Re: [BUG] Memory ordering between kmalloc() and kfree()? it's confusing!

[Harry Yoo <harry.yoo@oracle.com>]

On Thu, Feb 26, 2026 at 11:42:02AM -0500, Alan Stern wrote:
> On Fri, Feb 27, 2026 at 01:17:52AM +0900, Harry Yoo wrote:
> > On Thu, Feb 26, 2026 at 10:45:55AM -0500, Alan Stern wrote:
> > > On Thu, Feb 26, 2026 at 03:35:08PM +0900, Harry Yoo wrote:
> > > > Because the slab allocator itself doesn't guarantee that such
> > > > barriers are invoked within the allocator, it relies on users to
> > > > do this when needed.
> > > 
> > > It doesn't?  Then how does the slab allocator guarantee that two 
> > > different CPUs won't try to perform allocations or deallocations from 
> > > the same slab at the same time, messing everything up?
> > 
> > Ah, alloc/free slowpaths do use cmpxchg128 or spinlock and
> > don't mess things up.
> > 
> > But fastpath allocs/frees are served from percpu array that is protected
> > by a local_lock. local_lock has a compiler barrier in it, but that's
> > not enough.
> 
> If those things rely on a percpu array, how can one CPU possibly 
> manipulate a resource (slab or something else) that was changed by a 
> different CPU?

AFAICT that shouldn't happen within the slab allocator.

> The whole point of percpu data structures is that each 
> CPU gets its own copy.

Exactly.

But I'm not talking about what happens within the allocator,
but rather, about what slab expects to happen outside the allocator.

Something like this:

CPU X				CPU Y
ptr = kmalloc();
WRITE_ONCE(gp, ptr);
				if (p = READ_ONCE(gp))
					kfree(p);

Yes, it's a crazy thing to do. CPU Y isn't guaranteed to see
up-to-date version of object content or metadata.

Instead, the code should do:

CPU X				CPU Y
ptr = kmalloc();
gp = smp_store_release(&gp, ptr);

				if (p = smp_load_acquire(&gp))
					kfree(p);

One reason that I started this discussion was to argue that we should
have a well-defined a contract between the slab allocator and its users.

-- 
Cheers,
Harry / Hyeonggon