[Devel] Re: [RFC v17][PATCH 22/60] c/r: external checkpoint of a task other than ourself

[Serge E. Hallyn]

Quoting Oren Laadan (orenl at librato.com):
> +/* setup checkpoint-specific parts of ctx */
> +static int init_checkpoint_ctx(struct ckpt_ctx *ctx, pid_t pid)
> +{
> +	struct task_struct *task;
> +	struct nsproxy *nsproxy;
> +	int ret;
> +
> +	/*
> +	 * No need for explicit cleanup here, because if an error
> +	 * occurs then ckpt_ctx_free() is eventually called.
> +	 */
> +
> +	ctx->root_pid = pid;
> +
> +	/* root task */
> +	read_lock(&tasklist_lock);
> +	task = find_task_by_vpid(pid);
> +	if (task)
> +		get_task_struct(task);
> +	read_unlock(&tasklist_lock);
> +	if (!task)
> +		return -ESRCH;
> +	else
> +		ctx->root_task = task;
> +
> +	/* root nsproxy */
> +	rcu_read_lock();
> +	nsproxy = task_nsproxy(task);
> +	if (nsproxy)
> +		get_nsproxy(nsproxy);
> +	rcu_read_unlock();
> +	if (!nsproxy)
> +		return -ESRCH;
> +	else
> +		ctx->root_nsproxy = nsproxy;
> +
> +	/* root freezer */
> +	ctx->root_freezer = task;
> +	geT_task_struct(task);
> +
> +	ret = may_checkpoint_task(ctx, task);
> +	if (ret) {
> +		ckpt_write_err(ctx, NULL);
> +		put_task_struct(task);
> +		put_task_struct(task);
> +		put_nsproxy(nsproxy);

I don't think this is safe - the ckpt_ctx_free() will
free them a second time because you're not setting them
to NULL, right?

> +		return ret;
> +	}
> +
> +	return 0;
> +}
> +

-serge
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