* Stupid question about the plan9 kernel
@ 1997-01-19 0:49 philw
0 siblings, 0 replies; 4+ messages in thread
From: philw @ 1997-01-19 0:49 UTC (permalink / raw)
ok, so now rob corrected me. Where ever the boot loader
left it is wrong. _start in l.s moves the kernel stack
into the MACH area before calling C. So all the init
code gets run out of a stack in MACH. That is used as
a transition stack.
phil
^ permalink raw reply [flat|nested] 4+ messages in thread
* Stupid question about the plan9 kernel
@ 1997-01-19 0:13 philw
0 siblings, 0 replies; 4+ messages in thread
From: philw @ 1997-01-19 0:13 UTC (permalink / raw)
I realized I did not answer the stack
question. The answer is the kernel stack
is wherever the boot loader put it.
After the first sched the processor will
have made the transition to a mapped
stack with valid u-area.
phil
^ permalink raw reply [flat|nested] 4+ messages in thread
* Stupid question about the plan9 kernel
@ 1997-01-18 23:48 philw
0 siblings, 0 replies; 4+ messages in thread
From: philw @ 1997-01-18 23:48 UTC (permalink / raw)
>What does it means when u is 0? Where is (are?) the kernel stack(s?) stored?
When CPU's boot u == nil so if you are a new
cpu coming online you need to call schedinit
to install the labels, then sched to bring
a process up onto the cpu. these are initial
conditions.
phil
^ permalink raw reply [flat|nested] 4+ messages in thread
* Stupid question about the plan9 kernel
@ 1997-01-18 23:02 Jean
0 siblings, 0 replies; 4+ messages in thread
From: Jean @ 1997-01-18 23:02 UTC (permalink / raw)
In the schedinit() and sched() functions, there is a test :
if (u){ ... }
What does it means when u is 0? Where is (are?) the kernel stack(s?) stored?
^ permalink raw reply [flat|nested] 4+ messages in thread
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