I am trying to create a executor program for regular users on linux with SUID bit set so whatever commands, passed to the program as parameters, get executed with root permission. However when I try to implement this as a bash script, this does not work, where it works when implemented in C. I want to know what I am doing wrong for the shell script. The codes are below
Shell Script:
#! /bin/bash
if [ $# -lt 1 ]; then
echo "Usage: $0 <Command String>"
exit 1
fi
$@
#Also tried this, same result
#exec $@
Execution:
root#: chmod 755 exec.sh
root#: chmod u+s exec.sh
root#: ll exec.sh
-rwsr-xr-x 1 root root 75 Sep 19 16:55 exec.sh
regular_user$: ./exec.sh whoami
regular_user
C Program:
#include <stdlib.h>
#include <stdio.h>
int main ( int argc, char *argv[] )
{
if ( argc < 2 ) {
printf( "Usage: %s <Command String>n", argv[0] );
return 1;
}
else
{
argv[argc]=NULL;
//setuid(0); //Works without these
//setgid(0);
int exit=execvp(argv[1], argv+1);
return exit;
}
}
Execution:
root#: gcc exec.c -o exec.obj
root#: chmod 755 exec.obj
root#: chmod u+s exec.obj
root#: ll exec.obj
-rwsr-xr-x 1 root root 6979 Sep 19 17:03 exec.obj
regular_user$: ./exec.obj whoami
root
Both files have identical permissions
-rwsr-xr-x 1 root root 75 Sep 19 16:55 exec.sh
-rwsr-xr-x 1 root root 6979 Sep 19 17:03 exec.obj
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Answer
It is documented in execve(2) :
Linux ignores the set-user-ID and set-group-ID bits on scripts.
IIRC, setuid scripts would be a significant security hole
See this question
You could configure sudo
to avoid asking a password – see sudoers(5) (or use super
)
You could also write a simple C program wrapping your shell script, and make it setuid.