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DateSat, 24 Mar 2007 20:35:05 +0530
FromSrivatsa Vaddagiri <>
SubjectRe: [ckrm-tech] [PATCH 1/7] containers (V7): Generic container system abstracted from cpusets code
On Mon, Feb 12, 2007 at 12:15:22AM -0800, menage@google.com wrote:
> +static int attach_task(struct container *cont, char *pidbuf, char **ppathbuf)
> +{
> +	pid_t pid;
> +	struct task_struct *tsk;
> +	struct container *oldcont;
> +	int retval;
> +
> +	if (sscanf(pidbuf, "%d", &pid) != 1)
> +		return -EIO;
> +
> +	if (pid) {
> +		read_lock(&tasklist_lock);
> +
> +		tsk = find_task_by_pid(pid);
> +		if (!tsk || tsk->flags & PF_EXITING) {

This is probably carrying over code from cpuset.c, but :

/me thinks that there is a ugly race here with 'tsk' exiting.
What happens if the tsk is marked PF_EXITING just after this check?
If that happens, then:

> +			read_unlock(&tasklist_lock);
> +			return -ESRCH;
> +		}
> +
> +		get_task_struct(tsk);
> +		read_unlock(&tasklist_lock);
> +
> +		if ((current->euid) && (current->euid != tsk->uid)
> +		    && (current->euid != tsk->suid)) {
> +			put_task_struct(tsk);
> +			return -EACCES;
> +		}
> +	} else {
> +		tsk = current;
> +		get_task_struct(tsk);
> +	}
> +
> +	retval = security_task_setscheduler(tsk, 0, NULL);
> +	if (retval) {
> +		put_task_struct(tsk);
> +		return retval;
> +	}
> +
> +	mutex_lock(&callback_mutex);
> +
> +	task_lock(tsk);
> +	oldcont = tsk->container;
> +	if (!oldcont) {
> +		task_unlock(tsk);
> +		mutex_unlock(&callback_mutex);
> +		put_task_struct(tsk);
> +		return -ESRCH;
> +	}
> +	atomic_inc(&cont->count);
> +	rcu_assign_pointer(tsk->container, cont);

Above assignment A1 can race with below assignment A2 in container_exit() :

	tsk->container = &top_container; /* the_top_container_hack - see above */
What happens if A1 follows after A2? I feel very uncomfortable abt it.

IMO, we need to use task_lock() in container_exit() to avoid this race.

(I think this race already exists in mainline cpuset.c?)

P.S : cpuset.c checks for PF_EXITING twice in attach_task(), while this
patch seems to be checking only once. Is that fine?


-- 
Regards,
vatsa
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