Reputation: 32081
If I use a password as a command-line parameter it's public on the system using ps
.
But if I'm in a bash shell script and I do something like:
...
{ somecommand -p mypassword }
...
is this still going to show up in the process list? Or is this safe?
Upvotes: 7
Views: 13254
Reputation: 51
How about using a file descriptor approach:
env -i bash --norc # clean up environment
set +o history
read -s -p "Enter your password: " passwd
exec 3<<<"$passwd"
mycommand <&3 # cat /dev/stdin in mycommand
See:
Hiding secret from command line parameter on Unix
Upvotes: 5
Reputation: 9466
The called program can change its command line by simply overwriting argv
like this:
#include <stdlib.h>
#include <string.h>
int main(int argc, char** argv) {
int arglen = argv[argc-1]+strlen(argv[argc-1])+1 - argv[0];
memset(argv[0], arglen, 0);
strncpy(argv[0], "secret-program", arglen-1);
sleep(100);
}
Testing:
$ ./a.out mySuperPassword &
$ ps -f
UID PID PPID C STIME TTY TIME CMD
me 20398 18872 0 11:26 pts/3 00:00:00 bash
me 20633 20398 0 11:34 pts/3 00:00:00 secret-program
me 20645 20398 0 11:34 pts/3 00:00:00 ps -f
$
UPD: I know, it is not completely secure and may cause race conditions, but many programs that accept password from command line do this trick.
Upvotes: 3
Reputation: 14346
The only way to escape from being shown in the the process list is if you reimplement the entire functionality of the program you want to call in pure Bash functions. Function calls are not seperate processes. Usually this is not feasible, though.
Upvotes: 0
Reputation: 393104
Command lines will always be visible (if only through /proc).
So the only real solution is: don't. You might supply it on stdin, or a dedicated fd:
./my_secured_process some parameters 3<<< "b@dP2ssword"
with a script like (simplicity first)
#!/bin/bash
cat 0<&3
(this sample would just dump a bad password to stdout)
Now all you need to be concerned with is:
Upvotes: 9