Re: kernel 3.2.27 on arm: WARNING: at mm/page_alloc.c:2109 __alloc_pages_nodemask+0x1d4/0x68c()

Maxime Bizon
2012-10-05T10:49:54Z

On Fri, 2012-10-05 at 09:41 +0200, Eric Dumazet wrote:

> By the way, the commit you pointed has no effect on the reallocation
> performed by pskb_expand_head() :

The commit has a side effect, because the problem appeared after it was
merged (and goes away if I revert it)

> int size = nhead + skb_end_offset(skb) + ntail;
>
> So pskb_expand_head() always assumed the current head is fully used, and
> because we have some kmalloc-power-of-two contraints, each time
> pskb_expand_head() is called with a non zero (nhead + ntail) we double
> the skb->head ksize.

That is true, but only after the commit I mentioned.

Before that commit, we indeed reallocate skb->head to twice the size,
but skb->end is *not* positioned at the end of newly allocated data. So
on the next pskb_expand_head(), if head and tail are not big values, the
kmalloc() will be of the same size.

The commit adds this after allocation:

size = SKB_WITH_OVERHEAD(ksize(data))
[...]
skb->end = skb->head + size;

so on the next pskb_expand_head, we are going to allocate twice the size
for sure.

> So why are we using skb_end_offset(skb) here is the question.
>
> I guess it could be (skb_tail_pointer(skb) - skb->head) on some uses.

I think your patch is wrong, ntail is not the new tailroom size, it's
what missing to the current tailroom size, by adding ntail + nhead +
tail_offset we are removing previous tailroom.

We cannot shrink the skb that way here I guess, a caller may check
needed headroom & tailroom, calls with nhead=1/ntail=0 because only
headroom is missing, but after the call tailroom would be less than
before the call.

Why don't we juste reallocate to this size:

MAX(current_alloc_size, nhead + ntail + current_end - current_head)

--
Maxime