这个想法是，可以只有一个线程获取写锁(futex值0)，锁可以是打开的(futex值1)，或者可以有许多读取线程(futex值大于1)。因此，低于1 (只有一个)的值会阻塞futex上的读取器和写入器，而高于1的值只会阻塞写入器。解锁线程会唤醒其中一个等待线程，但您需要小心，不要消耗仅由写线程唤醒的读取器。

#define cpu_relax() __builtin_ia32_pause()
#define cmpxchg(P, O, N) __sync_val_compare_and_swap((P), (O), (N))

static unsigned _lock = 1; // read-write lock futex
const static unsigned _lock_open = 1;
const static unsigned _lock_wlocked = 0;

static void _unlock()
{
    unsigned current, wanted;
    do {
        current = _lock;
        if (current == _lock_open) return;
        if (current == _lock_wlocked) {
            wanted = _lock_open;
        } else {
            wanted = current - 1;
        }
    } while (cmpxchg(&_lock, current, wanted) != current);
    syscall(SYS_futex, &_lock, FUTEX_WAKE_PRIVATE, 1, NULL, NULL, 0);
}

static void _rlock()
{
    unsigned current;
    while ((current = _lock) == _lock_wlocked || cmpxchg(&_lock, current, current + 1) != current) {
        while (syscall(SYS_futex, &_lock, FUTEX_WAIT_PRIVATE, current, NULL, NULL, 0) != 0) {
            cpu_relax();
            if (_lock >= _lock_open) break;
        }
        // will be able to acquire rlock no matter what unlock woke us
    }
}

static void _wlock()
{
    unsigned current;
    while ((current = cmpxchg(&_lock, _lock_open, _lock_wlocked)) != _lock_open) {
        while (syscall(SYS_futex, &_lock, FUTEX_WAIT_PRIVATE, current, NULL, NULL, 0) != 0) {
            cpu_relax();
            if (_lock == _lock_open) break;
        }
        if (_lock != _lock_open) {
            // in rlock - won't be able to acquire lock - wake someone else
            syscall(SYS_futex, &_lock, FUTEX_WAKE_PRIVATE, 1, NULL, NULL, 0);
        }
    }
}

票数 1

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/3964776

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