如何在时间到期后给用户一定的时间回答和打印消息？ （蟒蛇）

Question

我尝试使用多线程来限制用户的输入时间，但是这段代码似乎不起作用。

# giving the user less than 5 seconds to enter a number

import time
from threading import Thread


def ask():
    start_time = time.time()
    a = float(input("Enter a number:\n"))
    time.sleep(0.001)


def timing():
    if time.time() - start_time > 5: 
        print("Time's Up")
        quit()
    time.sleep(0.001)


t1 = Thread(target=ask)
t2 = Thread(target=timing)
t1.start()
t2.start()
t1.join()
t2.join()

Answer 1

您需要将计时功能置于无限模式。 Bcz它只运行一次。

尝试以下代码：

import time
import os
from threading import Thread

start_time = time.time()
a = None

def ask():
    global start_time
    start_time = time.time()
    global a
    a = float(input("Enter a number:\n"))
    time.sleep(0.001)


def timing():
    while True:
        global a
        if a is not None:
            break
        if time.time() - start_time > 5: 
            print("Time's Up")
            os._exit(1)
        time.sleep(0.001)


t1 = Thread(target=ask)
t2 = Thread(target=timing)
t1.start()
t2.start()
t1.join()
t2.join()

Answer 2

我不喜欢全局变量，但是以下代码似乎可以很好地满足您的需求：

import time
from threading import Thread
import sys

start_time = 0
timeUp = False

def ask():
    global start_time, timeUp
    start_time = time.time()
    a = float(input("Enter a number:\n"))
    time.sleep(0.001)
    if a and (not timeUp):
        print('This is a: ', a)

def timing():
    global timeUp
    while True:
        if time.time() - start_time > 5: 
            print("Time's Up")
            timeUp = True
            sys.exit()
        time.sleep(0.001)


t1 = Thread(target=ask)
t2 = Thread(target=timing)
t1.start()
t2.start()
t1.join()
t2.join()