Пример #1
0
void
ThreadPoolWorker::run()
{
    // This is hokey in the extreme.  To compute the stack limit,
    // subtract the size of the stack from the address of a local
    // variable and give a 2k buffer.  Is there a better way?
    uintptr_t stackLimitOffset = WORKER_THREAD_STACK_SIZE - 2*1024;
    uintptr_t stackLimit = (((uintptr_t)&stackLimitOffset) +
                             stackLimitOffset * JS_STACK_GROWTH_DIRECTION);

    AutoLockMonitor lock(*this);
    JS_ASSERT(state_ == ACTIVE);
    for (;;) {
        while (!worklist_.empty()) {
            TaskExecutor *task = worklist_.popCopy();
            {
                AutoUnlockMonitor unlock(*this);
                task->executeFromWorker(workerId_, stackLimit);
            }
        }

        if (state_ == TERMINATING)
            break;

        lock.wait();
    }

    JS_ASSERT(worklist_.empty());
    state_ = TERMINATED;
    lock.notify();
}
Пример #2
0
/*
 * In the pathological cases that dominate much of the test case runtime,
 * rooting analysis spends tons of time scanning the stack during a tight-ish
 * loop. Since statically, everything is either rooted or it isn't, these scans
 * are almost certain to be worthless. Detect these cases by checking whether
 * the addresses of the top several rooters in the stack are recurring. Note
 * that there may be more than one CheckRoots call within the loop, so we may
 * alternate between a couple of stacks rather than just repeating the same one
 * over and over, so we need more than a depth-1 memory.
 */
static bool
SuppressCheckRoots(js::Vector<Rooter, 0, SystemAllocPolicy> &rooters)
{
    static const unsigned int NumStackMemories = 6;
    static const size_t StackCheckDepth = 10;

    static uint32_t stacks[NumStackMemories];
    static unsigned int numMemories = 0;
    static unsigned int oldestMemory = 0;

    // Ugh. Sort the rooters. This should really be an O(n) rank selection
    // followed by a sort. Interestingly, however, the overall scan goes a bit
    // *faster* with this sort. Better branch prediction of the later
    // partitioning pass, perhaps.
    qsort(rooters.begin(), rooters.length(), sizeof(Rooter), CompareRooters);

    // Forward-declare a variable so its address can be used to mark the
    // current top of the stack.
    unsigned int pos;

    // Compute the hash of the current stack.
    uint32_t hash = HashGeneric(&pos);
    for (unsigned int i = 0; i < Min(StackCheckDepth, rooters.length()); i++)
        hash = AddToHash(hash, rooters[rooters.length() - i - 1].rooter);

    // Scan through the remembered stacks to find the current stack.
    for (pos = 0; pos < numMemories; pos++) {
        if (stacks[pos] == hash) {
            // Skip this check. Technically, it is incorrect to not update the
            // LRU queue position, but it'll cost us at most one extra check
            // for every time a hot stack falls out of the window.
            return true;
        }
    }

    // Replace the oldest remembered stack with our current stack.
    stacks[oldestMemory] = hash;
    oldestMemory = (oldestMemory + 1) % NumStackMemories;
    if (numMemories < NumStackMemories)
        numMemories++;

    return false;
}
Пример #3
0
bool
ThreadPoolWorker::submit(TaskExecutor *task)
{
    AutoLockMonitor lock(*this);
    JS_ASSERT(state_ == ACTIVE);
    if (!worklist_.append(task))
        return false;
    lock.notify();
    return true;
}
Пример #4
0
static void
GatherRooters(js::Vector<Rooter, 0, SystemAllocPolicy> &rooters,
              Rooted<void*> **thingGCRooters,
              unsigned thingRootKind)
{
    Rooted<void*> *rooter = thingGCRooters[thingRootKind];
    while (rooter) {
        Rooter r = { rooter, ThingRootKind(thingRootKind) };
        JS_ALWAYS_TRUE(rooters.append(r));
        rooter = rooter->previous();
    }
}