문제

I use python 2.6

>>> hex(-199703103)
'-0xbe73a3f'

>>> hex(199703103)
'0xbe73a3f'

Positive and negative value are the same?

When I use calc, the value is FFFFFFFFF418C5C1.

도움이 되었습니까?

해결책

Python's integers can grow arbitrarily large. In order to compute the raw two's-complement the way you want it, you would need to specify the desired bit width. Your example shows -199703103 in 64-bit two's complement, but it just as well could have been 32-bit or 128-bit, resulting in a different number of 0xf's at the start.

hex() doesn't do that. I suggest the following as an alternative:

def tohex(val, nbits):
  return hex((val + (1 << nbits)) % (1 << nbits))

print tohex(-199703103, 64)
print tohex(199703103, 64)

This prints out:

0xfffffffff418c5c1L
0xbe73a3fL

다른 팁

Because Python integers are arbitrarily large, you have to mask the values to limit conversion to the number of bits you want for your 2s complement representation.

>>> hex(-199703103 & (2**32-1)) # 32-bit
'0xf418c5c1L'
>>> hex(-199703103 & (2**64-1)) # 64-bit
'0xfffffffff418c5c1L'

Python displays the simple case of hex(-199703103) as a negative hex value (-0xbe73a3f) because the 2s complement representation would have an infinite number of Fs in front of it for an arbitrary precision number. The mask value (2**32-1 == 0xFFFFFFFF) limits this:

FFF...FFFFFFFFFFFFFFFFFFFFFFFFF418c5c1
&                             FFFFFFFF
--------------------------------------
                              F418c5c1

Adding to Marks answer, if you want a different output format, use

'{:X}'.format(-199703103 & (2**32-1))
>>> import struct
>>> struct.pack("i", -4)
'\xfc\xff\xff\xff'
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