Read: swtch.S and proc.c (focus on the code that switches
between processes, specifically scheduler
and sched
).
Hand-In Procedure
You are to turn in this homework during lecture. Please write up your answers to the exercises below and hand them in to a 6.828 staff member at the beginning of lecture.
Introduction
In this homework you will investigate how the kernel switches between two processes.
Assignment:
Suppose a process that is running in the kernel
calls sched()
, which ends up jumping
into scheduler()
.
Turn in:
Where is the stack that sched()
executes on?
Turn in:
Where is the stack that scheduler()
executes on?
Turn in:
When sched()
calls swtch()
,
does that call to swtch()
ever return? If so, when?
Turn in:
Why does swtch()
copy %eip from the stack into the
context structure, only to copy it from the context
structure to the same place on the stack
when the process is re-activated?
What would go wrong if swtch()
just left the
%eip on the stack and didn't store it in the context structure?
Surround the call to swtch()
in schedule()
with calls
to cons_putc()
like this:
cons_putc('a'); swtch(&cpus[cpu()].context, &p->context); cons_putc('b');
Similarly,
surround the call to swtch()
in sched()
with calls
to cons_putc()
like this:
cons_putc('c'); swtch(&cp->context, &cpus[cpu()].context); cons_putc('d');
Rebuild your kernel and boot it on bochs. With a few exceptions you should see a regular four-character pattern repeated over and over.
Turn in: What is the four-character pattern?
Turn in: The very first characters are ac
. Why does
this happen?
Turn in: Near the start of the last line you should see
bc
. How could this happen?
This completes the homework.