有界信号量多放会怎样

本节的东西：

Gate(n)   最多 n 个同时在里面：Semaphore(n)；顺便用一把小锁维护 inside / peak（峰值）
run_gate(n, workers, hold)   workers 条线程各进一次 Gate 待 hold 秒，交回峰值

普通的和有界的各建一个 2 的信号量，都拿一次放两次，看谁报错、之后各自还能连拿几次：

import threading
import time


class Counter:
    def __init__(self):
        self.n = 0

    def inc(self):
        tmp = self.n
        time.sleep(0.001)
        self.n = tmp + 1


class SafeCounter(Counter):
    def __init__(self):
        super().__init__()
        self.lock = threading.Lock()

    def inc(self):
        with self.lock:
            tmp = self.n
            time.sleep(0.001)
            self.n = tmp + 1


def hammer(counter, workers=2, times=30):
    def job():
        for _ in range(times):
            counter.inc()
    ts = [threading.Thread(target=job) for _ in range(workers)]
    for t in ts:
        t.start()
    for t in ts:
        t.join()
    return counter.n


def probe(lock):
    """这把锁现在能不能立刻拿到？能就拿了再放回去，交回 True；拿不到交回 False。"""
    if lock.acquire(blocking=False):
        lock.release()
        return True
    return False


class Guard:
    """手写的 with 替身：进就 acquire，出就 release——不管是正常出还是异常出。"""
    def __init__(self, lock):
        self.lock = lock

    def __enter__(self):
        self.lock.acquire()
        return self

    def __exit__(self, exc_type, exc, tb):
        self.lock.release()
        return False

res = []
for S in (threading.Semaphore(2), threading.BoundedSemaphore(2)):
    S.acquire()
    S.release()
    try:
        S.release()
        err = "没报"
    except ValueError:
        err = "报了"
    k = 0
    while S.acquire(blocking=False):
        k += 1
    res.append(err + ":" + str(k))
print("/".join(res))