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I have the following number in C/C++, for example, 0x0202020202ULL and I'd like to print it out in binary form 1000000010000000100000001000000010.

Could you please help?

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2
  • I actually meant ULL. Commented Aug 12, 2013 at 1:39
  • I fail to see the relevance of C here. You can have 0x0202020202 in Python as well. Commented Aug 12, 2013 at 7:23

1 Answer 1

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You can use a combination of slicing(or str.rstrip), int and format.

>>> inp = '0x0202020202UL'
>>> format(int(inp[:-2], 16), 'b')
'1000000010000000100000001000000010'
# Using `str.rstrip`, This will work for any hex, not just UL
>>> format(int(inp.rstrip('UL'), 16), 'b')
'1000000010000000100000001000000010'

Update:

from itertools import islice
def formatted_bin(inp):
   output = format(int(inp.rstrip('UL'), 16), 'b')
   le = len(output)
   m = le % 4
   padd = 4 - m if m != 0 else 0
   output  = output.zfill(le + padd)
   it = iter(output)
   return ' '.join(''.join(islice(it, 4)) for _ in xrange((le+padd)/4))

print formatted_bin('0x0202020202UL')
print formatted_bin('0x10')
print formatted_bin('0x101010')
print formatted_bin('0xfff')

output:

0010 0000 0010 0000 0010 0000 0010 0000 0010
0001 0000
0001 0000 0001 0000 0001 0000
1111 1111 1111
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3 Comments

nice. just a little trick, how to separate it into 4 bits separated by a space, like 10 0000 0010 0000 0010 0000 0010 0000 0010?
still not correct. for example ·0x10· would produce wrong result
@Eugene Using the rstrip version I got '0001 0000' for 0x10. The slicing version is for unsigned long only, while rstrip version will work for any hex.

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