1. 2-Dining Puppies
Mutex Failure

2. Outline for Today Objective: Administrative details:
Reader-writer problem
Message Passing
Administrative details:
Bring forks and magnets

3. Readers/Writers Problem
Synchronizing access to a file or data record in a database such that any number of threads requesting read-only access are allowed but only one thread requesting write access is allowed, excluding all readers.

4. Template for Readers/Writers
{while (true) {
read }
Writer()
{while (true) {
write }
/*request r access*/
/*request w access*/
/*release r access*/
/*release w access*/

5. Template for Readers/Writers
{while (true) {
read }
Writer()
{while (true) {
write }
fd = open(foo, 0);
fd = open(foo, 1);
close(fd);
close(fd);

6. Template for Readers/Writers
{while (true) {
read }
Writer()
{while (true) {
write }
startRead();
startWrite();
endRead();
endWrite();

7. R/W - Monitor Style Boolean busy = false; int numReaders = 0;
Lock filesMutex;
Condition OKtoWrite, OKtoRead;
void startRead () {
filesMutex.Acquire( );
while ( busy )
OKtoRead.Wait(&filesMutex);
numReaders++;
filesMutex.Release( );}
void endRead () {
numReaders--;
if (numReaders == 0)
OKtoWrite.Signal(&filesMutex);
void startWrite() {
filesMutex.Acquire( );
while (busy || numReaders != 0) OKtoWrite.Wait(&filesMutex);
busy = true;
filesMutex.Release( );}
void endWrite() {
busy = false;
OKtoRead.Broadcast(&filesMutex); OKtoWrite.Signal(&filesMutex);

8. Guidelines for Choosing Lock Granularity
1. Keep critical sections short. Push “noncritical” statements outside of critical sections to reduce contention.
2. Limit lock overhead. Keep to a minimum the number of times mutexes are acquired and released.
Note tradeoff between contention and lock overhead.
3. Use as few mutexes as possible, but no fewer.
Choose lock scope carefully: if the operations on two different data structures can be separated, it may be more efficient to synchronize those structures with separate locks.
Add new locks only as needed to reduce contention. “Correctness first, performance second!”

9. Semaphore Solution int readCount = 0; semaphore mutex = 1;
semaphore writeBlock = 1;

10. Writer(){ while(TRUE){ other stuff; P(writeBlock); access resource;
Reader(){
while (TRUE) {
other stuff;
P(mutex);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex); }}
Writer(){
while(TRUE){
other stuff;
P(writeBlock);
access resource;
V(writeBlock); }}

11. Attempt at Writer Priority
int readCount = 0, writeCount = 0;
semaphore mutex1 = 1, mutex2 = 1;
semaphore readBlock = 1;
semaphore writeBlock = 1;

12. Reader2 Reader1 Writer1 Reader(){ while (TRUE) { other stuff;
P(readBlock);
P(mutex1);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex1);
V(readBlock);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex1); }}
Writer(){
while(TRUE){
other stuff;
P(mutex2);
writeCount = writeCount +1;
if (writeCount == 1)
P(readBlock);
V(mutex2);
P(writeBlock);
access resource;
V(writeBlock);
writeCount - writeCount - 1;
if (writeCount == 0)
V(readBlock); }}
Reader1
Writer1

13. Reader2 Writer1 Reader1 Reader(){ while (TRUE) { other stuff;
P(readBlock);
P(mutex1);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex1);
V(readBlock);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex1); }}
Writer(){
while(TRUE){
other stuff;
P(mutex2);
writeCount = writeCount +1;
if (writeCount == 1)
P(readBlock);
V(mutex2);
P(writeBlock);
access resource;
V(writeBlock);
writeCount - writeCount - 1;
if (writeCount == 0)
V(readBlock); }}
Writer1
Reader1

14. Reader(){
while (TRUE) {
other stuff;
P(writePending);
P(readBlock);
P(mutex1);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex1); V(readBlock);
V(writePending);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex1); }}
Writer(){
while(TRUE){
other stuff;
P(mutex2);
writeCount = writeCount +1;
if (writeCount == 1)
P(readBlock);
V(mutex2);
P(writeBlock);
access resource;
V(writeBlock);
writeCount - writeCount - 1;
if (writeCount == 0)
V(readBlock); }}

15. Reader2 Writer1 Reader1 Assumptions about semaphore semantics?
while (TRUE) {
other stuff;
P(writePending);
P(readBlock);
P(mutex1);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex1); V(readBlock);
V(writePending);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex1); }}
Writer(){
while(TRUE){
other stuff;
P(mutex2);
writeCount = writeCount +1;
if (writeCount == 1)
P(readBlock);
V(mutex2);
P(writeBlock);
access resource;
V(writeBlock);
writeCount - writeCount - 1;
if (writeCount == 0)
V(readBlock); }}
Writer1
Reader1
Assumptions about semaphore semantics?

