Re: [PATCH] net: Fix potential memory leak in proto_register()

[Cong Wang]

On Wed, Aug 12, 2020 at 2:21 AM linmiaohe <linmiaohe@xxxxxxxxxx> wrote:
>
> Hi all:
> David Miller <davem@xxxxxxxxxxxxx> wrote:
> >From: Cong Wang <xiyou.wangcong@xxxxxxxxx>
> >Date: Tue, 11 Aug 2020 16:02:51 -0700
> >
> >>> @@ -3406,6 +3406,16 @@ static void sock_inuse_add(struct net *net,
> >>> int val)  }  #endif
> >>>
> >>> +static void tw_prot_cleanup(struct timewait_sock_ops *twsk_prot) {
> >>> +       if (!twsk_prot)
> >>> +               return;
> >>> +       kfree(twsk_prot->twsk_slab_name);
> >>> +       twsk_prot->twsk_slab_name = NULL;
> >>> +       kmem_cache_destroy(twsk_prot->twsk_slab);
> >>
> >> Hmm, are you sure you can free the kmem cache name before
> >> kmem_cache_destroy()? To me, it seems kmem_cache_destroy() frees the
> >> name via slab_kmem_cache_release() via kfree_const().
> >> With your patch, we have a double-free on the name?
> >>
> >> Or am I missing anything?
> >
> >Yep, there is a double free here.
> >
> >Please fix this.
>
> Many thanks for both of you to point this issue out. But I'am not really understand, could you please explain it more?
> As far as I can see, the double free path is:
> 1. kfree(twsk_prot->twsk_slab_name)
> 2. kmem_cache_destroy
>         --> shutdown_memcg_caches
>                 --> shutdown_cache
>                         --> slab_kmem_cache_release
>                                 --> kfree_const(s->name)
> But twsk_prot->twsk_slab_name is allocated from kasprintf via kmalloc_track_caller while twsk_prot->twsk_slab->name is allocated
> via kstrdup_const. So I think twsk_prot->twsk_slab_name and twsk_prot->twsk_slab->name point to different memory, and there is no double free.
>

You are right. Since it is duplicated, then there is no need to keep
->twsk_slab_name, we can just use twsk_slab->name. I will send
a patch to get rid of it.

> Or am I missing anything?
>
> By the way, req_prot_cleanup() do the same things, i.e. free the slab_name before involve kmem_cache_destroy(). If there is a double
> free, so as here?

Ditto. Can be just removed.

Thanks.