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C代码逆向工程中的用户名和密码错误

[英]wrong user and password in C code reverse engineering

I have this piece of code from a binary file that asks for user and password, I managed to find the user "mari" and the password "luig" but it says wrong userName here is the code:我从一个二进制文件中获得了这段代码,它要求输入用户名和密码,我设法找到了用户“mari”和密码“luig”,但它说用户名错误,这里是代码:

undefined8 main(void)
{
  int iVar1;
  undefined4 local_96;
  undefined2 local_92;
  undefined local_90;
  undefined4 local_8f;
  undefined2 local_8b;
  undefined local_89;
  char local_88 [64];
  char local_48 [64];

  local_8f = 0x6769756c;
  local_8b = 0x3169;
  local_89 = 0;
  local_96 = 0x6972616d;
  local_92 = 0x316f;
  local_90 = 0;
  printf("enter username :");
  __isoc99_scanf(&DAT_00102019,local_48);
  iVar1 = strcmp(local_48,(char *)&local_96);
  if (iVar1 == 0) {
    printf("enter password :");
    __isoc99_scanf(&DAT_00102019,local_88);
    iVar1 = strcmp(local_88,(char *)&local_8f);
    if (iVar1 == 0) {
      printf("welldone use it to submit the flag :D");
    }
    else {
      printf("wrong password");
    }
  }
  else {
    printf("wrong username");
  }
  return 0;
}

why "mari" as user and "luig" as password doesn't work?为什么“mari”作为用户而“luig”作为密码不起作用?

The user name and password to compare with are stored in integer variables as hex:要比较的用户名和密码以十六进制形式存储在 integer 变量中:

  local_8f = 0x6769756c;  // 4 byte variable: g i u l
  local_8b = 0x3169;      // 2 byte variable: 1 i
  local_89 = 0;           // 1 byte variable: \0

  local_96 = 0x6972616d;  // 4 byte variable: i r a m
  local_92 = 0x316f;      // 2 byte variable: 1 o
  local_90 = 0;           // 1 byte variable: \0

These variables are stored in memory with the bytes reversed.这些变量存储在 memory 中,字节反转。

So variable local_8f when interpreted as a string in memory, actually reads luigi1\0 and variable local_96 when interpreted as a string in memory, actually reads mario1\0 .因此,变量local_8f在 memory 中解释为字符串时,实际上读取luigi1\0 ,而变量local_96在 memory 中解释为字符串时,实际上读取mario1\0 And those are the uid/pwd you must enter.这些是您必须输入的 uid/pwd。

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