IDF-CTF-简单的Elf逆向Writeup

ElfCrackMe1

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简单的Elf逆向Writeup

题目来源：IDF实验室 CTF训练营；题目链接http://ctf.idf.cn/index.php?g=game&m=article&a=index&id=39
题目下载：http://pan.baidu.com/s/1kTl5wxD

解法1：

IDA查看伪代码法：

1. 下载文件，现在Linux环境下运行，可以看到关键字符串u r wrong和plz enter the flag:
   
2. 把文件拖到IDA中，shift+F12查找字符串，双击u r right跳转到相应位置
3. 如下图，双击调用关键字符串u r right的函数，
   跳转后F5查看伪代码如下

    int __cdecl main(int argc, const char **argv, const char **envp)
    {
      int v3; // ebx@6
      const char **v4; // rdx@22
      __int64 v6; // [sp+0h] [bp-C0h]@1
      __int64 v7; // [sp+8h] [bp-B8h]@1
      __int64 v8; // [sp+10h] [bp-B0h]@1
      __int64 v9; // [sp+18h] [bp-A8h]@1
      __int64 v10; // [sp+20h] [bp-A0h]@1
      __int64 v11; // [sp+28h] [bp-98h]@1
      __int64 v12; // [sp+30h] [bp-90h]@1
      __int64 v13; // [sp+38h] [bp-88h]@1
      int v14; // [sp+40h] [bp-80h]@1
      char v15[]; // [sp+80h] [bp-40h]@2
      char v16; // [sp+91h] [bp-2Fh]@14
      char v17; // [sp+92h] [bp-2Eh]@15
      char v18; // [sp+93h] [bp-2Dh]@16
      char v19; // [sp+94h] [bp-2Ch]@17
      char v20; // [sp+95h] [bp-2Bh]@18
      int v21; // [sp+A4h] [bp-1Ch]@1
      int v22; // [sp+A8h] [bp-18h]@7
      int i; // [sp+ACh] [bp-14h]@9
   
      v21 = ;
      memset(&v6, , 0x58uLL);
      v6 = 854698492143LL;
      v7 = 880468295913LL;
      v8 = 597000454391LL;
      v9 = 605590388953LL;
      v10 = 932007903423LL;
      v11 = 760209211613LL;
      v12 = 579820585151LL;
      v13 = 940597838039LL;
      v14 = ;
      printf(
        "plz enter the flag:",
        argv,
        11LL,
        854698492143LL,
        880468295913LL,
        597000454391LL,
        605590388953LL,
        932007903423LL,
        760209211613LL,
        579820585151LL,
        940597838039LL,
        *(_QWORD *)&v14);
      while (  )
      {
        v3 = v21;
        v15[v3] = getch();
        if ( !v15[v3] || v15[v21] ==  )
          break;
        if ( v15[v21] ==  )
        {
          printf("\b\b", v6, v7, v8, v9, v10, v11, v12, v13, *(_QWORD *)&v14);
          --v21;
        }
        else
        {
          putchar(v15[v21++]);
        }
      }
      v22 = ;
      if ( v21 !=  )
        v22 = ;
      for ( i = ; i <= ; ++i )
      {
        if ( v15[i] != (*((_DWORD *)&v6 + i) - ) /  )
        {
          v22 = ;
          argv = (const char **)((*((_DWORD *)&v6 + i) - ) / );
          printf("%d", argv, v6, v7, v8, v9, v10, v11, v12, v13, *(_QWORD *)&v14);
        }
      }
      if ( v16 !=  || v17 !=  || v18 !=  || v19 !=  || v20 !=  )
        v22 = ;
      v15[v21] = ;
      puts("\r");
      if ( v22 )
      {
        puts("u r wrong\r\n\r");
        main((unsigned __int64)"u r wrong\r\n\r", argv, v4);
      }
      else
      {
        puts("u r right!\r");
      }
      return ;
    }

分析这段伪代码，可以得到关键部分如下：

• v22需为0，按'/'随手注释
  
• 如上图中所示，要使v22==0,所有v22=1的语句均不能运行，则需要：v21==22，69行判断均不进入，既要 v15[i] != (*((_DWORD *)&v6 + i) - 1) / 2，同时，v16~v20依次等于48,56，50,51,125，即字符0823｝（在相应数字上按r键把相应的ASCII码转换为字符）
  同样按'/'键随手注释
  
• 找到定义v15的代码处，则根据上述的关键条件，初步猜测需要输入22位字符，其中前17位存入字符串v15中，后5位覆盖v16~v20的取值，使v16~v20分别等于0823｝
  
• 继续向下分析，找到输入的代码块，经过分析可得到关键信息如上图，其中v21即为输入的长度，通过关键条件v21==22验证了猜测输入字符串长度为22，同时需要保证输入的后五位为0823｝
  

• 向下分析v15需要满足的条件，着重分析v15[i] != (*((_DWORD *)&v6 + i) - 1) / 2，

  &v6为取v6的地址；(_DWORD *)&v6强制转化为_DWORD型指针，即两个字节；((_DWORD *)&v6 + i)为从&v6向后取sizeof(_DWORD)*i个字节；*((_DWORD *)&v6 + i) - 1)为取从&v6向后取4i个字节的值
  关于地址与指针加减问题http://www.cnblogs.com/WangAoBo/p/6365114.html

  

• 如上，分析v6的赋值段代码，v6~v13为_int64型，占8个字节，int型为4个字节，化为16进制如上图
  
• v6~v14通过小端存储方式在内存中的存储情况如上

则可写出python脚本解得flag： wctf{ElF_lnX_Ckm_0823}

python脚本如下：

 v6=[0x0EF, 0x0C7, 0x0E9, 0x0CD, 0x0F7, 0x8B, 0x0D9,
     0x8D, 0x0BF, 0x0D9, 0x0DD, 0x0B1, 0x0BF, 0x87,
     0x0D7, 0x0DB, 0x0BF]

 L=[]
 for i in range(17):
     num = (v6[i] - 1)/2
     ans = chr(int(num))
     L.append(ans)

 flag = ''.join(L)
 flag+='0823}'

 print(flag)