Bounded buffer problem

In computing, the producer–consumer problem (also known as the bounded-buffer problem) is a classic example of a multi-process synchronization problem. The problem describes two processes, the producer and the consumer, who share a common, fixed-size buffer used as a queue. The producer's job is to generate data, put it into the buffer, and start again. At the same time, the consumer is consuming the data (i.e., removing it from the buffer), one piece at a time. The problem is to make sure that the producer won't try to add data into the buffer if it's full and that the consumer won't try to remove data from an empty buffer.

The solution for the producer is to either go to sleep or discard data if the buffer is full. The next time the consumer removes an item from the buffer, it notifies the producer, who starts to fill the buffer again. In the same way, the consumer can go to sleep if it finds the buffer to be empty. The next time the producer puts data into the buffer, it wakes up the sleeping consumer. The solution can be reached by means of inter-process communication, typically using semaphores. An inadequate solution could result in a deadlock where both processes are waiting to be awakened. The problem can also be generalized to have multiple producers and consumers.

To solve the problem, a less experienced programmer might come up with a solution shown below. In the solution two library routines are used, sleep and wakeup. When sleep is called, the caller is blocked until another process wakes it up by using the wakeup routine. The global variable itemCount holds the number of items in the buffer.

The problem with this solution is that it contains a race condition that can lead to a deadlock. Consider the following scenario:

Since both processes will sleep forever, we have run into a deadlock. This solution therefore is unsatisfactory.

An alternative analysis is that if the programming language does not define the semantics of concurrent accesses to shared variables (in this case itemCount) without use of synchronization, then the solution is unsatisfactory for that reason, without needing to explicitly demonstrate a race condition.

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