[U-Boot] [PATCH 7/8] mtd/nand/ubi: assortment of alignment fixes

[Scott Wood <scottwood@freescale.com>]

On Fri, 2015-07-03 at 15:44 +0200, Marek Vasut wrote:
> On Thursday, July 02, 2015 at 11:35:19 PM, Scott Wood wrote:
> > On Thu, 2015-07-02 at 07:53 +0200, Marek Vasut wrote:
> > > On Thursday, July 02, 2015 at 01:04:52 AM, Marcel Ziswiler wrote:
> > > > From: Marcel Ziswiler <marcel.ziswiler@toradex.com>
> > > > 
> > > > Various U-Boot adoptions/extensions to MTD/NAND/UBI did not take 
> > > > buffer
> > > > alignment into account which led to failures of the following form:
> > > > 
> > > > ERROR: v7_dcache_inval_range - start address is not aligned -
> > > > 0x1f7f0108 ERROR: v7_dcache_inval_range - stop address is not aligned
> > > > - 0x1f7f1108
> > > > 
> > > > Signed-off-by: Marcel Ziswiler <marcel.ziswiler@toradex.com>
> > > 
> > > What about using ALLOC_CACHE_ALIGN_BUFFER() and friends instead ? See
> > > include/common.h for their definition, this is what those functions are
> > > exactly for.
> > 
> > ALLOC_CACHE_ALIGN_BUFFER() is for statically allocating an aligned buffer.
> 
> You're confusing this with DEFINE_ALIGN_BUFFER, no ?

OK, not "statically", but on the stack.  It is not appropriate to turn 
dynamic allocations into stack allocations without considering how large the 
allocation can be.  It'd also be more intrusive a change than necessary, even 
if the sizes were small enough.

> > Dynamically allocating an aligned buffer is exactly what memalign() is 
> > for.
> 
> Isn't memalign()ed memory aligned only to the start address, while the end
> address (and thus the length) is not aligned ?

The end address is aligned if the size passed to memalign is aligned.  Maybe 
add a wrapper that calls memalign() with the size rounded up to 
ARCH_DMA_MINALIGN?

>  This is what memalign(3) has
> to say:
> 
> "
> The function posix_memalign() allocates size bytes and places the
> address of the allocated memory in *memptr.  The address of the
> allo? cated memory will be a multiple of alignment, which must
> be a power of two and a multiple of sizeof(void *).  If size is  0,
> then  the value placed in *memptr is either NULL, or a unique pointer
> value that can later be successfully passed to free(3).
> 
> The obsolete function memalign() allocates size bytes and returns a
> pointer to the allocated memory.  The memory address will be a mul?
> tiple of alignment, which must be a power of two.
> "

posix_memalign() does not exist in U-Boot, and it's not clear to me why 
memalign() should be considered obsolete.  Is the difference just the ability 
to return -EINVAL?

-Scott