Convert integer to hex-string with specific format

Question

I am new to python and have following problem: I need to convert an integer to a hex string with 6 bytes.

e.g. 281473900746245 --> \"\\xFF\\xFF\\xBF\\xDE\\x16\\x05\"

Answer 1

If you don't use Python 3.2 (I'm pretty sure you don't), consider the next approach:

>>> i = 281473900746245
>>> hex_repr = []
>>> while i:
...     hex_repr.append(struct.pack('B', i & 255))
...     i >>= 8
...
>>> ''.join(reversed(hex_repr))
'\xff\xff\xbf\xde\x16\x05'

Answer 2

In Python 3.2 or above, you can use the to_bytes() method of the interger.

>>> i = 281473900746245       
>>> i.to_bytes((i.bit_length() + 7) // 8, "big")
b'\xff\xff\xbf\xde\x16\x05'

Answer 3

There might be a better solution, but you can do this:

x = 281473900746245
decoded_x = hex(x)[2:].decode('hex') # value: '\xff\xff\xbf\xde\x16\x05'

Breakdown:

hex(x)                     # value: '0xffffbfde1605'
hex(x)[2:]                 # value: 'ffffbfde1605'
hex(x)[2:].decode('hex')   # value: '\xff\xff\xbf\xde\x16\x05'

Update:

Per @multipleinstances and @Sven's comments, since you might be dealing with long values, you might have to tweak the output of hex a little bit:

format(x, 'x')     # value: 'ffffbfde1605'

Sometimes, however, the output of hex might be an odd-length, which would break decode, so it'd probably be better to create a function to do this:

def convert(int_value):
   encoded = format(int_value, 'x')

   length = len(encoded)
   encoded = encoded.zfill(length+length%2)

   return encoded.decode('hex')