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SubjectRe: [PATCH v2 1/3] mm: rework memcg kernel stack accounting
On Thu, Aug 23, 2018 at 09:23:50AM -0700, Roman Gushchin wrote:
> On Wed, Aug 22, 2018 at 04:12:13PM +0200, Michal Hocko wrote:
> > On Tue 21-08-18 14:35:57, Roman Gushchin wrote:
> > > @@ -248,9 +253,20 @@ static unsigned long *alloc_thread_stack_node(struct task_struct *tsk, int node)
> > > static inline void free_thread_stack(struct task_struct *tsk)
> > > {
> > > #ifdef CONFIG_VMAP_STACK
> > > - if (task_stack_vm_area(tsk)) {
> > > + struct vm_struct *vm = task_stack_vm_area(tsk);
> > > +
> > > + if (vm) {
> > > int i;
> > >
> > > + for (i = 0; i < THREAD_SIZE / PAGE_SIZE; i++) {
> > > + mod_memcg_page_state(vm->pages[i],
> > > + MEMCG_KERNEL_STACK_KB,
> > > + -(int)(PAGE_SIZE / 1024));
> > > +
> > > + memcg_kmem_uncharge(vm->pages[i],
> > > + compound_order(vm->pages[i]));
> >
> > when do we have order > 0 here?
>
> I guess, it's not possible, but hard-coded 1 looked a bit crappy.
> Do you think it's better?

Yes, specifying the known value (order 0) is much better. I asked
myself the same question as Michal: we're walking through THREAD_SIZE
in PAGE_SIZE steps, how could it possibly be a higher order page?

It adds an unnecessary branch to the code and the reader's brain.

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