54 lines
1.5 KiB
Python
54 lines
1.5 KiB
Python
|
import struct
|
||
|
|
||
|
# Stolen from PyCypto (Public Domain)
|
||
|
|
||
|
|
||
|
def b(s):
|
||
|
return s.encode("latin-1") # utf-8 would cause some side-effects we don't want
|
||
|
|
||
|
|
||
|
def long_to_bytes(n, blocksize=0):
|
||
|
"""long_to_bytes(n:long, blocksize:int) : string
|
||
|
Convert a long integer to a byte string.
|
||
|
If optional blocksize is given and greater than zero, pad the front of the
|
||
|
byte string with binary zeros so that the length is a multiple of
|
||
|
blocksize.
|
||
|
"""
|
||
|
# after much testing, this algorithm was deemed to be the fastest
|
||
|
s = b('')
|
||
|
pack = struct.pack
|
||
|
while n > 0:
|
||
|
s = pack('>I', n & 0xffffffff) + s
|
||
|
n = n >> 32
|
||
|
# strip off leading zeros
|
||
|
for i in range(len(s)):
|
||
|
if s[i] != b('\000')[0]:
|
||
|
break
|
||
|
else:
|
||
|
# only happens when n == 0
|
||
|
s = b('\000')
|
||
|
i = 0
|
||
|
s = s[i:]
|
||
|
# add back some pad bytes. this could be done more efficiently w.r.t. the
|
||
|
# de-padding being done above, but sigh...
|
||
|
if blocksize > 0 and len(s) % blocksize:
|
||
|
s = (blocksize - len(s) % blocksize) * b('\000') + s
|
||
|
return s
|
||
|
|
||
|
|
||
|
def bytes_to_long(s):
|
||
|
"""bytes_to_long(string) : long
|
||
|
Convert a byte string to a long integer.
|
||
|
This is (essentially) the inverse of long_to_bytes().
|
||
|
"""
|
||
|
acc = 0
|
||
|
unpack = struct.unpack
|
||
|
length = len(s)
|
||
|
if length % 4:
|
||
|
extra = (4 - length % 4)
|
||
|
s = b('\000') * extra + s
|
||
|
length = length + extra
|
||
|
for i in range(0, length, 4):
|
||
|
acc = (acc << 32) + unpack('>I', s[i:i+4])[0]
|
||
|
return acc
|