I have an integer (67) that needs to be converted to hex and stored as a string like:
"\x00\x00\x00\x43"
How can I do this in Python?
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3have you tried anything?emecas– emecas2013-04-10 19:06:55 +00:00Commented Apr 10, 2013 at 19:06
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1Dang, my copy/paste buffer missed. I meant to mark this one: stackoverflow.com/questions/1708598/…000– 0002013-04-10 19:10:57 +00:00Commented Apr 10, 2013 at 19:10
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I tried hex(integer)[2:].zfill(8). But this only returns a hex without the "\x".user2267580– user22675802013-04-10 19:15:42 +00:00Commented Apr 10, 2013 at 19:15
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1 Answer
Updated due to ambiguity in OP.
Try...
def convert(i):
result = ''
for c in struct.pack('>i', 67):
c = hex(ord(c))[2:]
if len(c) < 2:
c = '0%s' % c
result += '\\x%s' % c
return result
>>> print convert(67)
\x00\x00\x00\x43
10 Comments
000
LOL I should start copying answers instead of marking as duplicates from now on. stackoverflow.com/questions/1708598/…
abarnert
I'm not sure why this was downvoted. Sure, it shouldn't be upvoted, because it was posted after the "possible duplicate" link to the exact same answer, and the question should be closed anyway… but it's not wrong, misleading, confusing, etc., so why downvote it?
user2267580
struct.pack('>i", 67) returns "C". I need it to return "\x00\x00\x00\x43"
Aya
Oops. I didn't notice you'd marked this as a dupe before I posted the answer. If you're going to mark as a dupe, should you not also close the question at the same time to avoid confusion?
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