Convert a Python int into a big-endian string of bytes

user_104474 0
Viewed 143807

I have a non-negative int and I would like to efficiently convert it to a big-endian string containing the same data. For example, the int 1245427 (which is 0x1300F3) should result in a string of length 3 containing three characters whose byte values are 0x13, 0x00, and 0xf3.

My ints are on the scale of 35 (base-10) digits.

How do I do this?

python
9 Answers

You can use the struct module:

import struct
print(struct.pack('>I', your_int))

'>I' is a format string. > means big endian and I means unsigned int. Check the documentation for more format chars.

user_400050

This is fast and works for small and (arbitrary) large ints:

def Dump(n): 
  s = '%x' % n
  if len(s) & 1:
    s = '0' + s
  return s.decode('hex')
print repr(Dump(1245427))  #: '\x13\x00\xf3'
user_972480

Probably the best way is via the built-in struct module:

>>> import struct
>>> x = 1245427
>>> struct.pack('>BH', x >> 16, x & 0xFFFF)
'\x13\x00\xf3'
>>> struct.pack('>L', x)[1:]  # could do it this way too
'\x13\x00\xf3'

Alternatively -- and I wouldn't usually recommend this, because it's mistake-prone -- you can do it "manually" by shifting and the chr() function:

>>> x = 1245427
>>> chr((x >> 16) & 0xFF) + chr((x >> 8) & 0xFF) + chr(x & 0xFF)
'\x13\x00\xf3'

Out of curiosity, why do you only want three bytes? Usually you'd pack such an integer into a full 32 bits (a C unsigned long), and use struct.pack('>L', 1245427) but skip the [1:] step?

user_687070 
def tost(i):
  result = []
  while i:
    result.append(chr(i&0xFF))
    i >>= 8
  result.reverse()
  return ''.join(result)
user_958100

Very easy with pwntools , the tools created for software hacking

(Un-ironically, I stumbled across this thread and tried solutions here, until I realised there exists conversion functionality in pwntools)

import pwntools

x2 = p32(x1)
user_122583120
Related