memory address generator in python

Question

I need to bruteforce some of the address space using python. At the moment this is my code:

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`

And the output last part of the output:

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

My question is if it is possible to get rid of the "offsets" array, and do it in a much cleaner way?

msvalkon · Accepted Answer · 2014-04-10 12:30:18Z

1

Sure is, just iterate over the xrange and call chr() on the current item:

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

Notice that it'll print the actual characters. If you just want to print the values, you can use the backticks, however a more pythonic approach is to use 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'

edited Apr 10, 2014 at 12:30

answered Apr 10, 2014 at 12:23

msvalkon

12.1k2 gold badges46 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sufian Latif · Accepted Answer · 2014-04-10 12:23:17Z

1

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

answered Apr 10, 2014 at 12:23

Sufian Latif

13.4k3 gold badges36 silver badges71 bronze badges

Comments

adrian.nicolau · Accepted Answer · 2014-04-10 12:26:54Z

1

You can generate the offsets array in a one-liner:

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

answered Apr 10, 2014 at 12:26

adrian.nicolau

5357 silver badges18 bronze badges

Comments

roippi · Accepted Answer · 2014-04-10 13:41:34Z

1

You could just do the math, and let struct construct the 4-byte sequence.

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

So that gives

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']

edited Apr 10, 2014 at 13:41

answered Apr 10, 2014 at 13:08

roippi

26k4 gold badges52 silver badges75 bronze badges

Collectives™ on Stack Overflow

memory address generator in python

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related