On Thu, Feb 7, 2013 at 8:23 AM, Michal Simek <monstr@xxxxxxxxx> wrote:#define iowrite16be(v, addr) iowrite16(be16_to_cpu(v), (addr))
#define iowrite16(v, addr) writew((v), (addr))
#define writew(b,addr) __raw_writew(__cpu_to_le16(b),addr)
static inline void __raw_writew(u16 b, volatile void __iomem *addr)
{
*(volatile u16 __force *) addr = b;
}
How is this suppose to work on Big Endian?
be16_to_cpu(v) is (v)
and
__cpu_to_le16(b) is swab16(v)
Yes.
But on native BE system ( I expect that v is in big endian)
iowrite16be(v, addr) should be just *(volatile u16 __force *) addr =
v; not *(volatile u16 __force *) addr = swab16(v);
What I would expect is
#define iowrite16be(v, addr) __raw_writew(__cpu_to_be16(v), addr)
Indeed, it should be "__cpu_to_be16(v)" instead of "be16_to_cpu(v)".
What do you mean by that?
Bummer, I missed that current iowrite16be() uses (the little endian)
iowrite16(),
not _raw_writew(), and thought the only difference between the original
and your version was the endianness conversion macro.
Yes,
#define iowrite16be(v, addr) __raw_writew(__cpu_to_be16(v), addr)
should be correct.