Question

我需要使用python强制执行一些地址空间。目前这是我的代码：

offsets = [
"\x00","\x01","\x02","\x03","\x04","\x05","\x06","\x07","\x08","\x09","\x0a","\x0b","\x0c","\x0d","\x0e","\x0f"
,"\x10","\x11","\x12","\x13","\x14","\x15","\x16","\x17","\x18","\x19","\x1a","\x1b","\x1c","\x1d","\x1e","\x1f"
,"\x20","\x21","\x22","\x23","\x24","\x25","\x26","\x27","\x28","\x29","\x2a","\x2b","\x2c","\x2d","\x2e","\x2f"
,"\x30","\x31","\x32","\x33","\x34","\x35","\x36","\x37","\x38","\x39","\x3a","\x3b","\x3c","\x3d","\x3e","\x3f"
,"\x40","\x41","\x42","\x43","\x44","\x45","\x46","\x47","\x48","\x49","\x4a","\x4b","\x4c","\x4d","\x4e","\x4f"
,"\x50","\x51","\x52","\x53","\x54","\x55","\x56","\x57","\x58","\x59","\x5a","\x5b","\x5c","\x5d","\x5e","\x5f"
,"\x60","\x61","\x62","\x63","\x64","\x65","\x66","\x67","\x68","\x69","\x6a","\x6b","\x6c","\x6d","\x6e","\x6f"
,"\x70","\x71","\x72","\x73","\x74","\x75","\x76","\x77","\x78","\x79","\x7a","\x7b","\x7c","\x7d","\x7e","\x7f"
,"\x80","\x81","\x82","\x83","\x84","\x85","\x86","\x87","\x88","\x89","\x8a","\x8b","\x8c","\x8d","\x8e","\x8f"
,"\x90","\x91","\x92","\x93","\x94","\x95","\x96","\x97","\x98","\x99","\x9a","\x9b","\x9c","\x9d","\x9e","\x9f"
,"\xa0","\xa1","\xa2","\xa3","\xa4","\xa5","\xa6","\xa7","\xa8","\xa9","\xaa","\xab","\xac","\xad","\xae","\xaf"
,"\xb0","\xb1","\xb2","\xb3","\xb4","\xb5","\xb6","\xb7","\xb8","\xb9","\xba","\xbb","\xbc","\xbd","\xbe","\xbf"
,"\xc0","\xc1","\xc2","\xc3","\xc4","\xc5","\xc6","\xc7","\xc8","\xc9","\xca","\xcb","\xcc","\xcd","\xce","\xcf"
,"\xd0","\xd1","\xd2","\xd3","\xd4","\xd5","\xd6","\xd7","\xd8","\xd9","\xda","\xdb","\xdc","\xdd","\xde","\xdf"
,"\xe0","\xe1","\xe2","\xe3","\xe4","\xe5","\xe6","\xe7","\xe8","\xe9","\xea","\xeb","\xec","\xed","\xee","\xef"
,"\xf0","\xf1","\xf2","\xf3","\xf4","\xf5","\xf6","\xf7","\xf8","\xf9","\xfa","\xfb","\xfc","\xfd","\xfe","\xff"]


for i in xrange(110, 256):
    num = offsets[i]
    address = "\xee" + num + "\xff\xbf"
    print `address`

输出的最后一部分输出：

'\xee\xfa\xff\xbf'
'\xee\xfb\xff\xbf'
'\xee\xfc\xff\xbf'
'\xee\xfd\xff\xbf'
'\xee\xfe\xff\xbf'
'\xee\xff\xff\xbf'

我的问题是，是否有可能摆脱“偏移”数组，并以更清洁的方式进行？

Answer 1

for i in xrange(110, 256):
    address = "\xee" + "\\" + hex(i)[1:] + "\xff\xbf"
    print `address`

Answer 2

当然，只需迭代xrange并在当前项目上调用chr()：

>>> for i in xrange(110, 256):
...     print "\xee" + chr(i) + "\xff\xbf"

请注意，它会打印实际字符。如果你只想打印这些值，你可以使用反引号，但是更加pythonic的方法是使用repr()：

>>> for i in xrange(110, 256):
...    print(repr("\xee" + chr(i) + "\xff\xbf"))
...
    -- SNIP --
'\xee\xfa\xff\xbf'
'\xee\xfb\xff\xbf'
'\xee\xfc\xff\xbf'
'\xee\xfd\xff\xbf'
'\xee\xfe\xff\xbf'
'\xee\xff\xff\xbf'

Answer 3

您可以在单行中生成偏移数组：

offsets = ["0x{:02x}".format(_) for _ in range(0x100)]

Answer 4

你可以做数学运算，让struct构造4字节序列。

base = b'\xee\x00\xff\xbf' #assuming little-endian for the math coming up

struct.unpack('I',base)
Out[89]: (3221160174,)

#showing they're the same, no magic involved
int('ee',16) + int('ff',16)*(16**4) + int('bf',16)*(16**6)
Out[90]: 3221160174

#equivalent to bitshifting, if you prefer
int('ee',16) + (int('ff',16) << 16) + (int('bf',16) << 24)
Out[91]: 3221160174

这样就给了

start = 3221160174

[struct.pack('I', start + (x<<8)) for x in range(256)]
Out[93]: 
[b'\xee\x00\xff\xbf',
 b'\xee\x01\xff\xbf',
 b'\xee\x02\xff\xbf',
 b'\xee\x03\xff\xbf',
 b'\xee\x04\xff\xbf',
 b'\xee\x05\xff\xbf',
 b'\xee\x06\xff\xbf',
 #snip...
 b'\xee\xff\xff\xbf']

python中的内存地址生成器

4 个答案: