Synchronization: semaphores and some more stuff 1 Operating Systems, 2014, Meni Adler, Danny Hendler...

Synchronization: semaphoresSynchronization: semaphoresand some more stuffand some more stuff

1Operating Systems, 2014, Meni Adler, Danny Hendler & Amnon Meisels

What's wrong with busy waiting?What's wrong with busy waiting?

Doesn't make sense for uni-processoro May cause priority inversion and deadlock

Wastes CPU timeo But is efficient if waiting-time is short

The mutual exclusion algorithms we saw used busy-waiting. What’s wrong with that?


What's wrong with busy waiting? What's wrong with busy waiting?

Busy waiting may cause priority-inversion and deadlockBusy waiting may cause priority-inversion and deadlock

Process A's priority is higher than process B's Process B enters the CS Process A needs to enter the CS, busy-waits for B to exit the CS Process B cannot execute as long as the higher-priority process A is executing/ready

Priority inversion and deadlock resultPriority inversion and deadlock result3Operating Systems, 2014, Meni Adler, Danny Hendler & Amnon Meisels

OutlineOutline

Semaphores and the producer/consumer problem Counting semaphores from binary semaphores Event counters and message passing synchronization


Semaphores

up(S) [the `v’ operation] If there are blocked

processes, wake-up one of them

Else S++

down(S) [the ‘p’ operation] If S≤0 the process is blocked.

It will resume execution only after it is woken-up

Else S--

Two atomic operations are supported by a semaphore S:

S is non-negative

Supported by Windows, Unix, …


Semaphores: is the following correct?

up(S) [the `v’ operation] S++ If there are blocked

processes, wake-up one of them

down(S) [the ‘p’ operation] If S≤0 the process is blocked.

It will resume execution only after it is woken-up

S--



7

Consider the following bad scneario:Consider the following bad scneario:

S=0 and process A performs down(S) down(S) – A is blocked Process B performs up(S)up(S) – S=1 S=1 A is ready Process C performs down(S) down(S) – S=0 & C proceedsS=0 & C proceeds Process A gets a time-slice and proceeds – S=0 S=0

A single up() freed 2 down()sA single up() freed 2 down()sOperating Systems, 2011, Danny Hendler & Amnon Meisels

Pseudo-code in previous slide is wrong

Implementing mutex with semaphores

Shared data: semaphore lock; /* initially lock = 1 */

down(lock)

Critical section

up(lock)

Does the algorithm satisfy mutex?Does it satisfy deadlock-freedom?Does it satisfy starvation-freedom?

YesYesDepends…


Semaphore as a General Synchronization ToolSemaphore as a General Synchronization Tool

Execute B in Pj only after A executed in Pi

Use semaphore flag initialized to 0 Code:

Pi Pj

… …A down(flag)

up(flag) B

Execute B in Pj only after A executed in Pi

Use semaphore flag initialized to 0 Code:

Pi Pj

… …A down(flag)

up(flag) B

0

time


More on synchronization using semaphoresMore on synchronization using semaphores

Three processes p1; p2; p3

semaphores s1 = 1, s2 = 0;p1p1 p2p2 p3p3down(s1s1); down(s2s2); down(s2s2);A B Cup(s2s2); up(s2s2); up(s1s1);

Which execution orders of A, B, C, are possible?

(A B* C)*10Operating Systems, 2014, Meni Adler, Danny Hendler & Amnon Meisels

P0 P1

down(S); down(Q); down(Q); down(S);

move1 move2up(S); up(Q);up(Q) up(S);

1

1

Example: move money between two accounts which are protected by semaphores S and Q

No guarantee for correct synchronization No guarantee for correct synchronization

Does this work?Deadlock!


Negative-valued semaphoresNegative-valued semaphores

up(S) S++ If there are blocked

processes (i.e. S≤0), wake-up one of them

-3

down(S) S-- If S<0 the process is blocked.

