Represent number as a bytes using 16-bit blocks

Question

I wish to convert a number like 683550 (0xA6E1E) to b'\x1e\x6e\x0a\x00', where the number of bytes in the array is a multiple of 2 and where the len of the bytes object is only so long as it needs to be to represent the number.

This is as far as I got:

"{0:0{1}x}".format(683550,8)

giving:

'000a6e1e'

What's up with the crazy byte order? That's not big-endian or little-endian. — user2357112
– user2357112, Commented Aug 24, 2016 at 21:01
@user2357112 I'm sorry. The reason it was messed up was because I was using "hexdump -x" to view the bytes in the file. Without the -x it makes more sense. — Baz
– Baz, Commented Aug 25, 2016 at 18:41

Daniel · Accepted Answer · 2016-08-25 19:05:04Z

2

Use the .tobytes-method:

num = 683550
bytes = num.to_bytes((num.bit_length()+15)//16*2, "little")

edited Aug 25, 2016 at 19:05

answered Aug 24, 2016 at 20:56

Daniel

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4 Comments

Baz Over a year ago

Outputting [hex(i) for i in t] for your answer gives ['0x1e', '0x6e', '0xa', '0x0']. I'm looking for ['0x6e', '0x1e', '0x0','0xa']

Daniel Over a year ago

so you really want this mixed little-big-endian stuff. So see my updated answer.

Baz Over a year ago

I'm sorry @Daniel, your original answer was indeed correct. I was using "hexdump -x" to view the file which was switching things around. I've edited my question with the correct bytes.

Daniel Over a year ago

so back again, again.

Eduardo Cuesta · Accepted Answer · 2016-08-24 23:09:16Z

0

Using python3:

def encode_to_my_hex_format(num, bytes_group_len=2, byteorder='little'):
  """
  @param byteorder can take the values 'little' or 'big'
  """
  bytes_needed = abs(-len(bin(num)[2: ]) // 8)

  if bytes_needed % bytes_group_len:
    bytes_needed += bytes_group_len - bytes_needed % bytes_group_len

  num_in_bytes = num.to_bytes(bytes_needed, byteorder)
  encoded_num_in_bytes = b''

  for index in range(0, len(num_in_bytes), bytes_group_len):
    bytes_group = num_in_bytes[index: index + bytes_group_len]

    if byteorder == 'little':
      bytes_group = bytes_group[-1: -len(bytes_group) -1 : -1]

    encoded_num_in_bytes += bytes_group

  encoded_num = ''

  for byte in encoded_num_in_bytes:
    encoded_num += r'\x' + hex(byte)[2: ].zfill(2)

  return encoded_num

print(encode_to_my_hex_format(683550))

answered Aug 24, 2016 at 23:09

Eduardo Cuesta

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2 Comments

Baz Over a year ago

You need to output bytes not a string.

Eduardo Cuesta Over a year ago

Then just return encoded_num_in_bytes instead of encoded_num in the function.

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