我正在尝试自学 C++ 中的缓冲区溢出和利用。我充其量是一个中级 C++ 人,所以请耐心等待。我遵循了一些教程,但这里有一些示例代码来说明我的问题:

#include <string>
#include <iostream>

using namespace std;

int main()
{
  begin:
  int authentication = 0;
  char cUsername[10], cPassword[10];
  char cUser[10], cPass[10];

  cout << "Username: ";
  cin >> cUser;

  cout << "Pass: ";
  cin >> cPass;

  strcpy(cUsername, cUser);
  strcpy(cPassword, cPass);

  if(strcmp(cUsername, "admin") == 0 && strcmp(cPassword, "adminpass") == 0)
  {
    authentication = 1;
  }
  if(authentication)
  {
    cout << "Access granted\n";
    cout << (char)authentication;
  }
  else
  {
    cout << "Wrong username and password\n";
  }

  system("pause");
  goto begin;
}

我知道这里有各种各样的坏 juju 和 cin << String 等......无论如何,当我在 AcUser 中输入太多字母(例如大量 cPass )时,我只是从 Visual Studio 收到访问冲突。但是,如果我输入 20ish A 's,然后是一个空格,然后是另一个 AcUser 中,它会跳过向我询问 cPass (假设因为它是在空格字符导致先前对 cin 的调用返回之后被填充)并授予我使用权。

在什么时候以及为什么数据溢出到“身份验证”中,为什么它只在我有空间时发生,而不是在我有一百万个 A 时发生......当我使用时,我从来没有遇到过“访问冲突” cUser 输入中的空格。

最佳答案

我稍微修改了您的程序以使其更具说明性:

#include <iostream>

int main( void )
{
 int authentication = 0;
 char cUsername[ 10 ];
 char cPassword[ 10 ];

 std::cout << "Username: ";
 std::cin >> cUsername;

 std::cout << "Pass: ";
 std::cin >> cPassword;

 if( std::strcmp( cUsername, "admin" ) == 0 && std::strcmp( cPassword, "adminpass" ) == 0 )
 {
  authentication = 1;
 }
 if( authentication )
 {
  std::cout << "Access granted\n";
  std::cout << ( char )authentication;
 }
 else
 {
  std::cout << "Wrong username and password\n";
 }

