So let's be polite and really take turns. None of this wanting stuff.

Initially turn=1

Code for P1                      Code for P2

Loop forever {                   Loop forever {
    while (turn = 2) {}              while (turn = 1) {}
    critical-section                 critical-section
    turn <-- 2                       turn <-- 1
    non-critical-section }           non-critical-section }

This one forces alternation, so is not general enough. Specifically, it does not satisfy condition three, which requires that no process in its non-critical section can stop another process from entering its critical section. With alternation, if one process is in its non-critical section (NCS) then the other can enter the CS once but not again.

In fact, it took years (way back when) to find a correct solution. Many earlier ``solutions'' were found and several were published, but all were wrong. The first correct solution was found by a mathematician named Dekker, who combined the ideas of turn and wants. The basic idea is that you take turns when there is contention, but when there is no contention, the requesting process can enter. It is very clever, but I am skipping it (I cover it when I teach distributed operating systems in V22.0480 or G22.2251). Subsequently, algorithms with better fairness properties were found (e.g., no task has to wait for another task to enter the CS twice).

What follows is Peterson's solution, which also combines turn and wants to force alternation only when there is contention. When Peterson's solution was published, it was a surprise to see such a simple soluntion. In fact Peterson gave a solution for any number of processes. A proof that the algorithm satisfies our properties (including a strong fairness condition) for any number of processes can be found in Operating Systems Review Jan 1990, pp. 18-22.

Initially P1wants=P2wants=false  and  turn=1

Code for P1                        Code for P2

Loop forever {                     Loop forever {
    P1wants <-- true                   P2wants <-- true
    turn <-- 2                         turn <-- 1
    while (P2wants and turn=2) {}      while (P1wants and turn=1) {}
    critical-section                   critical-section
    P1wants <-- false                  P2wants <-- false
    non-critical-section               non-critical-section

Hardware assist (test and set)

The wordatomically means that the two actions performed by TAS(x) (testing, i.e., returning the old value of x and setting , i.e., assigning true to x) are inseparable. Specifically it is not possible for two concurrent TAS(x) operations to both return false (unless there is also another concurrent statement that sets x to false).

With TAS available implementing a critical section for any number of processes is trivial.

loop forever {
    while (TAS(s)) {}   ENTRY
    CS
    s<--false           EXIT
    NCS

2.3.4: Sleep and Wakeup

Remark: Tanenbaum does both busy waiting (as above) and blocking (process switching) solutions. We will only do busy waiting, which is easier. Sleep and Wakeup are the simplest blocking primitives. Sleep voluntarily blocks the process and wakeup unblocks a sleeping process. We will not cover these.

Homework: Explain the difference between busy waiting and blocking.

2.3.5: Semaphores

Remark: Tannenbaum use the term semaphore only for blocking solutions. I will use the term for our busy waiting solutions. Others call our solutions spin locks.

P and V and Semaphores

The entry code is often called P and the exit code V. Thus the critical section problem is to write P and V so that

loop forever
    P
    critical-section
    V
    non-critical-section

1. Mutual exclusion.
2. No speed assumptions.
3. No blocking by processes in NCS.
4. Forward progress (my weakened version of Tanenbaum's last condition).

Note that I use indenting carefully and hence do not need (and sometimes omit) the braces {} used in languages like C or java.

A binary semaphore abstracts the TAS solution we gave for the critical section problem.

• A binary semaphore S takes on two possible values ``open'' and ``closed''.
• Two operations are supported
• P(S) is

      while (S=closed) {}
      S<--closed     <== This is NOT the body of the while
  

  where finding S=open and setting S<--closed is atomic
• That is, wait until the gate is open, then run through and atomically close the gate
• Said another way, it is not possible for two processes doing P(S) simultaneously to both see S=open (unless a V(S) is also simultaneous with both of them).
• V(S) is simply S<--open

The above code is not real, i.e., it is not an implementation of P. It is, instead, a definition of the effect P is to have.

To repeat: for any number of processes, the critical section problem can be solved by

loop forever
    P(S)
    CS
    V(S)
    NCS

The only specific solution we have seen for an arbitrary number of processes is the one just above with P(S) implemented via test and set.

Remark: Peterson's solution requires each process to know its processor number. The TAS soluton does not. Moreover the definition of P and V does not permit use of the processor number. Thus, strictly speaking Peterson did not provide an implementation of P and V. He did solve the critical section problem.

To solve other coordination problems we want to extend binary semaphores.

• With binary semaphores, two consecutive Vs do not permit two subsequent Ps to succeed (the gate cannot be doubly opened).
• We might want to limit the number of processes in the section to 3 or 4, not always just 1.

Both of the shortcomings can be overcome by not restricting ourselves to a binary variable, but instead define a generalized or counting semaphore.

• A counting semaphore S takes on non-negative integer values
• Two operations are supported
• P(S) is

      while (S=0) {}
      S--
  

  where finding S>0 and decrementing S is atomic
• That is, wait until the gate is open (positive), then run through and atomically close the gate one unit
• Another way to describe this atomicity is to say that it is not possible for the decrement to occur when S=0 and it is also not possible for two processes executing P(S) simultaneously to both see the same necessarily (positive) value of S unless a V(S) is also simultaneous.
• V(S) is simply S++

These counting semaphores can solve what I call the semi-critical-section problem, where you premit up to k processes in the section. When k=1 we have the original critical-section problem.

initially S=k

loop forever
    P(S)
    SCS   <== semi-critical-section
    V(S)
    NCS

Producer-consumer problem

• Two classes of processes
  • Producers, which produce times and insert them into a buffer.
  • Consumers, which remove items and consume them.
• What if the producer encounters a full buffer?
  Answer: It waits for the buffer to become non-full.
• What if the consumer encounters an empty buffer?
  Answer: It waits for the buffer to become non-empty.
• Also called the bounded buffer problem.
  • Another example of active entities being replaced by a data structure when viewed at a lower level (Finkel's level principle).

Initially e=k, f=0 (counting semaphore); b=open (binary semaphore)

Producer                         Consumer

loop forever                     loop forever
    produce-item                     P(f)
    P(e)                             P(b); take item from buf; V(b)
    P(b); add item to buf; V(b)      V(e)
    V(f)                             consume-item

• k is the size of the buffer
• e represents the number of empty buffer slots
• f represents the number of full buffer slots
• We assume the buffer itself is only serially accessible. That is, only one operation at a time.
  • This explains the P(b) V(b) around buffer operations
  • I use ; and put three statements on one line to suggest that a buffer insertion or removal is viewed as one atomic operation.
  • Of course this writing style is only a convention, the enforcement of atomicity is done by the P/V.
• The P(e), V(f) motif is used to force ``bounded alternation''. If k=1 it gives strict alternation.