[Question] devm_kmalloc() for DMA ?
From: Masahiro Yamada
Date: Wed Mar 08 2017 - 06:07:39 EST
Hi experts,
I have a question about
how to allocate DMA-safe buffer.
In my understanding, kmalloc() returns
memory with DMA safe alignment
in order to avoid cache-sharing problem when used for DMA.
The alignment is decided by ARCH_DMA_MINALIGN.
For example, on modern ARM 32bit boards, this value is typically 64.
So, memory returned by kmalloc() has
at least 64 byte alignment.
On the other hand, devm_kmalloc() does not return
enough-aligned memory.
On my board (ARM 32bit), devm_kmalloc() returns
(ARCH_DMA_MINALIGN aligned address) + 0x10.
The reason of the offset 0x10 is obvious.
struct devres {
struct devres_node node;
/* -- 3 pointers */
unsigned long long data[]; /* guarantee ull alignment */
};
Management data is located at the top of struct devres.
Then, devm_kmalloc() returns dr->data.
The "unsigned long long" guarantees
the returned memory has 0x10 alignment,
but I think this may not be enough for DMA.
I noticed this when I was seeing drivers/mtd/nand/denali.c
The code looks as follows:
denali->buf.buf = devm_kzalloc(denali->dev,
mtd->writesize + mtd->oobsize,
GFP_KERNEL);
if (!denali->buf.buf) {
ret = -ENOMEM;
goto failed_req_irq;
}
/* Is 32-bit DMA supported? */
ret = dma_set_mask(denali->dev, DMA_BIT_MASK(32));
if (ret) {
dev_err(denali->dev, "No usable DMA configuration\n");
goto failed_req_irq;
}
denali->buf.dma_buf = dma_map_single(denali->dev, denali->buf.buf,
mtd->writesize + mtd->oobsize,
DMA_BIDIRECTIONAL);
Memory buffer is allocated by devm_kzalloc(), then
passed to dma_map_single().
Could this be a potential problem in general?
Is devm_kmalloc() not recommended
for buffer that can be DMA-mapped?
Any advice is appreciated.
--
Best Regards
Masahiro Yamada