C中的setgid（）调用不起作用

Question

I am trying to drop privileges in a C program and whatever I do GID remains 0. After hours of research I am clueless :(

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <grp.h>

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

        initgroups("nobody", 65534);
        setresuid(65534,65534,65534);
        setresgid(65534,65534,65534);
        setegid(65534);
        printf("gid: %d\n", getegid());
        execv("/usr/bin/id", argv);

}

Returns this:

gid: 0
uid=65534(nobody) gid=0(root) groups=65534(nogroup)

Why is GID 0 when I explicitly set all three (R,E and SGID) to 65534?

Answer 1

To set the GID values, the process needs an effective UID of 0 (root), but your code carefully throws away the root privileges by setting the UID values to 65534 before calling setresgid() , so the call fails — as you would have known had you tested the return values of the functions.

Reverse the order of the calls to setresuid() and setresgid() (and remove the superfluous setegid() too, of course, and the extra printf() ).

#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>  // Archaic: shouldn't be needed (but Linux docs list it)
#include <grp.h>

int main(int argc, char **argv)
{
    if (initgroups("nobody", 65534) == 0 &&
        setresgid(65534, 65534, 65534) == 0 &&
        setresuid(65534, 65534, 65534))
        execv("/usr/bin/id", argv);
    fprintf(stderr, "Oops!\n");
    return 0;
}

You can be more careful in your error reporting if you wish. Remember, execv() does not return if it is successful; there is no need to test its return value.