It will resume execution only when S is non-negative


If S is negative, then there are –S blocked If S is negative, then there are –S blocked processesprocesses


type semaphore = recordvalue: integer; L: list of process;

end;

Negative semaphore ImplementationNegative semaphore Implementation

-3

atomic down(S):S.value--;if (S.value < 0) { add this process to S.L; sleep; }

atomic up(S): S.value++;if (S.value <= 0) { remove a process P from S.L; wakeup(P); }


L

Implementing a spin-lock with TSLImplementing a spin-lock with TSL

In user space, one can use TSL (test-set-lock)mutex_lock:mutex_lock:

TSL REG, mutexCMP REG, #0JZE okokCALL thread_yieldthread_yield

JMP mutex_lockok:ok:

RET

mutex_unlock:mutex_unlock:MOV mutex, #0RET


type semaphore = recordvalue, flag: integer; L: list of process;

end;

-3

down(S):repeat until test-and-set(S.flag) S.value--; if (S.value < 0) { add this process to S.L; S.flag=0 sleep; } else S.flag=0

Implementing a negative semaphore with TSLImplementing a negative semaphore with TSL

up(S):repeat until test-and-set(S.flag) S.value++; if (S.value <= 0) { remove a process P from S.L; wakeup(P); }

S.flag=0


L

Any problem with this code?In down(), resetting flag and sleeping should be atomic.

More on semaphore implementationMore on semaphore implementation On a uni-processor, disabling interrupts may be used

TSL implementation works for multi-processors

On a multi-processor, we can use spin-lock mutual exclusion to protect semaphore access

Why is this better than busy-waiting in the 1st place?

Busy-waiting is now guaranteed to be very short


Producer-Consumer Problem

Paradigm for cooperating processes, • producer process produces

information that is consumed by a consumer process

Two versions• unbounded-buffer places no

practical limit on the size of the buffer

• bounded-buffer assumes that there is a fixed buffer size

buff

er

in

out


2

6

Out

In

item1

item4

item3

item2

consumer

0buffer

producer

Bounded Buffer

1

2

3

4

5

7


Implementation using semaphoresImplementation using semaphores

Two processes or more use a shared buffer in memory

The buffer has finite size(i.e., it is boundedbounded)

The producer writes to the buffer and the consumer reads from it

A full bufferfull buffer stops the producer

An empty bufferempty buffer stops the consumer


Producer-ConsumerProducer-Consumer implementation with semaphores#define N 100 /* Buffer size */typedef int semaphoresemaphore;semaphoresemaphore mutex = 1; /* access control to critical section */semaphoresemaphore empty = N; /* counts empty buffer slots */semaphoresemaphore full = 0; /* counts full slots */

void producer(void) {int item;while(TRUE) {produce_item(&item); /* generate something... */downdown(&empty); /* decrement count of empty */downdown(&mutex); /* enter critical section */enter_item(item); /* insert into buffer */upup(&mutex); /* leave critical section */upup(&full); /* increment count of full slots */} }


void consumerconsumer(void){

int item;

while(TRUE) { downdown(&full); /* decrement count of full */ downdown(&mutex); /* enter critical section */ remove_item(&item); /* take item from buffer) */ upup(&mutex); /* leave critical section */ upup(&empty); /* update count of empty */ consume_item(item); /* do something... */

}}


Producer-ConsumerProducer-Consumer implementation with semaphores

Outline




Assumes only values 0 or 1

Wait blocks if semaphore=0

Signal (up operation) either wakes up a waiting process, if there is one, or sets value to 1 (if value is already 1, signal is “wasted”)

How can we implement a counting semaphore by using binary semaphores?

Binary SemaphoreBinary Semaphore

Implementing a counting semaphore with binary Implementing a counting semaphore with binary semaphores (user space): semaphores (user space): take 1take 1

down(S):down(S):down(S1);SS.value--;if(SS.value < 0){

up(S1); down(S2); }

else up(S1);

binary-semaphore S1 initially 1, S2 initially 0, S.value initially 1

up(S):up(S):down(S1);SS.value++;if(SS.value ≤ 0)

up(S2); up(S1)

This code does not work. Why?


L1:L2:

Race condition for counting semaphore Race condition for counting semaphore take 1take 1

1. Processes Q1 – Q4 perform down(S), Q2 – Q4 are preempted between lines

L1 and L2: the value of the counting semaphore is now -3

2. Processes Q5-Q7 now perform up(S): the value of the counting semaphore is

now 0

3. Now, Q2-Q4 wake-up in turn and perform line L2 (down S2)

4. Q2 runs but Q3-Q4 block.


There is a discrepancy between the value of S and the There is a discrepancy between the value of S and the number of processes waiting on itnumber of processes waiting on it

Implementing a counting semaphore with binary Implementing a counting semaphore with binary semaphores (user space): semaphores (user space): take 2take 2

down(S):down(S):down(S1);SS.value--;if(SS.value < 0){up(S1); //L1down(S2); } //L2up(S1);

up(S):up(S):down(S1);SS.value++;if(SS.value ≤ 0)

up(S2); else up(S1)

Does this code work?


