On Thu, 2023-01-26 at 17:31 +0000, David Laight wrote:
> Changing the size to kzalloc() doesn't help.
> The alignment depends on the allocator and is only required to have
> a relatively small alignment (ARCH_MINALIGN?) regardless of the size.
>
> IIRC one of the allocators adds a small header to every item.
> It won't return 16 byte aligned items at all.
I'm relying on the behaviour described in Documentation/core-
api/memory-allocation.rst:
The address of a chunk allocated with kmalloc is aligned to at
least ARCH_KMALLOC_MINALIGN bytes. For sizes which are a power of
two, the alignment is also guaranteed to be at least the respective
size.
Is this wrong?
Andrew
--
Andrew Donnellan OzLabs, ADL Canberra
ajd@xxxxxxxxxxxxx IBM Australia Limited
