Hand-In Procedure
You are to turn in this homework before the lecture begins. Please drop your homework in a cardboard box near to the lecture podium. Your homework must be hand-written (typed homeworks will not be accepted).
Read: spinlock.c In this part of the assignment we will explore some of the interaction between interrupts and locking.
Make sure you understand what would happen if the kernel executed the following code snippet:
struct spinlock lk; initlock(&lk, "test lock"); acquire(&lk); acquire(&lk);(Feel free to use QEMU to find out.
acquire
is in spinlock.c
.)
An acquire
ensures interrupts are off
on the local processor using the cli
instruction
(via pushcli()
),
and interrupts remain off until the release
of the last lock held by that processor
(at which point they are enabled using sti
).
Let's see what happens if we turn on interrupts while
holding the ide
lock.
In iderw
in ide.c
, add a call
to sti()
after the acquire()
,
and a call to cli()
just before the release()
.
Rebuild the kernel and boot it in QEMU.
Chances are the kernel will panic soon after boot; try booting QEMU a few times
if it doesn't.
Turn in:
explain in a few sentences why the kernel panicked.
You may find it useful to look up the stack trace
(the sequence of %eip
values printed by panic
)
in the kernel.asm
listing.
sti()
and cli()
you added,
rebuild the kernel, and make sure it works again.
Now let's see what happens if we turn on interrupts
while holding the file_table_lock
.
This lock protects the table of file descriptors,
which the kernel modifies when an application opens or closes
a file.
In filealloc()
in file.c
, add
a call to
sti()
after the call to acquire()
,
and a cli()
just before each of the
release()
es.
You will also need to add
#include "x86.h"
at the top of the file after
the other #include
lines.
Rebuild the kernel and boot it in QEMU.
It will not panic.
Turn in:
explain in a few sentences why the kernel didn't panic.
Why do file_table_lock
and ide_lock
have
different behavior in this respect?
You do not need to understand anything about the details of the IDE hardware to answer this question, but you may find it helpful to look at which functions acquire each lock, and then at when those functions get called.
(There is a very small but non-zero chance that the kernel will panic
with the extra sti()
in filealloc()
.
If the kernel does panic, make doubly sure that
you removed the sti()
call from
iderw
. If it continues to panic and the
only extra sti()
is in filealloc()
,
then email the staff
and think about buying a lottery ticket.)
Turn in:
Why does release()
clear
lk->pcs[0]
and lk->cpu
before clearing lk->locked
?
Why not wait until after?
xchg
to implement locks (or other forms of
synchronization). The concurrency on uni-processor systems
is due to pre-emption of threads by the timer interrupt.
Because the timer interrupt may pre-empt a thread at
any point in program execution, instructions of two
threads may interleave in any arbitrary order.
lock()
and
unlock()
primitives using interrupt enable and
disable mechanisms. Answer the associated questions.
lock(L) { cli(); //disable preemption while (L==0) continue; L = 0; sti(); //enable preemption } unlock(L) { L = 1; }Does this implementation of locks work on a uniprocessor? If not, why not?
lock(L) { int acquired = 0; while (!acquired) { cli(); if (L == 1) { acquired = 1; L = 0; } sti(); } } unlock(L) { L = 1; }Does this implementation of locks work on a uniprocessor? If not, why not?
Here is a version of the code for a single-queue, multiple-consumer, mutiple-producer problem.
struct pcq { void *ptr; struct spinlock lock; }; void* pcqread(struct pcq *q) { void *p; acquire(&q->lock); while(q->ptr == 0) sleep(q, &q->lock); p = q->ptr; q->ptr = 0; wakeup(q); /* wake pcqwrite */ release(&q->lock); return p; } void pcqwrite(struct pcq *q, void *p) { acquire(&q->lock); while(q->ptr != 0) sleep(q, &q->lock); q->ptr = p; wakeup(q); /* wake pcqread */ release(&q->lock); return p; }
Turn in:
Both producer (pcqwrite
) and consumer (pcqread
)
are sleeping on the same channel q
. Is this correct? Why or
why not? Should they sleep on different channels? For example, what happens
if the producer calls
wakeup(q)
? Can some unrelated part of the code call wakeup
a consumer thread?
Read: sysfile.c (create(), sys_unlink()), fs.c (readi(), writei(), dirlink(), ialloc(), iupdate(), iget(), ilock(), iunlock(), iput(), itrunc()), bio.c (bget(), bread(), bwrite(), brelse())
Add the following line at the beginning of the log_write() function in log.c
cprintf("log_write %d\n", b->sector);This will record all writes to the file system along with the sector number.
Start a new session on xv6 with a fresh disk (using make clean
followed
by make qemu
, and type the following command:
$ echo > aThis command creates a new file. You will see a series of disk writes (printed in log_write()).
Turn in:
Report the printed output and explain what is being written in each disk write.
What is the third disk write (to sector
29)? You may want to insert cprintf()
statements in xv6 code to
see where the writes are coming from.
Interrupt the previous command (leave the newly created file unchanged). Next, execute the following command to write data to this file:
$ echo x > a
Turn in:
Report the printed output and explain what is being written in each disk write.
Why do you see writes to block 4 (and to 426) twice? You may want to
insert cprintf()
statements in xv6 code to
see where the writes are coming from.
Next, delete the file by typing the following command:
$ rm a
Turn in: Report the printed output and explain what is being written in each disk write.
This homework is based on the ZCAV tool: see ZCAV documentation here.
To deal with this situation, modern disks divide their tracks into zones. The inner zones have lower angular sector density, while the outer zones have higher angular sector density. Effectively, this means that the disk throughput at the outer zones is higher. This is also called zoned constant angular velocity (ZCAV).
To take advantage of this, the disk controllers usually map the lower sector numbers (starting from sector number zero) on the outer tracks, while the higher sector numbers are mapped to inner tracks. The thinking being that the lower sector numbers are likely to have higher frequency of access. Notice that this is just an assumption and may not be true!
ZCAV program measures the disk throughput at different sector
numbers. To do this, it reads/writes 256MB (default) worth of data starting
at sector X
, where X
is varied from 0 to disk-size.
You can see the results of running ZCAV on some disk configurations here.
tar xzf bonnie++-1.03e.tgz
cd bonnie++-1.03e
./configure
make
sudo ./zcav /dev/sda
(replace sda
with sdb
, hda
, etc. depending on the disk being tested).
top
.
Also, you may want to check that no other IO-intensive operation is running
using iotop
.
./zcav filename
.