On Tue, Jan 20, 2026 at 05:13:43PM +0200, Eugen Hristev wrote:
> 
> 
> On 1/3/26 21:23, Mike Rapoport wrote:
> > On Sat, Jan 03, 2026 at 08:36:40AM +0200, Eugen Hristev wrote:
> >>
> >>
> >> On 12/29/25 08:56, Mike Rapoport wrote:
> >>> Hi Eugen,
> >>>
> >>> On Wed, Nov 19, 2025 at 05:44:19PM +0200, Eugen Hristev wrote:
> >>>> This memblock flag indicates that a specific block is registered
> >>>> into an inspection table.
> >>>> The block can be marked for inspection using memblock_mark_inspect()
> >>>> and cleared with memblock_clear_inspect()
> >>>
> >>> Can you explain why memblock should treat memory registered for inspection
> >>> differently?
> >>
> >> It should not, at a first glance.
> >>
> >> The purpose of the flag is to let memblock be aware of it.
> >> The flag is there to have a "memblock way" of registering the memory,
> >> which inside memblock , it can translate to a meminspect way of
> >> registering the memory. It's just an extra layer on top of meminspect.
> >> With this, it would be avoided to call meminspect all over the places it
> >> would be required, but rather use the memblock API.
> > 
> > memblock APIs are not available after boot on many architectures, most
> > notable being x86.
> > 
> > But regardless, I can't say I understand why using memblock APIs for
> > meminspect is better than using meminspect directly.
> > I'd imagine that using meminspect register APIs would actually make it more
> > consistent and it would be easier to identify what memory is registered
> > with meminspect.
> > 
> > In the end, memblock_alloc*() returns dynamically allocated memory, just
> > like kmalloc(), the difference is that memblock is active very early at
> > boot and disappears after core MM initialization.
> 
> Hi Mike,
> 
> Thanks for sharing your opinion.
> 
> David, what do you think, does it make sense to have this flag or we can
> ditch it and use meminspect directly ?
> 
> Also, for some memory blocks, they do not disappear ever, e.g. the
> printk log buffer, it's allocated early and never freed, so it's
> required to have some memblocks marked for inspection.

The allocated memory does not disappear, the memblock metadata does.

> Eugen

-- 
Sincerely yours,
Mike.

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