> For the complete documentation index, see [llms.txt](https://alham-rizvi.gitbook.io/alhamrizvi/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://alham-rizvi.gitbook.io/alhamrizvi/picoctf-writeups/picoctf/reverse-engineering/vault-door-4.md).

# vault-door-4

This challenge involves analyzing a Java program that validates a password using a byte array. The goal is to understand how the password is checked and reconstruct the correct input.

#### Source Code Analysis

Inside the `checkPassword()` function, the following logic is used:

```java
byte[] passBytes = password.getBytes();
byte[] myBytes = {
    106 , 85  , 53  , 116 , 95  , 52  , 95  , 98  ,
    0x55, 0x6e, 0x43, 0x68, 0x5f, 0x30, 0x66, 0x5f,
    0142, 0131, 0164, 063 , 0163, 0137, 0145, 060 ,
    '2' , '1' , '3' , '8' , '7' , '2' , '1' , '3' ,
};
```

* The user input is converted into a byte array using:

  ```java
  password.getBytes()
  ```
* The program defines another byte array (`myBytes`) which represents the correct password.
* It then compares both arrays.

So, the correct password is simply the ASCII representation of `myBytes`.

#### Understanding the Byte Array

The tricky part is that `myBytes` uses **mixed number formats**:

**1. Decimal values (base 10)**

```
106 → 'j'
85  → 'U'
53  → '5'
```

**2. Hexadecimal values (base 16, prefix `0x`)**

```
0x55 → 85  → 'U'
0x6e → 110 → 'n'
0x43 → 'C'
```

**3. Octal values (base 8, leading `0`)**

```
0142 → 98  → 'b'
0131 → 89  → 'Y'
0164 → 116 → 't'
```

**4. Character literals**

```
'2' → '2'
'1' → '1'
```

#### Decoding Process

To solve the challenge:

1. Convert all values (decimal, hex, octal) into decimal
2. Map each decimal value to its ASCII character
3. Combine all characters into a string

After decoding the entire array, we get:

```
jU5t_4_bUnCh_0f_bYt3s_e021387213
``'
```


---

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