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)