[Devel] Re: [PATCH 13/16] Switch to operating with pid_numbers instead of pids

sukadev at us.ibm.com sukadev at us.ibm.com
Wed Jul 25 12:13:34 PDT 2007


Pavel Emelianov [xemul at openvz.org] wrote:
| sukadev at us.ibm.com wrote:
| >Pavel Emelianov [xemul at openvz.org] wrote:
| >| Make alloc_pid() initialize pid_numbers and hash them
| >| into the hashtable, not the struct pid itself.
| >| 
| >| Signed-off-by: Pavel Emelianov <xemul at openvz.org>
| >| 
| >| ---
| >| 
| >|  pid.c |   47 +++++++++++++++++++++++++++++++++--------------
| >|  1 files changed, 33 insertions(+), 14 deletions(-)
| >| 
| >| --- ./kernel/pid.c.ve12	2007-07-05 11:06:41.000000000 +0400
| >| +++ ./kernel/pid.c	2007-07-05 11:08:23.000000000 +0400
| >| @@ -28,8 +28,10 @@
| >|  #include <linux/hash.h>
| >|  #include <linux/pid_namespace.h>
| >|  #include <linux/init_task.h>
| >| +#include <linux/proc_fs.h>
| >| 
| >| -#define pid_hashfn(nr) hash_long((unsigned long)nr, pidhash_shift)
| >| +#define pid_hashfn(nr, ns)	\
| >| +	hash_long((unsigned long)nr + (unsigned long)ns, pidhash_shift)
| >|  static struct hlist_head *pid_hash;
| >|  static int pidhash_shift;
| >|  struct pid init_struct_pid = INIT_STRUCT_PID;
| >| @@ -194,7 +198,7 @@ fastcall void put_pid(struct pid *pid)
| >|  	if (!pid)
| >|  		return;
| >| 
| >| -	ns = pid->numbers[0].ns;
| >| +	ns = pid->numbers[pid->level].ns;
| >|  	if ((atomic_read(&pid->count) == 1) ||
| >|  	     atomic_dec_and_test(&pid->count))
| >|  		kmem_cache_free(ns->pid_cachep, pid);
| >| @@ -210,13 +214,17 @@ static void delayed_put_pid(struct rcu_h
| >|  fastcall void free_pid(struct pid *pid)
| >|  {
| >|  	/* We can be called with write_lock_irq(&tasklist_lock) held */
| >| +	int i;
| >|  	unsigned long flags;
| >| 
| >|  	spin_lock_irqsave(&pidmap_lock, flags);
| >| -	hlist_del_rcu(&pid->pid_chain);
| >| +	for (i = 0; i <= pid->level; i++)
| >| +		hlist_del_rcu(&pid->numbers[i].pid_chain);
| >|  	spin_unlock_irqrestore(&pidmap_lock, flags);
| >| 
| >| -	free_pidmap(&init_pid_ns, pid->nr);
| >| +	for (i = 0; i <= pid->level; i++)
| >| +		free_pidmap(pid->numbers[i].ns, pid->numbers[i].nr);
| >| +
| >|  	call_rcu(&pid->rcu, delayed_put_pid);
| >|  }
| >| 
| >| @@ -224,30 +232,43 @@ struct pid *alloc_pid(struct pid_namespa
| >|  {
| >|  	struct pid *pid;
| >|  	enum pid_type type;
| >| -	int nr = -1;
| >| +	struct pid_namespace *ns;
| >| +	int i, nr;
| >| 
| >| -	pid = kmem_cache_alloc(init_pid_ns.pid_cachep, GFP_KERNEL);
| >| +	pid = kmem_cache_alloc(pid_ns->pid_cachep, GFP_KERNEL);
| >|  	if (!pid)
| >|  		goto out;
| >| 
| >| -	nr = alloc_pidmap(current->nsproxy->pid_ns);
| >| -	if (nr < 0)
| >| -		goto out_free;
| >| +	ns = pid_ns;
| >| +	for (i = pid_ns->level; i >= 0; i--) {
| >| +		nr = alloc_pidmap(ns);
| >| +		if (nr < 0)
| >| +			goto out_free;
| >
| >If pid_ns->level is say 3 and alloc_pidmap() succeeds when i=0,1
| 
| It cannot :) If level is 3, then we'll allocate for 3, 2, 1, 0 sequence.
| The loop is descending, not ascending...

Aah descending - thats right. But I still think there is a problem.

Here is your code that I am referring to:

        pid = kmem_cache_alloc(pid_ns->pid_cachep, GFP_KERNEL);
        if (!pid)
                goto out;

        ns = pid_ns;
        for (i = pid_ns->level; i >= 0; i--) {
                nr = alloc_pidmap(ns);
                if (nr < 0)
                        goto out_free;

                pid->numbers[i].nr = nr;
                pid->numbers[i].ns = ns;
                ns = ns->parent;
        }

	pid->level = pid_ns->level;

	<snip>

out:
	return pid;
out_free:
        for (i++; i <= pid->level; i++)
                free_pidmap(pid->numbers[i].ns, pid->numbers[i].nr);

        kmem_cache_free(pid_ns->pid_cachep, pid);
        pid = NULL;
	goto out;

<end code>

Lets say initially pid_ns->level = 3 and alloc_pidmap() succeeds for
i=3 and i=2 but fails for i=1 and we execute "goto out_free".

But pid->level is uninitialized at this point right ?

Even if it were set to zero (using kmem_cache_zalloc()), we may not
free the two pidmap entries we allocated for i=3 and i=2.

Suka
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