Advanced Reversing Mechanics 1

Very very advanced trust me

71, 66, 61, 42, 53, 45, 7A, 40, 51, 4C, 5E, 30, 79, 5E, 31, 5E, 64, 59, 5E, 38, 61, 36, 65, 37, 63, 7C,

Files:

• easy.o

Solution

Clearly, by the challenge name, the file is an ARM file. Opening it using Ghidra, the main function is decompiled to give:

undefined4 main(undefined4 param_1,int param_2)

{
  char *pcVar1;
  byte *pbVar2;
  byte *pbVar3;
  byte local_110 [256];
  
  pbVar2 = local_110;
  pcVar1 = stpcpy((char *)local_110,*(char **)(param_2 + 4));
  while (local_110[0] != 0) {
    *pbVar2 = local_110[0] - 1;
    pbVar2 = pbVar2 + 1;
    local_110[0] = *pbVar2;
  }
  if (pcVar1 + -(int)local_110 != (char *)0x0) {
    pbVar2 = local_110;
    do {
      pbVar3 = pbVar2 + 1;
      printf("%02X, ",(uint)*pbVar2);
      pbVar2 = pbVar3;
    } while (local_110 + (int)(pcVar1 + -(int)local_110) != pbVar3);
  }
  putchar(10);
  return 0;
}

Seems like all the program does is decrements the ASCII value of each character and prints the value in hexadecimal form. Once we know what is happening, reversing it is a piece of cake and the flag is easily found.

Flag

rgbCTF{ARM_1z_2_eZ_9b7f8d}