binary-semaphore S1 initially 1, S2 initially 0, S.value initially 1

up(S1) is performed by up(S) only if no process waits on S2only if no process waits on S2 Q5 leaves up(S) without releasing S1

Q6 cannot enter the critical section that protects the counter

It can only do so after one of Q2-Q4 releases S1

This generates a “lock-step” situation: an up(), a down(), an up()…

The critical section that protects the counter is entered The critical section that protects the counter is entered

alternately by a producer or a consumeralternately by a producer or a consumer

The effect of the added ‘else’The effect of the added ‘else’


Recall the bounded-buffer algorithmRecall the bounded-buffer algorithm#define N 100typedef int semaphoresemaphore;semaphoresemaphore mutex = 1;semaphoresemaphore empty = N;semaphoresemaphore full = 0;void producerproducer(void) {

int item;while(TRUE) {produce_item(&item);downdown(&empty);downdown(&mutex);enter_item(item);upup(&mutex);upup(&full);} }

void consumerconsumer(void){

intitem;

while(TRUE) { downdown(&full); downdown(&mutex); remove_item(&item); upup(&mutex); upup(&empty); consume_item(item);

}}


A Problematic Scheduling Scenario

Consider a Bounded buffer of 5 slots.Assume there are 6 processes each filling five slots in turn.

Empty.Value = 521

3 4

5 6



1. 1. five slots are filled by the first producer

Empty.Value = 021

3 4

5 6


A Problematic Scheduling Scenario1. 1. The second producer is blocked

Empty.Value = -121

3 4

5 6



1. 1. The third producer is blocked

Empty.Value = -221

3 4

5 6



11. The fourth producer is blocked

Empty.Value = -321

3 4

5 6



1. 1. The fifth producer is blocked

Empty.Value = -421

3 4

5 6



2. 2. All blocked producers are waiting on S2

Empty.Value = -521

3 4

5 6



3. 3. The consumer consumes an item and is blocked on Empty.S1 until a producer adds an item.

Empty.Value = -521

3 4

5 6



3. 3. The consumer consumes an item and is blocked on S1 , one producer adds an item.

Empty.Value = -421

3 4

5 6



44. Consumer must consume, only then another producer wakes up and produces an item

Empty.Value = -321

3 4

5 6



4. 4. Same as in step 3.

Empty.Value = -221

3 4

5 6



5. 5. And again…

Empty.Value = -121

3 4

5 6


Implementing a counting semaphore with binary Implementing a counting semaphore with binary semaphores (user space): semaphores (user space): take 3 take 3 (P.A. Kearns, 1988)(P.A. Kearns, 1988)

down(S)down(S)down(S1);SS.value--;if(SS.value < 0){

up(S1); //L1down(S2); //L2down(S1);SS.wake--; if(SS.wake > 0) then up(S2);} //L3

up(S1);

up(S):up(S):down(S1);SS.value++;if(SS.value <= 0) {

SS.wake++; up(S2); }up(S1);

binary-semaphore S1=1, S2=0, value initially 1, integer wake=0

Does THIS work?


Correctness arguments (Kearns)…

The counter S.wakeS.wake is used when processes performing down(S)down(S) are preempted between lines L1 and L2

In such a case, up(S2) performed by processes during up(S)up(S) has no effect

However, these processes accumulate their waking signals on the (protected) counter S.wakeS.wake

After preemption is over, any single process that wakes up from its block on down(S2) checks the value of S.wakeS.wake

The check is again protected For each count of the wake-up signals, the awakened process

performs the up(S2) (in line L3) Each re-scheduled process wakes up the next oneEach re-scheduled process wakes up the next one


Kearns' algorithm is wrongKearns' algorithm is wrong

Processes P0..P7 perform down(S),down(S), P0 goes through, P1..P7 are preempted just

after line L2 of the operation

Processes P8..P11 perform up(S)up(S) and their up(S2) operations release, say, P1..P4

Processes P5, P6, P7 are still waiting on S2 and S.wake = 4S.wake = 4

Processes P1..P4 are ready, just before line L3

Each of P1..P3 will decrement S.wakeS.wake in its turn, check that it's positive and

signal one of P5..P7

Four Four upup operations have released 7 operations have released 7 down down operations operations


