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：

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:
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

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)