0

我有这个非阻塞计时器:

import time
class Timer:
        def __init__(self):
        self.status = 0
    def update(self, data, tv):
        self.target = time.time() + data
        self.status = 1
    def set_status(self, state):
        self.status = state
    def get_status(self, tv):
        return self.status
    def get_left(self, tv):
        return int(self.target-time.time())

当我想要在主循环中使用计时器时,它非常有用,例如:

t = Timer()
while 1:
    #If timer is not active.
    if t.get_status() == 0:
        # Do some checks.
        # Start timer with 60 seconds.
        t.update(60)
    #If timer is active.
    if t.get_status() == 1:
         print "%s seconds left." % str(t.get_left())
         if t.get_left() <= 0:
             t.set_status(0)
             # Timer is done do some stuff.
time.sleep(1)

我希望它能够在不执行线程的情况下处理多个计时器。

4

1 回答 1

0

这是我自己的解决方案,如果有人有更好的想法,请告诉我。

import time
class Timer:
    def __init__(self):
        self.status = {}
        self.target = {}
    def update(self, tv, data):
        self.target[tv] = time.time() + data
        self.status[tv] = 1
    def set_status(self, tv, state):
        self.status[tv] = state
    def get_status(self, tv):
        return self.status[tv]
    def get_left(self, tv):
        return int(self.target[tv]-time.time())

t = Timer()
t.set_status('A', 0)
t.set_status('B', 0)
while 1:
    # Timer A checks
    if t.get_status('A') == 0:
        print "Timer A checks."
        # Start timer with 60 seconds.
        t.update('A', 60)
    if t.get_status('A') == 1:
         print "A has %s seconds left." % str(t.get_left('A'))
         if t.get_left('A') <= 0:
             t.set_status('A', 0)
             print "Timer A done."
    # Timer B checks
    if t.get_status('B') == 0:
        print "Timer B checks."
        # Start timer with 30 seconds
        t.update('B', 30)
    if t.get_status('B') == 1:
         print "B has %s seconds left." % str(t.get_left('B'))
         if t.get_left('B') <= 0:
             t.set_status('B', 0)
             print "Timer B done."
    time.sleep(1)
于 2012-12-11T10:55:15.320 回答