Implementing a counting semaphore with binary Implementing a counting semaphore with binary semaphores (user space): semaphores (user space): take 4 take 4 (Hemmendinger, 1989)(Hemmendinger, 1989)

down(S)down(S)down(S1);SS.value--;if(SS.value < 0){

up(S1);down(S2);down(S1);SS.wake--; if(SS.wake > 0) then up(S2);} // L3

up(S1);

up(S):up(S):down(S1);SS.value++;if(SS.value <= 0) {

SS.wake++; if (S.wake == 1) up(S2); }up(S1);

binary-semaphore S1=1, S2=0, integer wake=0

This works


Implementing a counting semaphore with binary Implementing a counting semaphore with binary semaphores (user space): semaphores (user space): take 5take 5 (Barz, 1983)(Barz, 1983)

down(S)down(S)down(S2);down(S1);

S.value--; if (S.value>0) then up(S2); up(S1);

up(S):up(S):down(S1);S.value++;if(S.value == 1) {

up(S2); }up(S1);

binary-semaphore S1=1, S2=min(1, init_value), value=init_value

This works, is simpler, and was published earlier …(!)

Can we switch the order of downs in down(S)? 45Operating Systems, 2014, Meni Adler, Danny Hendler & Amnon Meisels

Correctness arguments…Correctness arguments… The critical section is guarded by S1 and each of the operations

down(SS) and up(SS) uses it to correctly update the value of SS.value After updating (and inside the critical section) both operations

release the S2 semaphore only if value is positive SS.value is never negative, because any process performing

down(SS) is blocked at S2

Signals cannot be 'wasted' Signals cannot be 'wasted'


Fairness of semaphores

Order of releasing blocked processes:o Weak – up() performing process enters after enters after (one of the)

blocked processeso Strong – An upper bound on the number of entries of

process that performed up() if others are waiting Unfair:Unfair:

o No guarantee about the number of times the up() performing process enters before the blocked

o Open competition each time the lock is freeo Imitating the Java 'wait' 'notify' mechanismo Or the spin-lock of XV6…


Outline



Event CountersEvent Counters Integer counters with three operations:

o Advance(E): increment E by 1, wake up relevant sleepers

o Await(E,v): wait until E ≥ v. Sleep if E < vSleep if E < v

o Read(E): return the current value of E

Counter value is ever increasingCounter value is ever increasing

The Read() operation is not required for the bounded-buffer implementation in the next slide


producer-consumer with Event Countersproducer-consumer with Event Counters(for a single producer and a single consumer)(for a single producer and a single consumer)

#define N 100typedef int event_counter;event_counter in = 0; /* counts inserted items */event_counter out = 0; /* items removed from buffer */

void producer(void){int item, sequence = 0;

while(TRUE) {produce_item(&item);sequence = sequence + 1; /* counts items produced */await(out, sequence - N); /* wait for room in buffer */enter_item(item); /* insert into buffer */advance(&in); /* inform consumer */

}}


Event countersEvent counters (producer-consumer)

void consumer(void){ int item, sequence = 0;

while(TRUE) { sequence = sequence + 1; /* count items consumed */ await(in, sequence); /* wait for item */ remove_item(&item); /* take item from buffer */ advance(&out); /* inform producer */ consume_item(item);

}}


MessageMessage Passing Passing – no shared memory– no shared memory

In a multi-processor system without shared memory, synchronization can be implemented by message passing

Implementation issues:o Acknowledgements may be required (messages may be lost)o Message sequence numbers required to avoid message duplicationo Unique process addresses across CPUs (domains..)o Authentication (validate sender’s identity, a multi-machine environment…)

Two main functions:o send(destination, &message);o receive(source, &message) block while waiting...


Producer-consumer with Message PassingMessage Passing

#define N 100#define MSIZE 4 /* message size */

typedef int message(MSIZE);void producer(void){

int item;message m; /* message buffer */

while(TRUE) {produce_item(&item);receive(consumer, &m); /*wait for an empty */construct_message(&m, item);send(consumer, &m); /* send item */

}}


Message passingMessage passing (cont.) (cont.)

void consumer(void){int item, i;message m;

for(i = 0; i < N; i++) send(producer, &m); /* send N emptiesempties */

while(TRUE) {receive(producer, &m); /* get message with item */extract_item(&m, &item);send(producer, &m); /* send an empty reply */consume_item(item); }

}


Message passing variationsMessage passing variations

Messages can be addressed to a process address or to a mailbox

o Mailboxes are generated with some capacity. When sending a message to a full

mailbox, a process blocks

o Buffer management done by mailbox

Unix pipes - a generalization of messages … no fixed size message (blocking

receive)

If no buffer is maintained by the system, then send and receive must run in

lock-step. Example: Unix rendezvous


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