_hex is guaranteed to return an even-length string.

the trick then is to convert the integer into a bytestring. Python3.2 and higher has that built-in to int:

from binascii import b2a_hex

def hex2(integer):
    return b2a_hex(integer.to_bytes((integer.bit_length() + 7) // 8, 'big'))

Answer 4

Might want to look at the struct module, which is designed for byte-oriented i/o.

import struct
>>> struct.pack('>i',678)
'\x00\x00\x02\xa6'
#Use h instead of i for shorts
>>> struct.pack('>h',1043)
'\x04\x13'