Hi Tony,
On 9/27/2010 5:46 PM, Tony Lindgren wrote:
* Tony Lindgren<tony@xxxxxxxxxxx> [100924 17:14]:
* Benoit Cousson<b-cousson@xxxxxx> [100924 02:07]:
If we wanted to use only the offset in the data, we could initialize
the various sections separately. But then omapX_mux_init() would have
to know the partition number somehow..
After thinking about this a bit more, to me it seems like this is
the best way to go. Then there's no need to repeat the partition data
for each entry. How about this:
Earlier defines stay the same:
#define OMAP3_MUX(mode0, mux_value) \
{ \
.reg_offset = (OMAP3_CONTROL_PADCONF_##mode0##_OFFSET), \
.value = (mux_value), \
}
Then for omap4, we need to add a separate board mux table for each
partition:
#define OMAP4_MUX_PARTXXX(mode0, mux_value) \
{ \
.reg_offset = (OMAP4_PART_XXX_PADCONF_##mode0##_OFFSET), \
.value = (mux_value), \
}
#define OMAP4_MUX_PARTYYY(mode0, mux_value) \
{ \
.reg_offset = (OMAP4_PARTYYY_PADCONF_##mode0##_OFFSET), \
.value = (mux_value), \
}
OK for that one, that will save the extra id to store the partition in
each static data, but then you will still have to store it during the init?
...
For omap2 and 3, we just call omap_mux_init once with the mux_pbase
as we currently already do. Then for omap4, we call omap_mux_init for
each partition.
We also need to change omap_mux_read/write to allow specifying the
partition base address:
Then you need somehow a partition information from somewhere.
I don't see how we can avoid the id at that point? We can store the base
address instead, but then every mux entries will have it.
The caller of the omap_mux_read still have to figure out what base
address it has to use.
That move the issue to the upper layer, but we still need that.
u16 omap_mux_read(void __iomem *base, u16 offset)
{
if (cpu_is_omap24xx())
return __raw_readb(base + offset);
else
return __raw_readw(base + offset);
}
All the other mux interface functions can stay the same, we just need
to modify the mux.c code to look for signal names or GPIO number in
each registered partition.
OK, now I think I understand your point... Please ignore the previous
comments :-)
You will guess the partition by trying each array at a time, and the
first one will win.
That seems pretty good in fact.
I just have to do it now...
Thanks,
Benoit
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