如何在变量大于0时同时执行两个Python函数？

问题说明 #

想要实现的效果：在health大于0时，一边持续扣除health值，一边提示用户发起攻击。但当前代码中kidsenemyattack()和kidfight()无法同时执行，互相覆盖。

原代码 #

import time

health = 100
kidsHP = 10

def kidsenemyattack():
    global health
    print('The enemy deals damage!')
    health -= 4
    print(health)
    print('==========')
    time.sleep(3)
    kidsenemyattack()

def kidfight():
    global kidsHP
    ready = input('You can attack!')
    if ready in ['1','one','One']:
        kidsHP -= 1
    time.sleep(2.5)

while health > 0:
    kidsenemyattack() and kidfight()
    if kidsHP <= 0:
        print("You've defeated the evil kids")

问题根源 #

1. kidsenemyattack()用递归调用自己，一旦进入这个函数就会无限循环，永远不会执行到kidfight()。
2. and运算符不是用来同时执行两个函数的，它只会在第一个函数返回True时才执行第二个，而这里函数没有返回值（默认返回None），所以第二个函数根本不会运行。

解决方案：使用多线程 #

用Python的threading模块让两个任务并行执行，一个负责敌人持续攻击，一个负责接收用户输入发起攻击，主线程监控战斗状态。

修改后的代码 #

import time
import threading

health = 100
kidsHP = 10
stop_flag = False  # 全局标记控制线程停止

def kidsenemyattack():
    global health, stop_flag
    while health > 0 and not stop_flag:
        print('The enemy deals damage!')
        health -= 4
        print(f"Your current health: {health}")
        print('==========')
        time.sleep(3)
    if health <= 0:
        print("You've been defeated by the evil kids!")

def kidfight():
    global kidsHP, stop_flag
    while kidsHP > 0 and not stop_flag:
        try:
            ready = input('You can attack! Enter 1/one to attack: ')
            if ready.lower() in ['1', 'one']:
                kidsHP -= 1
                print(f"Evil kid's HP: {kidsHP}")
                if kidsHP <= 0:
                    print("You've defeated the evil kids!")
                    stop_flag = True
            time.sleep(0.5)  # 防止输入过于频繁
        except KeyboardInterrupt:
            stop_flag = True
            break

# 启动两个守护线程，主线程结束时自动退出
attack_thread = threading.Thread(target=kidsenemyattack, daemon=True)
fight_thread = threading.Thread(target=kidfight, daemon=True)

attack_thread.start()
fight_thread.start()

# 主线程监控战斗状态，直到结束
while not stop_flag and health > 0 and kidsHP > 0:
    time.sleep(1)

代码说明 #

• 用stop_flag统一控制两个线程的停止逻辑，避免线程无法正常退出。
• 把递归改成while循环，防止递归深度过大导致栈溢出。
• 守护线程（daemon=True）确保主线程结束时子线程自动终止。

替代方案：单循环轮询（无需多线程） #

如果不想用线程，可以在主循环里轮询敌人攻击时间和用户输入，通过非阻塞输入实现近似同时运行的效果：

import time
import sys
import select

health = 100
kidsHP = 10
last_enemy_attack = time.time()

while health > 0 and kidsHP > 0:
    # 每隔3秒触发一次敌人攻击
    if time.time() - last_enemy_attack >= 3:
        print('The enemy deals damage!')
        health -= 4
        print(f"Your current health: {health}")
        print('==========')
        last_enemy_attack = time.time()
    
    # 非阻塞检查用户输入
    if select.select([sys.stdin], [], [], 0.1)[0]:
        ready = sys.stdin.readline().strip()
        if ready.lower() in ['1', 'one']:
            kidsHP -= 1
            print(f"Evil kid's HP: {kidsHP}")
            if kidsHP <= 0:
                print("You've defeated the evil kids!")

if health <= 0:
    print("You've been defeated by the evil kids!")

内容的提问来源于stack exchange，提问作者Bulrunord