+#define rom_out_be16(addr,w) \
+ ({u16 __w, __v; \
+ __v = (w) >> 8; \
+ __w = ((*(__force volatile u16 *) ((addr) + 0x10000 + (__v<<1)))); \
^^^
Shouldn't this and __w be u8? I though the hardware only did 8-bit
accesses?
The 8 bit is the effective data path but the data written are read back on
the high byte of a 16 bit word. Not sure we really do need the dummy read
at all - in that case, it probably wouldn't matter.
+ __v = (w) & 0x0f; \
^^
ff?
Right you are :-( Didn't do hex calculation in a long while ...
+ __w = ((*(__force volatile u16 *) ((addr) + 0x10000 + (__v<<1)))); })
Woops, multiple use of `addr' and `w' in a macro can have nasty side-effects
(think e.g. `outw(addr++, value)'.
Oops - that I need to fix for sure.
Wouldn't it be better to implement rom_out_be16() to use rom_out_8(),
for easier understanding of the code?
I was worried about even more side effects there...
Michael
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