B> Hello Kernel Gurus, I'm wondering what is the best way to align a
B> chunk of kernel memory (from kmalloc, say) on a particular byte
B> boundary. 16 bytes, for instance.
I had the same problem; however, I noticed everything was
using a power-of-two chunk allocation scheme, so I did a dirty
function [__usb_kmalloc_align]:
#define MAX_LEVELS 16
void* __usb_kmalloc_align_r (size_t size, int alignment,
int prio, int depth)
{
void *ptr, *rptr;
if (depth > MAX_LEVELS)
return NULL;
ptr = kmalloc (size < alignment? alignment : size, prio);
if (((unsigned long)ptr & (alignment-1)) != 0)
{
rptr = __usb_kmalloc_align_r (size, alignment, prio, depth+1);
usb_kfree (ptr);
ptr = rptr;
}
return ptr;
}
void* __usb_kmalloc_align (size_t size, int alignment, int prio)
{
void* ptr;
void* rptr;
ptr = kmalloc (size < alignment? alignment : size, prio);
if (((unsigned long)ptr & (alignment-1)) != 0)
{
rptr = __usb_kmalloc_align_r (size, alignment, prio, 0);
usb_kfree (ptr);
ptr = rptr;
}
return ptr;
}
The point is, if the size of the object is smaller then the
alignment, ask for a chunk sized as the alignment, else the size. The
rounding to the next power of two will do the same when
allocating.
I've successfully allocated _everyone_ of them, even under
heavy load, but I am not sure it will always success. So I implemented
a workaround for the case it failed.
If not aligned, it will recursively allocate more chunks,
until one falls into the wanted alignment; then it will free them and
let you with the aligned one. It's ugly, but it works. There's a depth
limit, just in case ...
It's dirty, could be optimized, etc, etc ... but hey, I did it
in five minutes :)
Linux-USB! http://peloncho.fis.ucm.es/~inaky/USB.html -
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Inaky Perez Gonzalez -- PGP pubkey fingerprint -
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