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Linear-time implementations of {encode,decode}_long.
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1 changed files with 50 additions and 19 deletions
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@ -1285,10 +1285,12 @@ class Unpickler:
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class _EmptyClass:
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pass
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# Encode/decode longs.
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# Encode/decode longs in linear time.
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import binascii as _binascii
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def encode_long(x):
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r"""Encode a long to a two's complement little-ending binary string.
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r"""Encode a long to a two's complement little-endian binary string.
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>>> encode_long(255L)
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'\xff\x00'
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>>> encode_long(32767L)
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@ -1303,14 +1305,46 @@ def encode_long(x):
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'\x7f'
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>>>
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"""
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# XXX This is still a quadratic algorithm.
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# Should use hex() to get started.
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digits = []
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while not -128 <= x < 128:
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digits.append(x & 0xff)
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x >>= 8
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digits.append(x & 0xff)
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return "".join(map(chr, digits))
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if x == 0:
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return '\x00'
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if x > 0:
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ashex = hex(x)
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assert ashex.startswith("0x")
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njunkchars = 2 + ashex.endswith('L')
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nibbles = len(ashex) - njunkchars
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if nibbles & 1:
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# need an even # of nibbles for unhexlify
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ashex = "0x0" + ashex[2:]
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elif ashex[2] >= '8':
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# "looks negative", so need a byte of sign bits
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ashex = "0x00" + ashex[2:]
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else:
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# Build the 256's-complement: (1L << nbytes) + x. The trick is
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# to find the number of bytes in linear time (although that should
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# really be a constant-time task).
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ashex = hex(-x)
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assert ashex.startswith("0x")
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njunkchars = 2 + ashex.endswith('L')
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nibbles = len(ashex) - njunkchars
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if nibbles & 1:
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# need an even # of nibbles for unhexlify
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nibbles += 1
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nbytes = nibbles >> 1
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x += 1L << (nbytes * 8)
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assert x > 0
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ashex = hex(x)
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if x >> (nbytes * 8 - 1) == 0:
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# "looks positive", so need a byte of sign bits
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ashex = "0xff" + x[2:]
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if ashex.endswith('L'):
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ashex = ashex[2:-1]
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else:
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ashex = ashex[2:]
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assert len(ashex) & 1 == 0
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binary = _binascii.unhexlify(ashex)
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return binary[::-1]
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def decode_long(data):
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r"""Decode a long from a two's complement little-endian binary string.
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@ -1327,15 +1361,12 @@ def decode_long(data):
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>>> decode_long("\x7f")
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127L
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"""
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# XXX This is quadratic too.
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x = 0L
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i = 0L
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for c in data:
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x |= long(ord(c)) << i
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i += 8L
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if data and ord(c) >= 0x80:
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x -= 1L << i
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return x
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ashex = _binascii.hexlify(data[::-1])
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n = long(ashex, 16)
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if data[-1] >= '\x80':
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n -= 1L << (len(data) * 8)
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return n
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# Shorthands
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