Java中的阻塞信号量是什么？

Question

根据我的理解，阻塞信号量被初始化为零而不是一，其工作方式是任何执行p（S）操作的线程都将阻塞，直到V（S）首先释放为止

如果有人能举例说明这一机制，我们将不胜感激

Answer 1

a blocking semaphore

^{}类同时支持阻塞和非阻塞操作

a blocking semaphore is initialised to zero rather than one

您可以设置许可证的初始数量。如果将permits设置为1（new Semaphore(1)），则可以在release之前acquire一次

• 如果值为> 0，则某些acquire可能发生在release之前
• 如果该值为<= 0，则在调用acquire之前必须出现一些release

它记录在JavaDoc中：

This value may be negative, in which case releases must occur before any acquires will be granted.

---

class Example {

    public static void main(String[] args) throws InterruptedException {
        final Semaphore s1 = new Semaphore(0);
        s1.acquire(); 
        // the thread is disabled
        // no permit is available
        // a blocking operation

        final Semaphore s2 = new Semaphore(1);
        s2.acquire();
        // returns immediately
        // one permit was available

        final Semaphore s3 = new Semaphore(0);
        if (s3.tryAcquire()) {
            // a non-blocking operation
            // returns false immediately
        }
    }

}

Answer 2

引用Semaphare类的javadoc：

permits - the initial number of permits available. This value may be negative, in which case releases must occur before any acquires will be granted.

以及：

Each acquire() blocks if necessary until a permit is available, and then takes it. Each release() adds a permit, potentially releasing a blocking acquirer.

因此，假设您使用-1初始化，信号量将阻止acquire()调用，直到release()进入

从这个意义上说：您只需执行semaphore = new Sempahore(-1)，就会发现semaphore.acquire()将阻塞，直到其他线程执行了semaphore.release()调用

这就是一切