Windows WorkStation Remote BufferOverflow

Microsoft Windows WorkStation 服务(windows xp sp3)存在栈溢出漏洞。
a5这个参数,由于在执行wcscpy的字符串拷贝前，没有校验字符串的长度，因此会诱发栈缓冲区溢出（Stack Overflow） ，成功利用可以远程执行任意代码。

存在漏洞DLL 文件： wkssvc 或者 wkssvc.dll
DLL 名称： Network Workstation service library
描述：
wkssvc.dll是本地系统进行远程文件打印相关服务文件。

属于： Windows
系统 DLL文件： 是

分析如下（伪代码）：
/*
Found by Friddy 12.25
Email:qianyang@ssyeah.com
http://www.friddy.cn
*/
DWORD __userpurge sub_76854A96<eax>(int a1<eax>, HLOCAL *a2<esi>, int a3, wchar_t *a4,wchar_t *a5,int a6, int a7, int a8)
{
int v8; // eax@1
int v9; // ebx@1
HLOCAL v10; // eax@3
HLOCAL v11; // eax@4
HLOCAL v12; // eax@7
HLOCAL v13; // edi@7
int v15; // ecx@4
int v16; // edx@4
int v17; // eax@4
char v18; // zf@4
wchar_t *v19; // ST0C_4@5

v9 = a1;
v8 = 0;
if ( a4 )
v8 = *(_WORD *)(a7 + 2);
v10 = LocalAlloc(0x40u, v8 + ((2 * v9 + 39) & 0xFFFFFFFE));
*a2 = v10;
if ( v10 )
{
*(_DWORD *)v10 = 0;
v15 = a3;
v16 = a8;
*((_DWORD *)*a2 + 3) = v9;
*((_DWORD *)*a2 + 4) = 1;
*((_DWORD *)*a2 + 5) = v15;
v17 = dword_7686F588;
*((_DWORD *)*a2 + 6) = dword_7686F588;
v18 = a4 == 0;
*((_DWORD *)*a2 + 8) = v16;
dword_7686F588 = (v17 + 1) & 0x7FFFFFFF;
v11 = *a2;
if ( v18 )
{
*((_DWORD *)v11 + 2) = 0;
*((_DWORD *)*a2 + 7) = 0;
}
else
{
v19 = a4;
*((_DWORD *)v11 + 2) = (char *)v11 + 36;
wcscpy(*((wchar_t **)*a2 + 2), v19);
*((_DWORD *)*a2 + 7) = (unsigned int)(*a2 + 2 * v9 + 39) & 0xFFFFFFFE;
wcscpy(*((wchar_t **)*a2 + 7), *(const wchar_t **)(a7 + 4));
}
if ( !a5 )
return 0;
v12 = LocalAlloc(0x40u, 2 * a6 + 12);
v13 = v12;
if ( v12 )
{
wcscpy((wchar_t *)v12 + 4, a5);//栈溢出发生在这里
*((_DWORD *)v13 + 1) = a6;
*(_DWORD *)v13 = 1;
*((_DWORD *)*a2 + 1) = v13;
return 0;
}
LocalFree(*a2);
}
return GetLastError();
}

###############################################################################################################################################################################################################
//----- (7685499D) --------------------------------------------------------
signed int __stdcall sub_7685499D(int a1, int a2, wchar_t *a3, int a4, wchar_t *a5, int a6, int a7, int a8)
{
signed int v8; // edi@1
DWORD v9; // eax@2
wchar_t *v10; // ecx@7
int v12; // eax@21
int v13; // [sp+14h] [bp-4h]@1
int v14; // [sp+10h] [bp-8h]@1
int v15; // [sp+Ch] [bp-Ch]@2

v8 = 0;
v13 = 0;
v14 = 0;
if ( !(unsigned __int8)RtlAcquireResourceExclusive(&unk_7686F3E4, 1) )
{
v8 = 2140;
goto LABEL_18;
}
v9 = sub_76852B71((int)&dword_7686F3E0, a1, (int)&v15, 1);
if ( v9 )
goto LABEL_13;
if ( *(_DWORD *)(dword_7686F3E0 + 12 * v15) )
sub_76854B88(*(_DWORD *)(dword_7686F3E0 + 12 * v15), a5, (int)&v13, (int)&v14);
if ( v13 )
{
if ( !a3 && !*(_DWORD *)(v13 + 8) )
{
++*(_DWORD *)(v13 + 16);
++**(_DWORD **)(v13 + 4);
goto LABEL_17;
}
v9 = sub_76854A96(a4, (HLOCAL *)&a3, a2, a3, 0, 0, a7, a8);
if ( !v9 )
{
v12 = *(_DWORD *)(v13 + 4);
v10 = a3;
*((_DWORD *)a3 + 1) = *(_DWORD *)(v13 + 4);
++*(_DWORD *)v12;
goto LABEL_8;
}
LABEL_13:
v8 = v9;
LABEL_17:
RtlReleaseResource(&unk_7686F3E4);
LABEL_18:
NtClose(a2);
return v8;
}
v9 = sub_76854A96(a4, (HLOCAL *)&a3, a2, a3, a5, a6, a7, a8);//这里调用了漏洞，由此触发
if ( v9 )
goto LABEL_13;
v10 = a3;
LABEL_8:
if ( v14 )
*(_DWORD *)v14 = v10;
else
*(_DWORD *)(dword_7686F3E0 + 12 * v15) = v10;
RtlReleaseResource(&unk_7686F3E4);
return 0;
}

至于怎么用还得问下某人怎么构造 转某人的东西