Why can't I remove a cgroup?

[Jack Wang]

Nicky Chorley <ndchorley at gmail.com>于2022年8月9日 周二20:31写道：

> Hi folks,
>
> I wrote a small program to start a new process (a shell), create a
> cgroup and add the process to it. This all works fine - the process ID
> of the shell ends up in the cgroup.procs file and when my program
> exits, cgroup.procs ends up empty as expected. If I try and remove the
> cgroup directory, though, I get a load of "Operation not permitted"
> errors and I'm not sure why. I'm running on openSUSE (kernel 5.3.18)
> with cgroups v2.
>
> My program looks like:
>
> #define _GNU_SOURCE
> #include <sched.h>
> #include <signal.h>
> #include <stdio.h>
> #include <stdlib.h>
> #include <sys/stat.h>
> #include <sys/wait.h>
> #include <unistd.h>
>
> int startShell(void *argument) {
>     char* arguments[] = {"bash", NULL};
>     char* environment[] = {NULL};
>     execve("/bin/bash", arguments, environment);
>
>     return 0;
> }
>
> int main() {
>     mkdir(
>         "/sys/fs/cgroup/my_group",
>         S_IRWXO | S_IRGRP | S_IROTH
>     );
>
>     int stackSizeBytes = 65536;
>     char *stackStart = malloc(stackSizeBytes);
>     char *stackEnd = stackStart + stackSizeBytes;
>
>     pid_t childPid = clone(
>         startShell,
>         stackEnd,
>         SIGCHLD,
>         NULL
>     );
>
>     FILE* procsFile = fopen(
>         "/sys/fs/cgroup/my_group/cgroup.procs",
>         "w"
>     );
>
>     fprintf(procsFile, "%d\n", childPid);
>     fclose(procsFile);
>
>     waitpid(childPid, NULL, 0);
>
>     return 0;
> }
>
> and output looks like:
>
> $ su -c ./run
> Password:
>
> # echo $$
> 4382
> # cat /sys/fs/cgroup/my_group/cgroup.procs
> 4382
> 4522
> # exit
> $ cat /sys/fs/cgroup/my_group/cgroup.procs
> $ su -c "rm -rf /sys/fs/cgroup/my_group"

You have to use rmdir to remove

>
> Password:
> rm: cannot remove '/sys/fs/cgroup/my_group/cgroup.events': Operation
> not permitted
> rm: cannot remove '/sys/fs/cgroup/my_group/memory.events': Operation
> not permitted
> ... other lines omitted for brevity
>
> Could someone please help me understand what's going on here?
>
> Thanks!
>
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