Operating Systems

================ Start Lecture #6 ================

2.3.2: Critical sections

We must prevent interleaving sections of code that need to be atomic with respect to each other. That is, the conflicting sections need mutual exclusion. If process A is executing its critical section, it excludes process B from executing its critical section. Conversely if process B is executing is critical section, it excludes process A from executing its critical section.

Requirements for a critical section implementation.

  1. No two processes may be simultaneously inside their critical section.

  2. No assumption may be made about the speeds or the number of CPUs.

  3. No process outside its critical section (including the entry and exit code)may block other processes.

  4. No process should have to wait forever to enter its critical section.

2.3.3 Mutual exclusion with busy waiting

The operating system can choose not to preempt itself. That is, we do not preempt system processes (if the OS is client server) or processes running in system mode (if the OS is self service). Forbidding preemption for system processes would prevent the problem above where x<--x+1 not being atomic crashed the printer spooler if the spooler is part of the OS.

But simply forbidding preemption while in system mode is not sufficient.

Software solutions for two processes

Initially P1wants=P2wants=false

Code for P1                             Code for P2

Loop forever {                          Loop forever {
    P1wants <-- true         ENTRY          P2wants <-- true
    while (P2wants) {}       ENTRY          while (P1wants) {}
    critical-section                        critical-section
    P1wants <-- false        EXIT           P2wants <-- false
    non-critical-section }                  non-critical-section }

Explain why this works.

But it is wrong! Why?

Let's try again. The trouble was that setting want before the loop permitted us to get stuck. We had them in the wrong order!

Initially P1wants=P2wants=false

Code for P1                             Code for P2

Loop forever {                          Loop forever {
    while (P2wants) {}       ENTRY          while (P1wants) {}
    P1wants <-- true         ENTRY          P2wants <-- true
    critical-section                        critical-section
    P1wants <-- false        EXIT           P2wants <-- false
    non-critical-section }                  non-critical-section }

Explain why this works.

But it is wrong again! Why?

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.

The first example violated rule 4 (the whole system blocked). The second example violated rule 1 (both in the critical section. The third example violated rule 3 (one process in the NCS stopped another from entering its CS).

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)

TAS(b), where b is a binary variable, ATOMICALLY sets b<--true and returns the OLD value of b.
Of course it would be silly to return the new value of b since we know the new value is true.

The word atomically 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.