问题描述
我正在制作一个有2艘太空飞船战斗的程序。到目前为止,它仍然有效,但是当我们调用Attack函数时,会出现以下错误:
"TypeError: 'int' object is not callable on line 54"
我花了一些时间进行调试,但无法弄清楚哪里出了问题。这是代码:请提供一些建议使其生效
import turtle
import time
import random
drawer = turtle.Turtle()
drawer.shape("square")
drawer.speed(0)
attacker = turtle.Turtle()
attacker.shape("square")
attacker.speed(0.5)
attacker.penup()
color1 = (random.randint(0,255),random.randint(0,255))
color2 = (random.randint(0,255))
class spaceship:
def __init__(self,name,health,attack,sheild,x,y):
self.name = name
self.health = health
self.attack = attack
self.sheild = sheild
self.x = x
self.y = y
def draw_spaceship(self,color):
drawer.color(color)
drawer.penup()
drawer.goto(self.x,self.y)
drawer.begin_fill()
for i in range(0,4):
drawer.forward(100)
drawer.right(90)
drawer.end_fill()
def attack(self,color,enemy_name):
attacker.goto(self.x,self.y)
attacker.goto(enemy_name.x,enemy_name.y)
attacker.color(color)
where_attack = random.randrange(0,2)
if where_attack == 1:
where_attack = "health"
enemy_name.health -= self.attack
else:
where_attack = "sheild"
enemy_name.sheild -= self.attack
print(self.name,"is attacking the",where_attack,"of",enemy_name.name+".")
print("Now",enemy_name.name,"has",enemy_name.health,"health.")
# draws spaceships
Noob = spaceship("Noob",random.randrange(100,500),random.randrange(10,200),-250,50)
Noob.draw_spaceship(color1)
Pro = spaceship("Pro",1000),100,50)
Pro.draw_spaceship(color2)
drawer.forward(2000)
Pro.attack(color1,Noob)
解决方法
尝试重命名方法“攻击”,然后重试。似乎属性和方法使用相同的名称来调用。
,问题是您要声明spaceship
的属性(变量)attack
与方法attack()
相同。
由于Pro
是spaceship
的实例,因此当您在脚本末尾调用Pro.attack()
时,实际上是在尝试调用其类型为 int 的属性。 / p>
尝试将方法的名称更改为其他名称