 return ( 0 );
}

我用 x64 编译器命令行 MS 编译器编译它,没有优化。所以现在我们有一个我们想要“破解”的 exe。我们使用 WinDbg(非常好的调试器)加载程序并查看反汇编(注意,为了清晰起见,我提供了完整的调试信息):
00000001`3f1f1710 4883ec68        sub     rsp,68h
00000001`3f1f1714 488b0515db0300  mov     rax,qword ptr [Prototype_Console!__security_cookie (00000001`3f22f230)]
00000001`3f1f171b 4833c4          xor     rax,rsp
00000001`3f1f171e 4889442450      mov     qword ptr [rsp+50h],rax
00000001`3f1f1723 c744243800000000 mov     dword ptr [rsp+38h],0  // This gives us address of "authentication" on stack.
00000001`3f1f172b 488d156e1c0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0x78 (00000001`3f2233a0)]
00000001`3f1f1732 488d0d47f00300  lea     rcx,[Prototype_Console!std::cout (00000001`3f230780)]
00000001`3f1f1739 e8fdf9ffff      call    Prototype_Console!ILT+310(??$?6U?$char_traitsDstdstdYAAEAV?$basic_ostreamDU?$char_traitsDstd (00000001`3f1f113b)
00000001`3f1f173e 488d542428      lea     rdx,[rsp+28h] // This gives us address of "cUsername" on stack.
00000001`3f1f1743 488d0df6f00300  lea     rcx,[Prototype_Console!std::cin (00000001`3f230840)]
00000001`3f1f174a e823faffff      call    Prototype_Console!ILT+365(??$?5DU?$char_traitsDstdstdYAAEAV?$basic_istreamDU?$char_traitsDstd (00000001`3f1f1172)
00000001`3f1f174f 488d153e1c0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0x6c (00000001`3f223394)]
00000001`3f1f1756 488d0d23f00300  lea     rcx,[Prototype_Console!std::cout (00000001`3f230780)]
00000001`3f1f175d e8d9f9ffff      call    Prototype_Console!ILT+310(??$?6U?$char_traitsDstdstdYAAEAV?$basic_ostreamDU?$char_traitsDstd (00000001`3f1f113b)
00000001`3f1f1762 488d542440      lea     rdx,[rsp+40h] // This gives us address of "cPassword" on stack.
00000001`3f1f1767 488d0dd2f00300  lea     rcx,[Prototype_Console!std::cin (00000001`3f230840)]
00000001`3f1f176e e8fff9ffff      call    Prototype_Console!ILT+365(??$?5DU?$char_traitsDstdstdYAAEAV?$basic_istreamDU?$char_traitsDstd (00000001`3f1f1172)
00000001`3f1f1773 488d15321c0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0x84 (00000001`3f2233ac)]
00000001`3f1f177a 488d4c2428      lea     rcx,[rsp+28h]
00000001`3f1f177f e86c420000      call    Prototype_Console!strcmp (00000001`3f1f59f0)
00000001`3f1f1784 85c0            test    eax,eax
00000001`3f1f1786 751d            jne     Prototype_Console!main+0x95 (00000001`3f1f17a5)
00000001`3f1f1788 488d15291c0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0x90 (00000001`3f2233b8)]
00000001`3f1f178f 488d4c2440      lea     rcx,[rsp+40h]
00000001`3f1f1794 e857420000      call    Prototype_Console!strcmp (00000001`3f1f59f0)
00000001`3f1f1799 85c0            test    eax,eax
00000001`3f1f179b 7508            jne     Prototype_Console!main+0x95 (00000001`3f1f17a5)
00000001`3f1f179d c744243801000000 mov     dword ptr [rsp+38h],1
00000001`3f1f17a5 837c243800      cmp     dword ptr [rsp+38h],0
00000001`3f1f17aa 7426            je      Prototype_Console!main+0xc2 (00000001`3f1f17d2)
00000001`3f1f17ac 488d15151c0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0xa0 (00000001`3f2233c8)]
00000001`3f1f17b3 488d0dc6ef0300  lea     rcx,[Prototype_Console!std::cout (00000001`3f230780)]
00000001`3f1f17ba e87cf9ffff      call    Prototype_Console!ILT+310(??$?6U?$char_traitsDstdstdYAAEAV?$basic_ostreamDU?$char_traitsDstd (00000001`3f1f113b)
00000001`3f1f17bf 0fb6542438      movzx   edx,byte ptr [rsp+38h]
00000001`3f1f17c4 488d0db5ef0300  lea     rcx,[Prototype_Console!std::cout (00000001`3f230780)]
00000001`3f1f17cb e825f9ffff      call    Prototype_Console!ILT+240(??$?6U?$char_traitsDstdstdYAAEAV?$basic_ostreamDU?$char_traitsDstd (00000001`3f1f10f5)
00000001`3f1f17d0 eb13            jmp     Prototype_Console!main+0xd5 (00000001`3f1f17e5)
00000001`3f1f17d2 488d15ff1b0300  lea     rdx,[Prototype_Console!std::_Iosb<int>::end+0xb0 (00000001`3f2233d8)]
00000001`3f1f17d9 488d0da0ef0300  lea     rcx,[Prototype_Console!std::cout (00000001`3f230780)]
00000001`3f1f17e0 e856f9ffff      call    Prototype_Console!ILT+310(??$?6U?$char_traitsDstdstdYAAEAV?$basic_ostreamDU?$char_traitsDstd (00000001`3f1f113b)
00000001`3f1f17e5 33c0            xor     eax,eax
00000001`3f1f17e7 488b4c2450      mov     rcx,qword ptr [rsp+50h]
00000001`3f1f17ec 4833cc          xor     rcx,rsp
00000001`3f1f17ef e8bc420000      call    Prototype_Console!__security_check_cookie (00000001`3f1f5ab0)
00000001`3f1f17f4 4883c468        add     rsp,68h
00000001`3f1f17f8 c3              ret

现在,既然我们知道 x64 堆栈是如何工作的,我们就可以开始“黑客攻击”了。 RSP 是堆栈指针,函数堆栈是 RSP 值以上的地址(堆栈增长为更小的地址)。所以,我们看到 RSP+28hcUsernameRSP+38hauthenticationRSP+40hcPassword ,其中 28h、38h 和 40h 是十六进制偏移量。这是小图来说明:
-----> old RSP value // Stack frame of caller of `main` is above, stack frame of main is below

      16 bytes of
      "cPassword"
+40h
     8 bytes of "authentication"
+38h
      16 bytes of
      "cUsername"
+28h


-----> RSP value = old RSP-68h

我们从这里看到了什么?我们看到编译器在 8 字节边界上对齐数据:例如,我们要求为 cUsername 分配 10 个字节,但我们得到了 16 个字节 - x64 位堆栈自然地在 8 字节边界上对齐。这意味着为了写入 authentication 我们需要向 cUsername 写入更多 16 个字节(符号)。另请注意,该编译器将 cPassword 置于高于 authentication 的位置 - 我们无法使用 authentication 覆盖 cPassword ,只能使用 cUsername

所以现在我们运行我们的程序并输入 Username: 0123456789abcdef10123456789abcdef = 16 字节,下一个 1 将被放入 authentication 的低字节 - 对我们来说已经足够了:
Username: 0123456789abcdef1
Pass: whatever
Access granted
1

关于C++ 缓冲区溢出,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/8782852/

10-13 06:28