On Jun 18, 2008, at 11:10 AM, Jon Loeliger wrote:
Eric Witcher wrote:
The answer is that mpic_assign_isu(mpic, 1, paddr + 0x11000) places
the initial base register
for isu 1 on a reserved location in the PIC register map (see *).
I guess you can infer from this
that no badness occurs when
Eric Witcher wrote:
The answer is that mpic_assign_isu(mpic, 1, paddr + 0x11000) places the initial
base register
for isu 1 on a reserved location in the PIC register map (see *). I guess you
can infer from this
that no badness occurs when you write to a reserved location on the PIC.
See? S
The answer is that mpic_assign_isu(mpic, 1, paddr + 0x11000) places the initial
base register
for isu 1 on a reserved location in the PIC register map (see *). I guess you
can infer from this
that no badness occurs when you write to a reserved location on the PIC.
That aside, when you walk thro
Can anyone explain why the comment preceding mpic_assign_isu() on line 145 says
that the I2C registers are at 0x11020 yet the code on line 146 shows 0x11000?
The MPC8245 user manual shows the external interrupt 0 registers at 0x10200
(paddr=4) and the I2C registers at 0x11020.
<2.6.25.6>
