Lock-free “decrement if not zero”

问题

I'm currently reinventing the wheel of a thread pool in C++. I've eliminated almost all locks from the code, except for multiple instances of the following construct:

std::atomic_size_t counter;

void produce() {
    ++counter;
}

void try_consume() {
    if (counter != 0) {
        --counter;
        // ...
    }
    else {
        // ...
    }
}

So, I need a thread-safe lock-free version of this function:

bool take(std::atomic_size_t& value) {
    if (value != 0) {
        --value;
        return true;
    }
    return false;
}

There is one solution that I know of: use boost::lockfree::queue<std::monostate>, where pop() does the job. Is there any better/faster solution?

回答1:

The construct you're implementing is a counting lock, or counting semaphore. Use one from a library that has a trylock version instead of rolling your own, so you can get optimized OS-supported sleep / wakeup. Or do you always have useful work to do if the trylock (aka take) fails?

Note that you can avoid using any traditional locks while implementing your own, but "lock free" is a technical term with a different meaning than lockless. A consumer almost by definition can't be lock-free in the computer-science sense, because it can be stuck waiting forever if the producer thread(s) are blocked. Related: Lock-free Progress Guarantees

CAS is good. Just make sure that your function doesn't run compare_exchange_weak at all if it sees the counter already 0 with a pure load (typically with memory_order_relaxed). You don't want your CPU hammering on the location while other threads are trying to increment it so your thread will see a non-zero value.

Another option is a signed counter, and change the compare to >= 0. Check for overshoot in the result of fetch_add(-1), and if so correct it. (Threads that see the counter as temporarily negative just see it "locked"). But this is typically not better than a CAS retry loop; failed CAS is rare unless contention is very high. And extra atomic ops to correct overshoot will probably cost about as much (or more) than CAS retries.

来源：https://stackoverflow.com/questions/47650290/lock-free-decrement-if-not-zero

标签

c++

thread-safety

atomic

lock-free