Multithreaded Programming Guide

"Shared-Memory" Multiprocessors

Consider the purported solution to the producer/consumer problem shown in Example 10-5.

Although this program works on current SPARC-based multiprocessors, it assumes that all multiprocessors have strongly ordered memory. This program is therefore not portable.

Example 10-5 The Producer/Consumer Problem--Shared Memory Multiprocessors

                    char buffer[BSIZE];
                    unsigned int in = 0;
                    unsigned int out = 0;

void                             char
producer(char item) {              consumer(void) {
                                    char item;
        ;/* nothing */                do  
    while                                 ;/* nothing */
        (in - out == BSIZE);           while
                                         (in - out == 0);
    buffer[in%BSIZE] = item;            item = buffer[out%BSIZE];
    in++;                             out++;
}                                }

When this program has exactly one producer and exactly one consumer and is run on a shared-memory multiprocessor, it appears to be correct. The difference between in and out is the number of items in the buffer.

The producer waits (by repeatedly computing this difference) until there is room for a new item, and the consumer waits until there is an item in the buffer.

For memory that is strongly ordered (for instance, a modification to memory on one processor is immediately available to the other processors), this solution is correct (it is correct even taking into account that in and out will eventually overflow, as long as BSIZE is less than the largest integer that can be represented in a word).

Shared-memory multiprocessors do not necessarily have strongly ordered memory. A change to memory by one processor is not necessarily available immediately to the other processors. When two changes to different memory locations are made by one processor, the other processors do not necessarily see the changes in the order in which they were made because changes to memory don't happen immediately.

First the changes are stored in store buffers that are not visible to the cache.

The processor looks at these store buffers to ensure that a program has a consistent view, but because store buffers are not visible to other processors, a write by one processor doesn't become visible until it is written to cache.

The synchronization primitives (see Chapter 4, Programming with Synchronization Objects") use special instructions that flush the store buffers to cache. So, using locks around your shared data ensures memory consistency.

When memory ordering is very relaxed, Example 10-5has a problem because the consumer might see that in has been incremented by the producer before it sees the change to the corresponding buffer slot.

This is called weak ordering because stores made by one processor can appear to happen out of order by another processor (memory, however, is always consistent from the same processor). To fix this, the code should use mutexes to flush the cache.

The trend is toward relaxing memory order. Because of this, programmers are becoming increasingly careful to use locks around all global or shared data.

As demonstrated by Example 10-5and Example 10-6, locking is essential.