Integer with ASCII values to string

Question

I have an integer containing four ASCII codes for four chars:

0x31323334

I need to convert this integer to a string:

"1234" ~ "\x31\x32\x33\x34"

Is there better solution than this?

mystring = '%0*x' % (8, 0x31323334) # "31323334"
mystring.decode('hex') # "1234"

I think your way is OK. I would use instead str('0x31323334')[2:].decode('hex') as I think it's more readable, but the basic idea is the same. — Eduardo Ivanec
– Eduardo Ivanec, Commented Jan 2, 2012 at 21:27
@Eduardo Perhaps you mean hex(0x31323334)[2:].decode('hex')? — Dunes
– Dunes, Commented Jan 2, 2012 at 22:00
@Dunes: Almost - works for this case, but not in general. If the number starts with zeros then it either loses characters of fails with a TypeError. — Scott Griffiths
– Scott Griffiths, Commented Jan 2, 2012 at 23:21
@Scott Issue with zeroes is very important for me, I have to consider that. — Meloun
– Meloun, Commented Jan 3, 2012 at 8:41

Marco Mariani · Accepted Answer · 2012-01-02 21:23:33Z

1

Not sure it's better, but :)

>>> import struct
>>> struct.pack('>L', 0x31323334)
'1234'

edited Jan 2, 2012 at 21:23

answered Jan 2, 2012 at 21:16

Marco Mariani

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5 Comments

Scott Griffiths Over a year ago

In Python 2.6 / 2.7 I get '\x00\x00\x00\x001234'. Works for Python 2.5 though.

Marco Mariani Over a year ago

64 bit, I suppose... never used that package. can you try as I edited, with struct.pack('>L', 0x31323334)? it should be big-endian, 4 bytes wide.

Scott Griffiths Over a year ago

Yes it's a 32 vs. 64 bit thing, nothing to do with the Python version - my misunderstanding. It makes no difference with the 'L' format, on 64 bit struct.calcsize('L') is 8.

Dietrich Epp Over a year ago

Using '>I' will be more portable. Most 64-bit systems have 64-bit longs, the only exception is Windows.

Scott Griffiths Over a year ago

@DietrichEpp: I don't think there's a difference between '>L' and '>I' - according to the docs (and my tests) they are both the 'standard' 4 bytes long. It's only different if you use the native size, which might be 8 for a 'L' on non-Windows 64-bit Python.

Scott Griffiths · Accepted Answer · 2012-01-02 23:35:44Z

1

I don't think you'll get simpler than a format string and then a decode (needs Python 2.6+):

>>> "{0:08x}".format(0x31323334).decode('hex')
'1234'

answered Jan 2, 2012 at 23:35

Scott Griffiths

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