Herbie Murray
The NotifyAll method is used to wake up a thread. However, there are instances where the NotifyAll method does not wake up a thread, which is usually due to a timing issue. For example, if the first notify is sent too soon while the waiting thread is not in the waiting state, the NotifyAll method will not work.
| Characteristics | Values |
|---|---|
| Waking a sleeping thread | Takes a significant amount of CPU cycles |
| notifyAll method | Does not wake up a thread |
| Issue | Timing |
| Solution | Use a semaphore or something similar |
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Timing issues
On multicore machines, waiting threads are usually put into a polling loop first for a short time, to allow other threads to fulfil the condition. This can help to optimise CPU cycles.
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Using a semaphore
The NotifyAll method does not always wake up a thread. This is usually a timing issue, where one thread isn't actually waiting yet. In such cases, it is recommended to use a semaphore or something similar.
A semaphore is an integer variable, shared among multiple processes. The main aim of using a semaphore is process synchronization and access control for a common resource in a concurrent environment. The initial value of a semaphore depends on the problem at hand. Usually, we use the number of resources available as the initial value. For example, if there are three identical resources, the semaphore would be initialized to three.
Semaphores are typically used to coordinate access to resources, with the semaphore count initialized to the number of free resources. Threads then atomically increment the count when resources are added and atomically decrement the count when resources are removed. This helps to ensure that only one process can access a shared resource at a time.
Binary semaphores, also known as mutex locks, can have only two values – 0 and 1. They are used to implement the solution of critical section problems with multiple processes and a single resource. Counting semaphores can be used to control access to a given resource consisting of a finite number of instances.
It is important to note that a semaphore cannot solve a multiple identical resource problem on its own. It requires other state information, which is typically a shared resource protected via a separate mutex.
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Polling loops
On multicore machines, waiting threads are usually put into a polling loop for a short time to allow other threads to fulfil the condition. This means that arriving threads can proceed immediately, skipping the sleeping phase altogether.
However, there can be issues with timing. For example, if the first notify is sent too soon while the waiting thread is not in the waiting state, the NotifyAll method may not wake up a thread. This can be fixed by using variables to check if the thread is ready to be used, or by using a semaphore.
In some cases, placing the try { ... } catch ( InterruptedException e ) { ...} directly within the methods around this.wait and Thread.currentThread.sleep(1000) can result in deadlock behaviour. However, if the exception is caught inside the lambdas in the main method that makes the calls to the methods, it runs without any problems.
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CPU cycles
Putting a thread to sleep and waking it up takes a significant amount of CPU cycles. This is why it is an optimisation to allow an arriving thread to proceed immediately, skipping the sleeping phase altogether.
On multicore machines, waiting threads are usually put into a polling loop first for a short time, to allow other threads to fulfil the condition. This means that they are only put to sleep when no notification happens during the polling loop.
In Java, the NotifyAll method does not always wake up a thread. This is usually a timing issue, where the first notify is sent too soon while the waiting thread is not in the waiting state. This can be fixed by using a semaphore or something similar.
> public synchronized void methodB() {
> try {
> this.notifyAll();
> Thread.sleep(1000);
> this.notifyAll();
> System.out.println("Notified...");
> } catch (InterruptedException ex) {
> Logger.getLogger(NotifyAllTesting.class.getName()).log(Level.SEVERE, null, ex);
> System.out.println(ex);
> }
> }
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Deadlock behaviour
The NotifyAll method does not always wake up a thread. This is usually due to a timing issue, where the first notify is sent too soon while the waiting thread is not in the waiting state. This can be fixed by using a variable to check if the thread is ready to be used. For example, a semaphore is a good solution.
On multicore machines, waiting threads are usually put into a polling loop first for a short time, to allow other threads to fulfil the condition. This is an optimisation that will pay off, as putting a thread to sleep and waking it up takes a significant amount of CPU cycles.
If you place the try { ... } catch ( InterruptedException e ) { ... } directly within methodA and methodB around this.wait and Thread.currentThread.sleep(1000), you may get a deadlock behaviour. However, if you catch the exception inside the lambdas in the main method that makes the calls to methodA and methodB, it will run without any problems.
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Frequently asked questions
No, NotifyAll cannot wake a sleeping thread.
It is likely a timing issue. One thread isn't actually waiting yet.
You could make use of variables and check if the thread you started before is ready to be used. A semaphore is a good solution.
A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system.
