Computer Science Grade 12 20 min

Concurrency Control: Locks, Semaphores, and Monitors

Learn about concurrency control mechanisms like locks, semaphores, and monitors to manage concurrent access to shared resources in a distributed environment.

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1

Introduction & Learning Objectives

Learning Objectives Define concurrency, race conditions, and the critical section problem. Implement mutual exclusion using locks (mutexes) to prevent data corruption. Differentiate between binary and counting semaphores and apply them to resource allocation problems. Describe how monitors encapsulate shared data and synchronization logic to simplify concurrent programming. Analyze code snippets to identify potential race conditions and propose a correct synchronization solution. Design a solution to the classic Producer-Consumer problem using semaphores or monitors. Compare and contrast the use cases, advantages, and disadvantages of locks, semaphores, and monitors. Ever wondered how you and a friend can edit the same Google Doc simultaneously without overwriting each oth...
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Key Concepts & Vocabulary

TermDefinitionExample Race ConditionAn error condition where the outcome of a program depends on the unpredictable sequence or timing of operations from multiple threads accessing shared data.Two threads try to increment a shared counter `count` (initially 5). Thread A reads `count` (5), Thread B reads `count` (5). A computes `5+1=6` and writes `6` to `count`. B then computes `5+1=6` and also writes `6`. The final result is 6, when it should have been 7. Critical SectionA segment of code that accesses a shared resource and must not be executed by more than one thread at a time to prevent race conditions.In the code `count++`, the sequence of reading the value of `count`, incrementing it, and writing it back is the critical section. Mutual Exclusion (Mutex)A property that ensures only one...
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Core Syntax & Patterns

Lock/Mutex Pattern for Mutual Exclusion lock.acquire(); // --- Critical Section Start --- // Access shared resource // --- Critical Section End --- lock.release(); This is the fundamental pattern for protecting a block of code. A thread must acquire the lock before entering the critical section. Crucially, it must release the lock when it leaves, allowing other waiting threads to proceed. Semaphore P/V (Wait/Signal) Operations wait(S): Decrements semaphore S. If S becomes negative, the thread blocks. signal(S): Increments semaphore S. If S is not positive, a blocked thread is unblocked. Used to control access to a pool of N resources. `wait()` is called to request a resource, and `signal()` is called to release one. This pattern is the foundation for solving many complex syn...

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Sample Practice Questions

Challenging
In a monitor-based solution to the Producer-Consumer problem, what is the correct logic for the producer thread when it finds the buffer is full?
A.It should repeatedly check the buffer status in a tight loop until a slot is free (busy-wait).
B.It should call `notFull.wait()` on a condition variable, causing it to block and release the monitor lock.
C.It should exit with an error, as a full buffer is a critical failure.
D.It should call `notEmpty.signal()` to wake up a consumer, hoping it will free up space.
Challenging
When comparing semaphore-based and monitor-based solutions to the Producer-Consumer problem, what is a key advantage of the monitor-based approach?
A.The monitor solution is always faster due to fewer context switches.
B.The monitor solution is less prone to programmer error because mutual exclusion and condition signaling are encapsulated and more structured.
C.The semaphore solution cannot handle buffers of variable size, whereas the monitor solution can.
D.The semaphore solution requires three semaphores, while the monitor solution requires only one condition variable.
Challenging
A developer tries to fix a race condition on a shared object `data` by having two threads, T1 and T2, use locks. T1 uses `lock1.acquire()` and T2 uses `lock2.acquire()` before accessing `data`. Why is this solution fundamentally flawed?
A.Using two different locks is less efficient than using a single lock.
B.This will cause a deadlock because the threads will wait for each other's locks.
C.It fails to provide mutual exclusion because the threads are not locking on the same object.
D.The solution is correct, as each thread has its own lock to manage its access.

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