ELF ARM crackme 1337

Root-me challenge: If the binary file sends you 1337 you got the right password.

  1. ARM, ELF 32-bit LSB

  2. Decompile the code

int __cdecl main(int argc, const char **argv, const char **envp)
{
 int v4; // [sp+4h] [bp-30h]@2
 const char **v5; // [sp+8h] [bp-2Ch]@1
 int v6; // [sp+14h] [bp-20h]@3
 int v7; // [sp+18h] [bp-1Ch]@3
 int v8; // [sp+18h] [bp-1Ch]@6
 int i; // [sp+18h] [bp-1Ch]@9
 signed int v10; // [sp+1Ch] [bp-18h]@6

 v5 = argv;
 if ( argc > 1 )
 {
   v7 = 0;
   v6 = xmalloc(32);
   while ( v7 != 8 )
   {
     *(_DWORD *)(v6 + 4 * v7) = xmalloc(32);
     memset(*(_DWORD *)(v6 + 4 * v7++), 10, 32);
   }
   *(_DWORD *)(v6 + 32) = 0;
   v8 = 0;
   v10 = 65;
   while ( v8 != 31 )
     *(_BYTE *)(*(_DWORD *)(v6 + 12) + v8++) = v10++;
   *(_BYTE *)(*(_DWORD *)(v6 + 12) + 31) = 0;
   for ( i = 0; v5[1][i]; ++i )
   {
     if ( v5[1][i] != *(_BYTE *)(*(_DWORD *)(v6 + 12) + i) )
     return -1;
   }
   v4 = 1337;
 }
 else
 {
   v4 = -1;
 }
 return v4;
}

  1. Analysis

  • The program receives the code from argv.

  • Uses v5 to compare the password with.

  • Space was reserved for 32 characters.

  • v6 is used to store the password.

  • 65 is A in ascii code.

  • The password is 32 letters from A up to 32 characters.

Resources