Question

程序应该响应并在时间变量指定的时间内打印val。

import timeit
from threading import Thread

val = 0
time = 3.05274241722e-05

def function1():
    global val
    for i in range(0,10000):
        val = val+i
    return val

def function2():
    global val
    for i in range(0,10000):
        val = val+i
    return val

if __name__ == '__main__':
    start_time = timeit.default_timer()
    Thread(target = function1).start()
    Thread(target = function2).start()
    while time-1 <=(timeit.default_timer() - start_time):
        if time-1 >(timeit.default_timer() - start_time):
            break
        print time-1
        print timeit.default_timer() - start_time
        print val
        pass

    print val
    Thread(target = function1).exit()
    Thread(target = function2).exit()

还请让我知道我是否正确使用了线程。谢谢

Answer 1

您的time-1是一个负值，它总是小于(timeit.default_timer() - start_time)，因此它是一个无限循环，if语句更加冗余。另外，你使用线程的方式是错误的。

尝试以下代码

import timeit
from threading import Thread
val = 0
time = 2.38282775879e-05
flag = False

def function1():
    global val
    global flag
    for i in range(0,1000):
        if not flag:
            val = val+i
            print val
        else:
            break

def function2():
    global val
    global flag
    for i in range(0,1000):
        if not flag:
            val = val+i
            print val
        else:
            break

if __name__ == '__main__':
    f1 = Thread(target = function1)
    f2 = Thread(target = function2)
    start_time = timeit.default_timer()
    f1.start()
    f2.start()
    f1.join(time)
    f2.join(time)
    flag = True
    end_time = timeit.default_timer()
    f1.join()
    f2.join()
    if time <= end_time - start_time:
        print "Timeout"
    else:
        print "Finished before timeout"
    print end_time - start_time
    print val

计时器不在python代码中工作

1 个答案: