How to hide a password passed as command line argument?
I'm running a software daemon that requires for certain actions to enter a passphrase to unlock some features which looks for example like that:
$ darkcoind masternode start <mypassphrase>
Now I got some security concerns on my headless debian server.
Whenever I search my bash history for example with Ctrl+R I can see this super strong password. Now I imagine my server is compromized and some intruder has shell access and can simply Ctrl+R to find my passphrase in the history.
Is there a way to enter the passphrase without it to be shown in bash history, ps, /proc or anywhere else?
Update 1: Passing no password to the daemon throws an error. This is no option.
Update 2: Don't tell me to delete the software or other helpful hints like hanging the developers. I know this is not a best-practice example but this software is based on bitcoin and all bitcoin based clients are some kind of json rpc server which listens to these commands and its a known security issue still being discussed (a, b, c).
Update 3: The daemon is already started and running with the command
$ darkcoind -daemon
Doing ps shows only the startup command.
$ ps aux | grep darkcoin
user 12337 0.0 0.0 10916 1084 pts/4 S+ 09:19 0:00 grep darkcoin
user 21626 0.6 0.3 1849716 130292 ? SLl May02 6:48 darkcoind -daemon
So passing the commands with the passphrase does not show up in ps or /proc at all.
$ darkcoind masternode start <mypassphrase>
$ ps aux | grep darkcoin
user 12929 0.0 0.0 10916 1088 pts/4 S+ 09:23 0:00 grep darkcoin
user 21626 0.6 0.3 1849716 130292 ? SLl May02 6:49 darkcoind -daemon
This leaves the question where does the history show up? Only in .bash_history?
Really, this should be fixed in the application itself. And such applications should be open source, so that fixing the issue in the app itself should be an option. A security related application which makes this kind of mistake might make other mistakes as well, so I wouldn't trust it.
Simple interposer
But you were asking for a different way, so here is one:
#define _GNU_SOURCE
#include <dlfcn.h>
int __libc_start_main(
int (*main) (int, char * *, char * *),
int argc, char * * ubp_av,
void (*init) (void),
void (*fini) (void),
void (*rtld_fini) (void),
void (* stack_end)
)
{
int (*next)(
int (*main) (int, char * *, char * *),
int argc, char * * ubp_av,
void (*init) (void),
void (*fini) (void),
void (*rtld_fini) (void),
void (* stack_end)
) = dlsym(RTLD_NEXT, "__libc_start_main");
ubp_av[argc - 1] = "secret password";
return next(main, argc, ubp_av, init, fini, rtld_fini, stack_end);
}
Compile this with
gcc -O2 -fPIC -shared -o injectpassword.so injectpassword.c -ldl
then run your process with
LD_PRELOAD=$PWD/injectpassword.so darkcoind masternode start fakepasshrase
The interposer library will run this code before the main function from your application gets executed. It will replace the last command line argument by the actual password in the call to main. The command line as printed in /proc/*/cmdline (and therefore seen by tools such as ps) will still contain the fake argument, though. Obviously you'd have to make the source code and the library you compile from it readable only to yourself, so best operate in a chmod 0700 directory. And since the password isn't part of the command invocation, your bash history is safe as well.
More advanced interposer
If you want to do anything more elaborate, you should keep in mind that __libc_start_main gets executed before the runtime library has been properly initialized. So I'd suggest avoiding any function calls unless they are absolutely essential. If you want to be able to call functions to your heart's content, make sure you do so just before main itself gets invoked, after all initialization is done. For the following example I have to thank Grubermensch who pointed out how to hide a password passed as command line argument which brought getpass to my attention.
#define _GNU_SOURCE
#include <dlfcn.h>
#include <unistd.h>
static int (*real_main) (int, char * *, char * *);
static int my_main(int argc, char * * argv, char * * env) {
char *pass = getpass(argv[argc - 1]);
if (pass == NULL) return 1;
argv[argc - 1] = pass;
return real_main(argc, argv, env);
}
int __libc_start_main(
int (*main) (int, char * *, char * *),
int argc, char * * ubp_av,
void (*init) (void),
void (*fini) (void),
void (*rtld_fini) (void),
void (* stack_end)
)
{
int (*next)(
int (*main) (int, char * *, char * *),
int argc, char * * ubp_av,
void (*init) (void),
void (*fini) (void),
void (*rtld_fini) (void),
void (* stack_end)
) = dlsym(RTLD_NEXT, "__libc_start_main");
real_main = main;
return next(my_main, argc, ubp_av, init, fini, rtld_fini, stack_end);
}
This prompts for the password, so you no longer have to keep the interposer library a secret. The placeholder argument is reused as password prompt, so invoke this like
LD_PRELOAD=$PWD/injectpassword.so darkcoind masternode start "Password: "
Another alternative would read the password from a file descriptor (like e.g. gpg --passphrase-fd does), or from x11-ssh-askpass, or whatever.
It's not just the history. It is going to show up in ps output as well.
Whoever wrote that piece of software should be hung, drawn and quartered. It is an absolute NO to have to supply a password on the command-line regardless whatever software it is.
For a daemon process it is even MORE unforgivable...
Besides rm -f on the software itself I don't know any solution for this. Honestly: Find other software to get the job done. Don't use such junk.
This will clear the ps output.
BE VERY AWARE: This could break the application. You are duly warned that here be dragons.
- Foreign processes shouldn't be fiddling around in a processes memory.
- If the process relies on this region for the password, you may break your application.
- Doing this could corrupt any working data you have in that process.
- This is an insane hack.
Now you are duly notified of these dire warnings. This will clear the output displayed in ps. It will not clear your history, nor will it clear the bash job history (such as running the process like myprocess myargs &). But ps will no longer show the arguments.
#!/usr/bin/python
import os, sys
import re
PAGESIZE=4096
if __name__ == "__main__":
if len(sys.argv) < 2:
sys.stderr.write("Must provide a pid\n")
sys.exit(1)
pid = sys.argv[1]
try:
cmdline = open("/proc/{0}/cmdline".format(pid)).read(8192)
## On linux, at least, argv is located in the stack. This is likely o/s
## independent.
## Open the maps file and obtain the stack address.
maps = open("/proc/{0}/maps".format(pid)).read(65536)
m = re.search('([0-9a-f]+)-([0-9a-f]+)\s+rw.+\[stack\]\n', maps)
if not m:
sys.stderr.write("Could not find stack in process\n");
sys.exit(1)
start = int("0x"+m.group(1), 0)
end = int("0x"+m.group(2), 0)
## Open the mem file
mem = open('/proc/{0}/mem'.format(pid), 'r+')
## As the stack grows downwards, start at the end. It is expected
## that the value we are looking for will be at the top of the stack
## somewhere
## Seek to the end of the stack minus a couple of pages.
mem.seek(end-(2*PAGESIZE))
## Read this buffer to the end of the stack
stackportion = mem.read(8192)
## look for a string matching cmdline. This is pretty dangerous.
## HERE BE DRAGONS
m = re.search(cmdline, stackportion)
if not m:
## cause this is an example dont try to search exhaustively, just give up
sys.stderr.write("Could not find command line in the stack. Giving up.")
sys.exit(1)
## Else, we got a hit. Rewind our file descriptor, plus where we found the first argument.
mem.seek(end-(2*PAGESIZE)+m.start())
## Additionally, we'll keep arg0, as thats the program name.
arg0len = len(cmdline.split("\x00")[0]) + 1
mem.seek(arg0len, 1)
## lastly overwrite the remaining region with nulls.
writeover = "\x00" * (len(cmdline)-arg0len)
mem.write(writeover)
## cleanup
mem.close()
except OSError, IOError:
sys.stderr.write("Cannot find pid\n")
sys.exit(1)
Invoke the program by saving it, chmod +x it. Then doing ./whatever <pidoftarget>
If this works, it will produce no output. If it fails, it will complain about something and quit.
Can you pass the argument from a file, accessible only by root or the required user?
It's a HUGE no-no to type passwords in the console, but last recourse...begin your line with a space so it doesn't appear in the history.