线程同时打印会弄乱文本输出

我在应用程序中使用 4 个线程，它们返回我想要打印给用户的文本。 由于我想避免线程独立打印这些文本，因此我创建了一个类来管理它。

我不知道我在这里做错了什么，但它仍然不起作用。

您可以在下面看到代码：

from threading import Thread
import time
import random

class Creature:
    def __init__(self, name, melee, shielding, health, mana):
        self.name = name
        self.melee = melee
        self.shielding = shielding
        self.health = health
        self.mana = mana

    def attack(self, attacker, opponent, echo):
        while 0 != 1:
            time.sleep(1)
            power = random.randint(1, attacker.melee)
            resistance = random.randint(1, opponent.shielding)
            resultant = power - resistance
            if resistance > 0:
                opponent.health -= resistance
                if opponent.health < 0:
                    msg = opponent.name, " is dead"
                    echo.message(msg)
                    quit()
                else:
                    msg = opponent.name, " lost ", resistance, " hit points due to an attack by ", attacker.name
                    echo.message(msg)

    def healing(self, healed, echo):
        while 0 != 1:
            time.sleep(1)
            if self.mana >= 25:
                if healed.health >= 0:
                    if healed.health < 50:
                        life = random.randint(1, 50)
                        self.mana -= 25
                        healed.health += life
                        if healed.health > 100:
                            healed.health = 100
                        msg = healed.name, " has generated himself and now has ", self.health, " hit points"
                        echo.message(msg)
                else:
                    quit()

class echo:
    def message(self, msg):
        print msg

myEcho = echo()

Monster = Creature("Wasp", 30, 15, 100, 100)
Player = Creature("Knight", 25, 20, 100, 100)

t1 = Thread(target = Player.attack, args = (Monster, Player, myEcho))
t1.start()
t2 = Thread(target = Monster.attack, args = (Player, Monster, myEcho))
t2.start()
t3 = Thread(target=Player.healing(Player, myEcho), args=())
t3.start()
t4 = Thread(target=Monster.healing(Monster, myEcho), args=())
t4.start()

在这里你可以看到混乱的输出：

*('Wasp'('Knight', ' l, ' lost ', ost 13, ' hit points ', 4, due to an attack by '' hi, 'Waspt poi')nts d
ue to an attack by ', 'Knight')
('Wasp', ' lost ', 12, ' hit points due to an attack by ', 'Knight')
('Knight', ' lost ', 17, ' hit points due to an attack by ', 'Wasp')
('Wasp', ' lost ', 6, ' hit points due to an attack by ', 'Knight'('Knight')
, ' lost ', 1, ' hit points due to an attack by ', 'Wasp')
('Wasp', ' lost ', 5, ' hit points due to an attack by ', 'Knight')
('Knight', ' lost ', 13, ' hit points due to an attack by ', 'Wasp')
(('Wa'Knighsp't', , ' los' lostt ' ', , 32, ' hit points due to an attack by ', 'Knight')
, ' hit points due to an attack by ', 'Wasp')*

你们知道如何解决这个问题吗？

Use a threading.Semaphore https://docs.python.org/2/library/threading.html#semaphore-objects确保不会发生任何冲突：

screenlock = Semaphore(value=1)   # You'll need to add this to the import statement.

然后，在你打电话之前echo.message，插入这一行以获得输出权：

screenlock.acquire()

然后这一行after你打电话echo.message以便允许另一个线程打印：

screenlock.release()