How to display a byte array as hex values

How to display a byte array as hex values

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>>> struct.pack('2I',12, 30) b'\x0c\x00\x00\x00\x1e\x00\x00\x00' >>> struct.pack('2I',12, 31) b'\x0c\x00\x00\x00\x1f\x00\x00\x00' >>> struct.pack('2I',12, 32) b'\x0c\x00\x00\x00 \x00\x00\x00' ^in question >>> struct.pack('2I',12, 33) b'\x0c\x00\x00\x00!\x00\x00\x00' ^in question I'd like all values to display as hex python share improve this question edited Dec 27 '13 at 19:27 asked  Dec 27 '13 at 19:23 Siavash 1,965 1 18 39

1   ' ' === '\x20', '!' === '\x21' –  Jan Dvorak  Dec 27 '13 at 19:25  I know but how can I get it display as a hex number, so everything is uniform? –  Siavash  Dec 27 '13 at 19:26 Do you mean, in the interactive console? I'd say it makes everything to not use hex codes, as strings are not supposed to hold unprintable characters, so you'd need to format them as hex codes manually; but then they'll show with double backslashes in the console. –  Jan Dvorak  Dec 27 '13 at 19:27  Im writing code to create a special packet, and for debug purposes I want to see where things are getting packed so I'm printing it out. I could use print from my code –  Siavash  Dec 27 '13 at 19:29 add a comment

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How about his? >>> data = struct.pack('2I',12, 30) >>> [hex(ord(c)) for c in data] ['0xc', '0x0', '0x0', '0x0', '0x1e', '0x0', '0x0', '0x0'] The expression [item for item in sequence] is a so called list comprehension. It's basically a very compact way of writing a simple for loop, and creating a list from the result. The ord() builtin function takes a string, and turns it into an integer that's its corresponding unicode code point (for characters in the ASCII character set that's the same as their value in the ASCII table). Its counterpart, chr() for 8bit strings or unichr() for unicode objects do the opposite. The hex() builtin then simply converts the integers into their hex representation. As pointed out by @TimPeters, in Python 3 you would need to lose the ord(), because iterating over a bytes object will (already) yield integers: Python 3.4.0a3 (default, Nov 8 2013, 18:33:56) >>> import struct >>> data = struct.pack('2I',12, 30) >>> type(data) <class 'bytes'> >>> type(data[1]) <class 'int'> >>> >>> [hex(i) for i in data] ['0xc', '0x0', '0x0', '0x0', '0x1e', '0x0', '0x0', '0x0'] share improve this answer edited Dec 27 '13 at 20:01 answered  Dec 27 '13 at 19:32 Lukas Graf 10.2k 2 24 46

3   Note: in Python3, you have to lose the ord() - iterating over a bytes object gives integers. –  Tim Peters Dec 27 '13 at 19:44 1   I like this answer, but could you explain how this is working please? what do the brackets do? –  Siavash Dec 27 '13 at 19:44 @TimPeters thank you, going to update the answer accordingly. –  Lukas Graf  Dec 27 '13 at 19:48 @Siavash sure! See my updated answer. –  Lukas Graf  Dec 27 '13 at 20:01 add a comment

up vote 5 down vote	You have to reformat it yourself if you want \x escapes everywhere; e.g., >>> import struct >>> r = struct.pack('2I',12, 33) >>> r b'\x0c\x00\x00\x00!\x00\x00\x00' >>> list(r) [12, 0, 0, 0, 33, 0, 0, 0] >>> print("".join("\\x%02x" % i for i in r)) \x0c\x00\x00\x00\x21\x00\x00\x00 share improve this answer answered  Dec 27 '13 at 19:30 Tim Peters 24.2k 4 39 55
	Given the desired usage, I guess joining just the hexes with spaces will do –  Jan Dvorak  Dec 27 '13 at 19:34 @JanDvorak, if that's what the OP wants, it's trivial to change this to do that. I'm just answering the question they asked ;-) –  Tim Peters  Dec 27 '13 at 19:35 add a comment

up vote 2 down vote

Try binascii.hexlify: >>> import binascii >>> import struct >>> binascii.hexlify(struct.pack('2I',12,30)) b'0c0000001e000000' >>> binascii.hexlify(struct.pack('2I',12,31)) b'0c0000001f000000' >>> binascii.hexlify(struct.pack('2I',12,32)) b'0c00000020000000' >>> binascii.hexlify(struct.pack('2I',12,33)) b'0c00000021000000' Or if you want spaces to make it more readable: >>> ' '.join(format(n,'02X') for n in struct.pack('2I',12,33)) '0C 00 00 00 21 00 00 00'