16. Reader(){
while (TRUE) {
other stuff;
P(writePending);
P(readBlock);
P(mutex1);
readCount = readCount +1;
if(readCount == 1)
P(writeBlock);
V(mutex1); V(readBlock);
V(writePending);
access resource;
readCount = readCount -1;
if(readCount == 0)
V(writeBlock);
V(mutex1); }}
Writer(){
while(TRUE){
other stuff;
P(mutex2);
writeCount = writeCount +1;
if (writeCount == 1)
P(readBlock);
V(mutex2);
P(writeBlock);
access resource;
V(writeBlock);
writeCount - writeCount - 1;
if (writeCount == 0)
V(readBlock); }}
Reader2
Writer1
Reader1
Assume the writePending semaphore was omitted. What would happen?

17. Assume the writePending semaphore was omitted in the solution just given. What would happen?
This is supposed to give writers priority. However, consider the following sequence:
Reader 1 arrives, executes thro’ P(readBlock);
Reader 1 executes P(mutex1);
Writer 1 arrives, waits at P(readBlock);
Reader 2 arrives, waits at P(readBlock);
Reader 1 executes V(mutex1); then V(readBlock);
Reader 2 may now proceed…wrong

18. Linux Reader/Writer Spinlocks
Class of reader/writer problems
Multiple readers OK
Mutual exclusion for writers
No upgrade from reader lock to writer lock
Favors readers – starvation of writers possible
rwlock_t
read_lock,read_unlock
read_lock_irq // also unlock
read_lock_irqsave
read_unlock_irqrestore
write_lock,write_unlock
//_irq,_irqsave,_irqrestore
write_trylock
rw_is_locked

19. Linux Reader/Writer Semaphores
All reader / writer semaphores are mutexes (usage count 1)
Multiple readers, solo writer
Uninterruptible sleep
Possible to downgrade writer to reader
down_read
down_write
up_read
up_write
downgrade_writer
down_read_trylock
down_write_trylock

20. Message-Passing

21. Interprocess Communication - Messages
Assume no explicit sharing of data elements in the address spaces of processes wishing to cooperate/communicate.
Essence of message-passing is copying (although implementations may avoid actual copies whenever possible).
Problem-solving with messages - has a feel of more active involvement by participants.

22. Issues
System calls for sending and receiving messages with the OS(s) acting as courier.
Variations on exact semantics of primitives and in the definition of what comprises a message.
Naming - direct (to/from pids), indirect (to distinct objects - e.g., mailboxes, ports, sockets)
How do unrelated processes “find” each other?
Buffering - capacity and blocking semantics.
Guarantees - in-order delivery? no lost messages?

23. Send and Receive

24. UNIX Sockets for Client/ Server Message Passing
1. Create a named socket syscalls: sfd = socket(…) bind (sfd, ptr,len)
2. Listen for clients listen(sfd,numpend)
4. Connection made and continue listening cfd=accept(sfd, …)
5. Exchange data write(cfd, …)
6. Done: close(cfd);
7. Really done: close(sfd);
Client
3. Create unnamed socket & ask for connection syscalls: cfd=socket(…) err=connect(cfd, ptr, …)
5. Exchange data read(cfd, …)
6. Done: close(cfd);
name
Sockets have domain (single unix machine, internet), type (stream-sequenced, reliable, 2-way connection, variable length streams of bytes; or dgram-connectionless, unreliable, fix-length msgs), and protocols
Bind associates a string with the sfd (creates a pathname in file system)
Ptr is addr of struct containing pathname string (can’t already exist – unlink a name first) if single machine
If internet socket it contains port number and internet addr
Accept creates a new unnamed socket
name
In a child process of server

25. 5 DP – Direct Send/Receive Message Passing Between Philosophers
Fork please?
Philosopher 0 (thinking)
Philosopher 4
Philosopher 1
Philosopher 3
(eating)
Philosopher 2

26. Umm. Oh yeah. Philosopher 0 (thinking) Philosopher 4 Philosopher 1
(eating)
Philosopher 2

27. Fork please? Philosopher 0 (thinking) Philosopher 4 Philosopher 1
(eating)
Philosopher 2

28. Fork please? Philosopher 0 (thinking) Philosopher 4 I’ll ignore
that request
until I’m done
Philosopher 1
Philosopher 3
(eating)
Philosopher 2

29. Fork please? Fork please? Philosopher 0 (thinking) Philosopher 4
(eating)
Fork please?
Philosopher 2

30. Start here
Client / Server
server->

31. Example: Time Service

32. Example: Time Service via Messages

33. Client / Server with Threads

34. Hiding Message-Passing: RPC

35. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
Receive
result = foo(param);

36. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
please do
foo for P0
with param
Receive
result = foo(param);
blocked
here

37. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
please do
foo for P0
with param
Receive
result = foo(param);
r = foo(param);
// actual call
blocked
here

38. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
Receive
result = foo(param);
r = foo(param);
// actual call
blocked
here
returning
r to P0
Reply

39. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
Receive
result = foo(param);
r = foo(param);
// actual call
returning
r to P0
Reply

40. Remote Procedure Call - RPC
Looks like a nice familiar procedure call P1 P0
Receive
result = foo(param);
r = foo(param);
// actual call
Reply

41. 5DP via RPC with Fork Manager
Looks like a nice familiar procedure call
Fork Server
Philosopher0
Receive
result = PickupForks (0);
r = proc(param);
// explicit queuing
when necessary
Reply

42. Example: Time Service via RPC

43. RPC Issues