Peterson算法

main.py

do_flag = [False, False]
turn = 0
# 以上是Peterson’s Solution的結構

exp = 0  # 共同存取的變數，預期他應該在執行完後等於20


def pro_0():  # 程序0
    global do_flag, turn, exp
    do_flag[0] = True
    yield
    if do_flag[1] and turn == 1:
        yield  # 允許在等待時運行另一個程序
    for i in range(10):
        exp = exp + 1
        print("PO 正在跑")
    return None


def pro_1():  # 程序1
    global do_flag, turn, exp
    do_flag[1] = True
    yield
    if do_flag[0] and turn == 0:
        yield  # 允許在等待時運行另一個程序
    for i in range(10):
        exp = exp + 1
        print("P1 正在跑")
    return None


if __name__ == "__main__":
    a = pro_0()
    b = pro_1()
    stack = [a, b]
    while True:
        for task in stack:
            try:
                next(task)
            except StopIteration:
                pass
        print(exp)
        if exp == 20:
            break